# NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines

## NCERT Solutions Class 11 Maths Chapter 10 â€“ Free PDF Download

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines provided here can be extremely beneficial for the students in their Maths board exam preparation. Hence, students aspiring to outshine in the Maths exam can refer to the NCERT Solutions given below. The solutions at BYJUâ€™S for Straight Lines contain comprehensive and straightforward answers to the questions given in the textbook. Experienced teachers of Maths develop these solutions using which students can practise and improve their understanding of essential topics.

We know that Chapter 10 contains structures that are an indispensable part of the latest Class 11 Maths CBSE Syllabus 2023-24. The important topics covered in the chapter are the slope of a line, Horizontal and vertical Lines, Point-slope form, Two-point form, Slope-intercept form, and Intercept â€“ form as well as Normal form. Thus, to assist students in mastering all these topics, our experts have created NCERT Solutions to makeÂ sure that they solve all the questions using simple methods and formulas. The NCERT Solutions for Class 11 for this chapter can be accessed from the link which is provided below.

## NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

### Access Exercise-wise NCERT Solutions for Class 11 Maths Chapter 10 â€“ Straight Lines

Exercise 10.1 SolutionsÂ â€“ 14 Questions

Exercise 10.2 SolutionsÂ â€“ 20 Questions

Exercise 10.3 SolutionsÂ â€“ 18 Questions

Miscellaneous Exercise on Chapter 10 SolutionsÂ â€“ 24 Questions

### Access NCERT Solutions for Class 11 Maths Chapter 10

EXERCISE 10.1 PAGE NO: 211

1. Draw a quadrilateral in the Cartesian plane whose vertices are (â€“ 4, 5), (0, 7), (5, â€“ 5) and (â€“ 4, â€“2). Also, find its area.

Solution:

Let ABCD be the given quadrilateral with vertices A (-4,5), B (0,7), C (5.-5) and D (-4,-2).

Now, let us plot the points on the Cartesian plane by joining the points AB, BC, CD, and AD, which give us the required quadrilateral.

To find the area, draw diagonal AC.

So, area (ABCD) = area (âˆ†ABC) + area (âˆ†ADC)

Then, area of triangle with vertices (x1,y1) , (x2, y2) and (x3,y3) is

Are of âˆ†Â ABC = Â½ [x1 (y2 â€“ y3) + x2 (y3 â€“ y1) + x3 (y1 â€“ y2)]

= Â½ [-4 (7 + 5) + 0 (-5 â€“ 5) + 5 (5 â€“ 7)] unit2

= Â½ [-4 (12) + 5 (-2)] unit2

= Â½ (58) unit2

= 29 unit2

Are of âˆ†Â ACD = Â½ [x1 (y2 â€“ y3) + x2 (y3 â€“ y1) + x3 (y1 â€“ y2)]

= Â½ [-4 (-5 + 2) + 5 (-2 â€“ 5) + (-4) (5 â€“ (-5))] unit2

= Â½ [-4 (-3) + 5 (-7) â€“ 4 (10)] unit2

= Â½ (-63) unit2

= -63/2 unit2

Since area cannot be negative, area âˆ† ACD = 63/2 unit2

Area (ABCD) = 29 + 63/2

= 121/2 unit2

2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Solution:

Let us consider ABC, the given equilateral triangle with side 2a.

Where, AB = BC = AC = 2a

In the above figure, assuming that the base BC lies on the x-axis such that the mid-point of BC is at the origin, i.e., BO = OC = a, where O is the origin.

The coordinates of point C are (0, a) and that of B are (0,-a).

The line joining a vertex of an equilateral âˆ† with the mid-point of its opposite side is perpendicular.

So, vertex A lies on the y â€“axis.

By applying Pythagorasâ€™ theorem,

(AC)2Â = OA2Â + OC2

(2a)2= a2Â +Â OC2

4a2Â â€“ a2Â =Â OC2

3a2 =Â OC2

OC =âˆš3a

Co-ordinates of point C = Â± âˆš3a, 0

âˆ´ The vertices of the given equilateral triangle are (0, a), (0, -a), (âˆš3a, 0)

Or (0, a), (0, -a) and (-âˆš3a, 0)

3. Find the distance between P (x1, y1) and Q (x2, y2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Solution:

Given:

Points P (x1, y1) and Q(x2, y2)

(i) When PQ is parallel to the y-axis, then x1Â = x2

So, the distance between P and Q isÂ given by

= |y2 â€“ y1|

(ii) When PQ is parallel to the x-axis, then y1Â = y2

So, the distance between P and Q is given by =

=Â

= |x2 â€“ x1|

4. Find a point on the x-axis which is equidistant from points (7, 6) and (3, 4).

Solution:

Let us consider (a, 0) to be the point on the x-axis that is equidistant from the point (7, 6) and (3, 4).

So,

Now, let us square on both sides; we get,

a2 â€“ 14a + 85 = a2 â€“ 6a + 25

-8a = -60

a = 60/8

= 15/2

âˆ´ The required point is (15/2, 0)

5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, â€“ 4) and B (8, 0).

Solution:

The co-ordinates of the mid-point of the line segment joining the points P (0, â€“ 4) and B (8, 0) are (0+8)/2, (-4+0)/2 = (4, -2)

The slope â€˜mâ€™ of the line non-vertical line passing through the point (x1, y1) and

(x2, y2) is given by m = (y2 â€“ y1)/(x2 â€“ x1) where, x â‰  x1

The slope of the line passing through (0, 0) and (4, -2) is (-2-0)/(4-0) = -1/2

âˆ´ The required slope is -1/2.

6. Without using Pythagorasâ€™ theorem, show that the points (4, 4), (3, 5) and (â€“1, â€“1) are the vertices of a right-angled triangle.

Solution:

The vertices of the given triangle are (4, 4), (3, 5) and (â€“1, â€“1).

The slope (m) of the line non-vertical line passing through the point (x1, y1) and

(x2, y2) is given by m = (y2 â€“ y1)/(x2 â€“ x1) where, x â‰  x1

So, the slope of the line AB (m1) = (5-4)/(3-4) = 1/-1 = -1

The slope of the line BC (m2) = (-1-5)/(-1-3) = -6/-4 = 3/2

The slope of the line CA (m3) = (4+1)/(4+1) = 5/5 = 1

It is observed that m1.m3Â = -1.1 = -1

Hence, the lines AB and CA are perpendicular to each other.

âˆ´Â given triangle is right-angled at A (4, 4)

And the vertices of the right-angled âˆ† are (4, 4), (3, 5) and (-1, -1)

7. Find the slope of the line, which makes an angle of 30Â° with the positive direction of the y-axis measured anticlockwise.

Solution:

We know that if a line makes an angle of 30Â° with the positive direction of the y-axis measured anti-clock-wise, then the angle made by the line with the positive direction of the x-axis measured anti-clock-wise is 90Â° + 30Â° = 120Â°

âˆ´ The slope of the given line is tan 120Â° = tan (180Â° â€“ 60Â°)

= â€“ tan 60Â°

= â€“âˆš3

8. Find the value of x for which the points (x, â€“ 1), (2, 1) and (4, 5) are collinear.

Solution:

If the points (x, â€“ 1), (2, 1) and (4, 5) are collinear, then the Slope of AB = Slope of BC

Then, (1+1)/(2-x) = (5-1)/(4-2)

2/(2-x) = 4/2

2/(2-x) = 2

2 = 2(2-x)

2 = 4 â€“ 2x

2x = 4 â€“ 2

2x = 2

x = 2/2

= 1

âˆ´ The required value of x is 1.

9. Without using the distance formula, show that points (â€“ 2, â€“ 1), (4, 0), (3, 3) and (â€“3, 2) are the vertices of a parallelogram.

Solution:

Let the given point be A (-2, -1) , B (4, 0) , C ( 3, 3) and D ( -3, 2)

So now, the slope of AB = (0+1)/(4+2) = 1/6

The slope of CD = (3-2)/(3+3) = 1/6

Hence, the Slope of AB = Slope of CD

âˆ´ ABÂ âˆ¥Â CD

Now,

The slope of BC = (3-0)/(3-4) = 3/-1 = -3

The slope of AD = (2+1)/(-3+2) = 3/-1 = -3

Hence, the Slope of BC = Slope of AD

Thus, the pair of opposite sides are quadrilateral are parallel, so we can say that ABCD is a parallelogram.

Hence, the given vertices, A (-2, -1), B (4, 0), C(3, 3) and D(-3, 2) are vertices of a parallelogram.

10. Find the angle between the x-axis and the line joining the points (3, â€“1) and (4, â€“2).

Solution:

The Slope of the line joining the points (3, -1) and (4, -2) is given by

m = (y2 â€“ y1)/(x2 â€“ x1) where, x â‰  x1

m = (-2 â€“(-1))/(4-3)

= (-2+1)/(4-3)

= -1/1

= -1

The angle of inclination of the line joining the points (3, -1) and (4, -2) is given by

tan Î¸ = -1

Î¸ = (90Â° + 45Â°) = 135Â°

âˆ´ The angle between the x-axis and the line joining the points (3, â€“1) and (4, â€“2) is 135Â°.

11. The slope of a line is double the slope of another line. If the tangent of the angle between them is 1/3, find the slopes of the lines.

Solution:

Let us consider â€˜m1â€™Â and â€˜mâ€™ be the slope of the two given lines such that m1 = 2m

We know that if Î¸ is the angle between the lines l1 and l2 with slope m1Â and m2, then

1+2m2Â = -3m

2m2Â +1 +3m = 0

2m (m+1) + 1(m+1) = 0

(2m+1) (m+1)= 0

m = -1 or -1/2

If m = -1, then the slope of the lines are -1 and -2

If m =Â -1/2, then the slope of the lines areÂ -1/2Â and -1

Case 2:

2m2Â â€“ 3m + 1 = 0

2m2Â â€“ 2m â€“ m + 1 = 0

2m (m â€“ 1) â€“ 1(m â€“ 1) = 0

m = 1 orÂ 1/2

If m = 1, then the slope of the lines are 1 and 2

If m =Â 1/2, then the slope of the lines areÂ 1/2Â and 1

âˆ´ The slope of the lines are [-1 and -2] orÂ [-1/2Â and -1] or [1 and 2] orÂ [1/2Â and 1]

12. A line passes through (x1, y1) and (h, k). If the slope of the line is m, show that k â€“ y1Â = m (h â€“ x1).

Solution:

Given: the slope of the line is â€˜mâ€™.

The slope of the line passing through (x1, y1) and (h, k) is (k â€“ y1)/(h â€“ x1)

So,

(k â€“ y1)/(h â€“ x1)Â  = m

(k â€“ y1) = m (h â€“ x1)

Hence, proved.

13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that a/h + b/k = 1Â

Solution:

Let us consider if the given points A (h, 0), B (a, b) and C (0, k) lie on a line.

Then, the slope of AB = slope of BC

(b â€“ 0)/(a â€“ h) = (k â€“ b)/(0 â€“ a)

By simplifying, we get

-ab = (k-b) (a-h)

-ab = ka- kh â€“ab +bh

ka +bh = kh

Divide both sides by kh; we get

ka/kh + bh/kh = kh/kh

a/h + b/k = 1

Hence, proved.

14. Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?

Solution:

We know that line AB passes through points A (1985, 92) and B (1995, 97).

Its slope will be (97 â€“ 92)/(1995 â€“ 1985) = 5/10 = 1/2

Let â€˜yâ€™ be the population in the year 2010. Then, according to the given graph, AB must pass through point C (2010, y)

So now, slope of AB = slope of BC

15/2 = y â€“ 97

y = 7.5 + 97 = 104.5

âˆ´ The slope of line AB is 1/2, while in the year 2010, the population will be 104.5 crores.

EXERCISE 10.2 PAGE NO: 219

In Exercises 1 to 8, find the equation of the line which satisfies the given conditions.

1. Write the equations for the x-and y-axes.

Solution:

The y-coordinate of every point on the x-axis is 0.

âˆ´ The equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

âˆ´ The equation of the y-axis is y = 0.

2. Passing through the point (â€“ 4, 3) with slope 1/2

Solution:

Given:

Point (-4, 3) and slope, m = 1/2

We know thatÂ the point (x, y) lies on the line with slope m through the fixed point (x0, y0) only if its coordinates satisfy the equation y â€“ y0Â = m (x â€“ x0)

So, y â€“ 3 = 1/2 (x â€“ (-4))

y â€“ 3 = 1/2 (x + 4)

2(y â€“ 3) = x + 4

2y â€“ 6 = x + 4

x + 4 â€“ (2y â€“ 6) = 0

x + 4 â€“ 2y + 6 = 0

x â€“ 2y + 10 = 0

âˆ´ The equation of the line is x â€“ 2y + 10 = 0

3. Passing through (0, 0) with slope m.

Solution:

Given:

Point (0, 0) and slope, m = m

We know thatÂ the point (x, y) lies on the line with slope m through the fixed point (x0, y0) only if its coordinates satisfy the equation y â€“ y0Â = m (x â€“ x0)

So, y â€“ 0 = m (x â€“ 0)

y = mx

y â€“ mx = 0

âˆ´ The equation of the line is y â€“ mx = 0

4. Passing through (2, 2âˆš3)Â and inclined with the x-axis at an angle of 75o.

Solution:

Given: point (2, 2âˆš3) and Î¸ = 75Â°

Equation of line: (y â€“ y1) = m (x â€“ x1)

where, m = slope of line = tanÂ Î¸ and (x1, y1) are the points through which line passes

âˆ´Â m = tan 75Â°

75Â° = 45Â° + 30Â°

Applying the formula:

We know that the point (x, y) lies on the line with slope m through the fixed point (x1, y1), only if its coordinates satisfy the equation y â€“ y1Â = m (x â€“ x1)

Then, yÂ â€“Â 2âˆš3 = (2 + âˆš3) (x â€“ 2)

yÂ â€“Â 2âˆš3 = 2 x â€“ 4 +Â âˆš3 x â€“ 2Â âˆš3

y = 2 x â€“ 4 +Â âˆš3 x

(2 +Â âˆš3) xÂ â€“Â y â€“ 4 = 0

âˆ´ The equation of the line is (2 + âˆš3) x â€“ y â€“ 4 = 0

5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope â€“2.

Solution:

Given:

Slope, m = -2

We know that if a line L with slope m makes x-intercept d, then the equation of L is

y = m(x âˆ’ d).

If the distance is 3 units to the left of the origin, then d = -3

So, y = (-2) (xÂ â€“Â (-3))

y = (-2) (x + 3)

y = -2xÂ â€“Â 6

2x + y + 6 = 0

âˆ´ The equation of the line is 2x + y + 6 = 0

6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30o with the positive direction of the x-axis.

Solution:

Given: Î¸ = 30Â°

We know that slope, m = tan Î¸

m = tan30Â°Â = (1/âˆš3)

We know that the point (x, y) on the line with slope m and y-intercept c lies on the line only if y = mx + c

If the distance is 2 units above the origin, c = +2

So, y = (1/âˆš3)x + 2

y = (x + 2âˆš3) /Â âˆš3

âˆš3 y = x + 2âˆš3

x â€“Â âˆš3 y + 2âˆš3 = 0

âˆ´ The equation of the line is x â€“ âˆš3 y + 2âˆš3 = 0

7. Passing through the points (â€“1, 1) and (2, â€“ 4).

Solution:

Given:

Points (-1, 1) and (2, -4)

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

y â€“ 1 = -5/3 (x + 1)

3 (yÂ â€“Â 1) = (-5) (x + 1)

3yÂ â€“Â 3 = -5xÂ â€“Â 5

3yÂ â€“Â 3 + 5x + 5 = 0

5xÂ +Â 3y + 2 = 0

âˆ´ The equation of the line is 5x + 3y + 2 = 0

8. Perpendicular distance from the origin is 5 units, and the angle made by the perpendicular with the positive x-axis is 30o.

Solution:

Given: p = 5 and Ï‰ = 30Â°

We know that the equation of the line having normal distance p from the origin and angle Ï‰, which the normal makes with the positive direction of the x-axis, is given by x cos Ï‰ + y sin Ï‰ = p.

Substituting the values in the equation, we get

x cos30Â° + y sin30Â° = 5

x(âˆš3 / 2) + y(Â 1/2Â ) = 5

âˆš3 x + y = 5(2) = 10

âˆš3 x + yÂ â€“Â 10 = 0

âˆ´ The equation of the line is âˆš3 x + y â€“ 10 = 0

9. The vertices ofÂ Î”PQR are P (2, 1), Q (â€“2, 3) and R (4, 5). Find the equation of the median through the vertex R.

Solution:

Given:

Vertices of Î”PQR, i.e., P (2, 1), Q (-2, 3) and R (4, 5)

Let RL be the median of vertex R.

So, L is a midpoint of PQ.

We know that the midpoint formula is given by
.

âˆ´Â L =Â = (0, 2)

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

y â€“ 5 = -3/-4 (x-4)

(-4) (yÂ â€“Â 5) = (-3) (x â€“ 4)

-4y + 20 = -3x + 12

-4y + 20 + 3xÂ â€“Â 12 = 0

3xÂ â€“Â 4y + 8 = 0

âˆ´ The equation of median through the vertex R is 3x â€“ 4y + 8 = 0

10. Find the equation of the line passing through (â€“3, 5) and perpendicular to the line through the points (2, 5) and (â€“3, 6).

Solution:

Given:

Points are (2, 5) and (-3, 6).

We know that slope, m = (y2 â€“ y1)/(x2 â€“ x1)

= (6 â€“ 5)/(-3 â€“ 2)

= 1/-5 = -1/5

We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other.

Then, m = (-1/m)

= -1/(-1/5)

= 5

We know thatÂ the point (x, y) lies on the line with slope m through the fixed point (x0, y0), only if its coordinates satisfy the equation y â€“ y0Â = m (x â€“ x0)

Then, yÂ â€“Â 5 = 5(xÂ â€“Â (-3))

yÂ â€“Â 5 = 5x + 15

5x + 15Â â€“Â y + 5 = 0

5xÂ â€“Â y + 20 = 0

âˆ´ The equation of the line is 5x â€“ y + 20 = 0

11. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.

Solution:

We know thatÂ the coordinates of a point dividing the line segment joining the points (x1, y1) and (x2, y2) internally in the ratio m: n are

We know that slope, m = (y2 â€“ y1)/(x2 â€“ x1)

= (3 â€“ 0)/(2 â€“ 1)

= 3/1

= 3

We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other.

Then, m = (-1/m) =Â -1/3

We know thatÂ the point (x, y) lies on the line with slope m through the fixed point (x0, y0), only if its coordinates satisfy the equation y â€“ y0Â = m (x â€“ x0)

Here, the point is

3((1 + n) yÂ â€“Â 3) = (-(1 + n) x + 2 + n)

3(1 + n) yÂ â€“Â 9 = â€“ (1 + n) x + 2 + n

(1 + n) x + 3(1 + n) yÂ â€“Â nÂ â€“Â 9Â â€“Â 2 = 0

(1 + n) x + 3(1 + n) yÂ â€“Â nÂ â€“Â 11 = 0

âˆ´ The equation of the line is (1 + n) x + 3(1 + n) y â€“ n â€“ 11 = 0

12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Solution:

Given: the line cuts off equal intercepts on the coordinate axes, i.e., a = b

We know that equation of the line intercepts a and b on the x-and the y-axis, respectively, which is

x/a + y/b = 1

So, x/a + y/a = 1

x + y = aÂ â€¦Â (1)

Given: point (2, 3)

2 + 3 = a

a = 5

Substitute value of â€˜aâ€™ in (1), we get

x + y = 5

x + yÂ â€“Â 5 = 0

âˆ´ The equation of the line is x + y â€“ 5 = 0

13. Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Solution:

We know that equation of the line-making intercepts a and b on the x-and the y-axis, respectively, is x/a + y/b = 1 . â€¦ (1)

Given: sum of intercepts = 9

a + b = 9

b = 9 â€“ a

Now, substitute the value of b in the above equation, and we get

x/a + y/(9 â€“ a) = 1

Given: the line passes through point (2, 2)

So, 2/a + 2/(9 â€“ a) = 1

[2(9 â€“ a) + 2a] / a(9 â€“ a) = 1 [18 â€“ 2a + 2a] / a(9 â€“ a) = 1

18/a(9 â€“ a) = 1

18 = a (9Â â€“Â a)

18 = 9aÂ â€“Â a2

a2Â â€“ 9a + 18 = 0

Upon factorising, we get

a2Â â€“ 3a â€“ 6a + 18 = 0

a (aÂ â€“Â 3)Â â€“Â 6 (aÂ â€“Â 3) = 0

(aÂ â€“Â 3) (aÂ â€“Â 6) = 0

a = 3 or a = 6

Let us substitute in (1)

Case 1 (a = 3):

Then b = 9 â€“ 3 = 6

x/3 + y/6 = 1

2x + y = 6

2x + yÂ â€“Â 6 = 0

Case 2 (a = 6):

Then b = 9 â€“ 6 = 3

x/6 + y/3 = 1

x + 2y = 6

x + 2yÂ â€“Â 6 = 0

âˆ´ The equation of the line is 2x + y â€“ 6 = 0 or x + 2y â€“ 6 = 0

14. Find the equation of the line through the point (0, 2), making an angle 2Ï€/3 with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Solution:

Given:

Point (0, 2) and Î¸ = 2Ï€/3

We know that m = tan Î¸

m = tan (2Ï€/3) = -âˆš3

We know thatÂ the point (x, y) lies on the line with slope m through the fixed point (x0, y0), only if its coordinates satisfy the equation y â€“ y0Â = m (x â€“ x0)

yÂ â€“Â 2 = -âˆš3 (xÂ â€“Â 0)

yÂ â€“Â 2 = -âˆš3 x

âˆš3 x + y â€“ 2 = 0

Given, the equation of the line parallel to the above-obtained equation crosses the y-axis at a distance of 2 units below the origin.

So, the point = (0, -2) and m = -âˆš3

From point slope form equation,

yÂ â€“Â (-2) = -âˆš3 (xÂ â€“Â 0)

y + 2 = -âˆš3 x

âˆš3 x + y + 2 = 0

âˆ´ The equation of the line is âˆš3 x + y â€“ 2 = 0, and the line parallel to it is âˆš3 x + y + 2 = 0

15. The perpendicular from the origin to a line meets it at the point (â€“2, 9). Find the equation of the line.

Solution:

Given:

Points are origin (0, 0) and (-2, 9).

We know that slope, m = (y2 â€“ y1)/(x2 â€“ x1)

= (9 â€“ 0)/(-2-0)

= -9/2

We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other.

m = (-1/m) = -1/(-9/2) = 2/9

We know thatÂ the point (x, y) lies on the line with slope m through the fixed point (x0, y0) only if its coordinates satisfy the equation y â€“ y0Â = m (x â€“ x0)

yÂ â€“Â 9 = (2/9) (xÂ â€“Â (-2))

9(yÂ â€“Â 9) = 2(x + 2)

9yÂ â€“Â 81 = 2x + 4

2x + 4Â â€“Â 9y + 81 = 0

2xÂ â€“Â 9y + 85 = 0

âˆ´ The equation of the line is 2x â€“ 9y + 85 = 0

16. The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Solution:

Let us assume â€˜Lâ€™ along X-axis and â€˜Câ€™ along Y-axis; we have two points (124.942, 20) and (125.134, 110) in XY-plane.

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

17. The owner of a milk store finds that he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between the selling price and demand, how many litres could he sell weekly at Rs. 17/litre?

Solution:

Assuming the relationship between the selling price and demand is linear.

Let us assume the selling price per litre along X-axis and demand along Y-axis, we have two points (14, 980) and (16, 1220) in XY-plane.

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

yÂ â€“Â 980 = 120 (xÂ â€“Â 14)

y = 120 (xÂ â€“Â 14) + 980

When x = Rs 17/litre,

y = 120 (17Â â€“Â 14) + 980

y = 120(3) + 980

y = 360 + 980 = 1340

âˆ´ The owner can sell 1340 litres weekly at Rs. 17/litre.

18. P (a, b) is the mid-point of a line segment between axes. Show that the equation of the line is x/a + y/b = 2

Solution:

Let AB be a line segment whose midpoint is P (a, b).

Let the coordinates of A and B be (0, y) and (x, 0), respectively.

a (yÂ â€“Â 2b) = -bx

ayÂ â€“Â 2ab = -bx

bx + ay = 2ab

Divide both sides with ab, then

Hence, proved.

19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find the equation of the line.

Solution:

Let us consider AB to be the line segment, such that r (h, k) divides it in the ratio 1: 2.

So, the coordinates of A and B be (0, y) and (x, 0), respectively.

We know that the coordinates of a point dividing the line segment join the points (x1, y1) and (x2, y2) internally in the ratio m: n is

h = 2x/3 and k = y/3

x = 3h/2 and y = 3k

âˆ´Â A = (0, 3k) and B = (3h/2, 0)

We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by

3h(y â€“ 3k) = -6kx

3hy â€“ 9hk = -6kx

6kx + 3hy = 9hk

Let us divide both sides by 9hk, and we get,

2x/3h + y/3k = 1

âˆ´ The equation of the line is given by 2x/3h + y/3k = 1

20. By using the concept of the equation of a line, prove that the three points (3, 0), (â€“ 2, â€“ 2) and (8, 2) are collinear.

Solution:

According to the question,

If we have to prove that the given three points (3, 0), (â€“ 2, â€“ 2) and (8, 2) are collinear, then we have to also prove that the line passing through the points (3, 0) and (â€“ 2, â€“ 2) also passes through the point (8, 2).

By using the formula,

The equation of the line passing through the points (x1, y1) and (x2, y2) is given by

-5y = -2 (xÂ â€“Â 3)

-5y = -2x + 6

2xÂ â€“Â 5y = 6

If 2x â€“ 5y = 6 passes through (8, 2),

2x â€“ 5y = 2(8) â€“ 5(2)

= 16 â€“ 10

= 6

= RHS

The line passing through points (3, 0) and (â€“ 2, â€“ 2) also passes through the point (8, 2).

Hence, proved. The given three points are collinear.

EXERCISE 10.3 PAGE NO: 227

1. Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0

(ii) 6x + 3y â€“ 5 = 0
(iii) y = 0

Solution:

(i) x + 7y = 0

Given:

The equation is x + 7y = 0

The slope-intercept form is represented in the form â€˜y = mx + câ€™, where m is the slope and c is the y-intercept.

So, the above equation can be expressed as

y = -1/7x + 0

âˆ´ The above equation is of the form y = mx + c, where m = -1/7 and c = 0

(ii) 6x + 3y â€“ 5 = 0

Given:

The equation is 6x + 3y â€“ 5 = 0

The slope-intercept form is represented in the form â€˜y = mx + câ€™, where m is the slope and c is the y-intercept.

So, the above equation can be expressed as

3y = -6x + 5

y = -6/3x + 5/3

= -2x + 5/3

âˆ´ The above equation is of the form y = mx + c, where m = -2 and c = 5/3

(iii) y = 0

Given:

The equation is y = 0

The slope-intercept form is given by â€˜y = mx + câ€™, where m is the slope and c is the y-intercept.

y = 0 Ã— x + 0

âˆ´ The above equation is of the form y = mx + c, where m = 0 and c = 0

2. Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y â€“ 12 = 0

(ii) 4x â€“ 3y = 6

(iii) 3y + 2 = 0

Solution:

(i) 3x + 2y â€“ 12 = 0

Given:

The equation is 3x + 2y â€“ 12 = 0

The equation of the line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepted on the x-axis and the y-axis, respectively.

So, 3x + 2y = 12

Now, let us divide both sides by 12; we get

3x/12 + 2y/12 = 12/12

x/4 + y/6 = 1

âˆ´ The above equation is of the form x/a + y/b = 1, where a = 4, b = 6

The intercept on the x-axis is 4.

The intercept on the y-axis is 6.

(ii) 4x â€“ 3y = 6

Given:

The equation is 4x â€“ 3y = 6

The equation of the line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepted on the x-axis and the y-axis, respectively.

So, 4x â€“ 3y = 6

Now, let us divide both sides by 6; we get

4x/6 â€“ 3y/6 = 6/6

2x/3 â€“ y/2 = 1

x/(3/2) + y/(-2) = 1

âˆ´ The above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2

The intercept on the x-axis is 3/2.

The intercept on the y-axis is -2.

(iii) 3y + 2 = 0

Given:

The equation is 3y + 2 = 0

The equation of the line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepted on the x-axis and the y-axis, respectively.

So, 3y = -2

Now, let us divide both sides by -2; we get

3y/-2 = -2/-2

3y/-2 = 1

y/(-2/3) = 1

âˆ´ The above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3

The intercept on the x-axis is 0.

The intercept on the y-axis is -2/3.

3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and the angle between the perpendicular and the positive x-axis.

(i) x â€“ âˆš3y + 8 = 0

(ii) y â€“ 2 = 0

(iii) x â€“ y = 4

Solution:

(i) x â€“ âˆš3y + 8 = 0

Given:

The equation is x â€“ âˆš3y + 8 = 0

The equation of the line in normal form is given by x cos Î¸ + y sin Î¸ = p where â€˜Î¸â€™ is the angle between the perpendicular and the positive x-axis and â€˜pâ€™ is the perpendicular distance from the origin.

So now, x â€“ âˆš3y + 8 = 0

x â€“ âˆš3y = -8

Divide both the sides by âˆš(12 + (âˆš3)2) = âˆš(1 + 3) = âˆš4 = 2

x/2 â€“ âˆš3y/2 = -8/2

(-1/2)x + âˆš3/2y = 4

This is in the form of: x cos 120o + y sin 120o = 4

âˆ´ The above equation is of the form x cos Î¸ + y sin Î¸ = p, where Î¸ = 120Â° and p = 4.

Perpendicular distance of the line from origin = 4

The angle between the perpendicular and positive x-axis = 120Â°

(ii) y â€“ 2 = 0

Given:

The equation is y â€“ 2 = 0

The equation of the line in normal form is given by x cos Î¸ + y sin Î¸ = p where â€˜Î¸â€™ is the angle between the perpendicular and the positive x-axis and â€˜pâ€™ is the perpendicular distance from the origin.

So now, 0 Ã— x + 1 Ã— y = 2

Divide both sides by âˆš(02 + 12) = âˆš1 = 1

0 (x) + 1 (y) = 2

This is in the form of: x cos 90o + y sin 90o = 2

âˆ´ The above equation is of the form x cos Î¸ + y sin Î¸ = p, where Î¸ = 90Â° and p = 2.

Perpendicular distance of the line from origin = 2

The angle between the perpendicular and positive x-axis = 90Â°

(iii) x â€“ y = 4

Given:

The equation is x â€“Â y + 4 = 0

The equation of the line in normal form is given by x cos Î¸ + y sin Î¸ = p where â€˜Î¸â€™ is the angle between the perpendicular and the positive x-axis and â€˜pâ€™ is the perpendicular distance from the origin.

So now, x â€“ y = 4

Divide both the sides by âˆš(12 + 12) = âˆš(1+1) = âˆš2

x/âˆš2 â€“ y/âˆš2 = 4/âˆš2

(1/âˆš2)x + (-1/âˆš2)y = 2âˆš2

This is in the form: x cos 315o + y sin 315o = 2âˆš2

âˆ´ The above equation is of the form x cos Î¸ + y sin Î¸ = p, where Î¸ = 315Â° and p = 2âˆš2.

Perpendicular distance of the line from origin = 2âˆš2

The angle between the perpendicular and the positive x-axis = 315Â°

4. Find the distance of the point (â€“1, 1) from the line 12(x + 6) = 5(y â€“ 2).

Solution:

Given:

The equation of the line is 12(x + 6) = 5(y â€“ 2).

12x + 72 = 5y â€“ 10

12x â€“ 5y + 82 = 0 â€¦ (1)

Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 12, B = â€“5, and C = 82

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

âˆ´ The distance is 5 units.

5. Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units.

Solution:

Given:

The equation of the line is x/3 + y/4 = 1

4x + 3y = 12

4x + 3y â€“ 12 = 0 â€¦. (1)

Now, compare equation (1) with the general equation of line Ax + By + C = 0, where A = 4, B = 3, and C = -12

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

So, the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

|4a â€“ 12| = 4 Ã— 5

Â± (4a â€“ 12) = 20

4a â€“ 12 = 20 or â€“ (4a â€“ 12) = 20

4a = 20 + 12 or 4a = -20 + 12

a = 32/4 or a = -8/4

a = 8 or a = -2

âˆ´ The required points on the x-axis are (-2, 0) and (8, 0)

6. Find the distance between parallel lines.
(i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0

(ii) l(x + y) + p = 0 and l (x + y) â€“ r = 0

Solution:

(i) 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0

Given:

The parallel lines are 15x + 8y â€“ 34 = 0 and 15x + 8y + 31 = 0.

By using the formula,

The distance (d) between parallel lines Ax + By + C1Â = 0 and Ax + By + C2Â = 0 is given by

âˆ´Â The distance between parallel lines is 65/17

(ii) l(x + y) + p = 0 and l (x + y) â€“ r = 0

Given:

The parallel lines are l (x + y) + p = 0 and l (x + y) â€“ r = 0

lx + ly + p = 0 and lx + ly â€“ r = 0

By using the formula,

The distance (d) between parallel lines Ax + By + C1Â = 0 and Ax + By + C2Â = 0 is given by

âˆ´Â The distance between parallel lines is |p+r|/lâˆš2

7. Find the equation of the line parallel to the line 3x âˆ’ 4y + 2 = 0 and passing through the point (â€“2, 3).

Solution:

Given:

The line is 3x â€“ 4y + 2 = 0

So, y = 3x/4 + 2/4

= 3x/4 + Â½

Which is of the form y = mx + c, where m is the slope of the given line.

The slope of the given line is 3/4

We know that parallel lines have the same slope.

âˆ´Â Slope of other line = m = 3/4

The equation of line having slope m and passing through (x1, y1) is given by

y â€“ y1 = m (x â€“ x1)

âˆ´ The equation of the line having slope 3/4 and passing through (-2, 3) is

y â€“ 3 = Â¾ (x â€“ (-2))

4y â€“ 3 Ã— 4 = 3x + 3 Ã— 2

3x â€“ 4y = 18

âˆ´Â The equation is 3x â€“ 4y = 18

8. Find equation of the line perpendicular to the line x â€“ 7y + 5 = 0 and having x intercept 3.

Solution:

Given:

The equation of line is x â€“ 7y + 5 = 0

So, y = 1/7x + 5/7 [which is of the form y = mx + c, where m is the slope of the given line.]

The slope of the given line is 1/7

The slope of the line perpendicular to the line having slope m is -1/m

The slope of the line perpendicular to the line having a slope of 1/7 is -1/(1/7) = -7

So, the equation of the line with slope -7 and the x-intercept 3 is given by y = m(x â€“ d)

y = -7 (x â€“ 3)

y = -7x + 21

7x + y = 21

âˆ´Â The equation is 7x + y = 21

9. Find angles between the lines âˆš3x + y = 1 andÂ x +Â âˆš3y = 1.

Solution:

Given:

The lines are âˆš3x + y = 1 andÂ x +Â âˆš3y = 1

So, y = -âˆš3x + 1 â€¦ (1) and

y = -1/âˆš3x + 1/âˆš3 â€¦. (2)

The slope of the line (1) is m1 = -âˆš3, while the slope of the line (2) is m2Â = -1/âˆš3

Let Î¸ be the angle between two lines.

So,

Î¸ = 30Â°

âˆ´ The angle between the given lines is either 30Â° or 180Â°- 30Â° = 150Â°

10. The line through the points (h, 3) and (4, 1) intersects the line 7x âˆ’ 9y âˆ’19 = 0. At the right angle. Find the value of h.

Solution:

Let the slope of the line passing through (h, 3) and (4, 1) be m1

Then, m1 = (1-3)/(4-h) = -2/(4-h)

Let the slope of line 7x â€“ 9y â€“ 19 = 0 be m2

7x â€“ 9y â€“ 19 = 0

So, y = 7/9x â€“ 19/9

m2Â = 7/9

Since the given lines are perpendicular,

m1Â Ã— m2Â = -1

-2/(4-h)Â Ã— 7/9 = -1

-14/(36-9h) = -1

-14 = -1 Ã— (36 â€“ 9h)

36 â€“ 9h = 14

9h = 36 â€“ 14

h = 22/9

âˆ´ The value of h is 22/9

11. Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x â€“ x1) + B (y â€“ y1) = 0.

Solution:

Let the slope of line Ax + By + C = 0 be m

Ax + By + C = 0

So, y = -A/Bx â€“ C/B

m = -A/B

By using the formula,

Equation of the line passing through point (x1, y1) and having slope m = -A/B is

y â€“ y1 = m (x â€“ x1)

y â€“ y1= -A/B (x â€“ x1)

B (y â€“ y1) = -A (x â€“ x1)

âˆ´Â A(x â€“ x1) + B(y â€“ y1) = 0

So, the line through point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x â€“ x1) + B (y â€“ y1) = 0

Hence, proved.

12. Two lines passing through point (2, 3) intersects each other at an angle of 60o. If the slope of one line is 2, find the equation of the other line.

Solution:

Given: m1Â = 2

Let the slope of the first line be m1

And let the slope of the other line be m2.

The angle between the two lines is 60Â°.

So,

13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (â€“1, 2).

Solution:

Given:

The right bisector of a line segment bisects the line segment at 90Â°.

End-points of the line segment AB are given as A (3, 4) and B (â€“1, 2).

Let the mid-point of AB be (x, y).

x = (3-1)/2= 2/2 = 1

y = (4+2)/2 = 6/2 = 3

(x, y) = (1, 3)

Let the slope of line AB be m1

m1Â = (2 â€“ 4)/(-1 â€“ 3)

= -2/(-4)

= 1/2

And let the slope of the line perpendicular to AB be m2

m2 = -1/(1/2)

= -2

The equation of the line passing through (1, 3) and having a slope of â€“2 is

(y â€“ 3) = -2 (x â€“ 1)

y â€“ 3 = â€“ 2x + 2

2x + y = 5

âˆ´ The required equation of the line is 2x + y = 5

14. Find the coordinates of the foot of the perpendicular from the point (â€“1, 3) to the line 3x â€“ 4y â€“ 16 = 0.

Solution:

Let us consider the coordinates of the foot of the perpendicular from (-1, 3) to the line 3x â€“ 4y â€“ 16 = 0 be (a, b)

So, let the slope of the line joining (-1, 3) and (a, b) be m1

m1 = (b-3)/(a+1)

And let the slope of the line 3x â€“ 4y â€“ 16 = 0 be m2

y = 3/4x â€“ 4

m2 = 3/4

Since these two lines are perpendicular, m1Â Ã— m2Â = -1

(b-3)/(a+1) Ã— (3/4) = -1

(3b-9)/(4a+4) = -1

3b â€“ 9 = -4a â€“ 4

4a + 3b = 5 â€¦â€¦.(1)

Point (a, b) lies on the line 3x â€“ 4y = 16

3a â€“ 4b = 16 â€¦â€¦..(2)

Solving equations (1) and (2), we get

a = 68/25 and b = -49/25

âˆ´ The coordinates of the foot of perpendicular are (68/25, -49/25)

15. The perpendicular from the origin to the line y = mx + c meets it at the point (â€“1, 2). Find the values of m and c.

Solution:

Given:

The perpendicular from the origin meets the given line at (â€“1, 2).

The equation of the line is y = mx + c

The line joining the points (0, 0) and (â€“1, 2) is perpendicular to the given line.

So, the slope of the line joining (0, 0) and (â€“1, 2) = 2/(-1) = -2

The slope of the given line is m.

m Ã— (-2) = -1

m = 1/2

Since point (-1, 2) lies on the given line,

y = mx + c

2 = 1/2 Ã— (-1) + c

c = 2 + 1/2 = 5/2

âˆ´ The values of m and c are 1/2 and 5/2, respectively.

16. If p and q are the lengths of perpendiculars from the origin to the lines x cos Î¸ âˆ’ y sin Î¸ = k cos 2Î¸ and x sec Î¸ + y cosec Î¸ = k, respectively, prove that p2Â + 4q2Â = k2

Solution:

Given:

The equations of the given lines are

x cos Î¸ â€“ y sin Î¸ = k cos 2Î¸ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

x sec Î¸ + y cosec Î¸ = k â€¦â€¦â€¦â€¦â€¦â€¦.â€¦ (2)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

q = k cos Î¸ sin Î¸

Multiply both sides by 2, and we get

2q = 2k cos Î¸ sin Î¸ = k Ã— 2sin Î¸ cos Î¸

2q = k sin 2Î¸

Squaring both sides, we get

4q2Â = k2Â sin22Î¸ â€¦â€¦â€¦â€¦â€¦â€¦â€¦(4)

Now add (3) and (4); we get

p2Â + 4q2Â = k2Â cos2Â 2Î¸ + k2Â sin2Â 2Î¸

p2Â + 4q2Â = k2Â (cos2Â 2Î¸ + sin2Â 2Î¸) [Since, cos2Â 2Î¸ + sin2Â 2Î¸ = 1]

âˆ´Â p2Â + 4q2Â = k2

Hence proved.

17. In the triangle ABC with vertices A (2, 3), B (4, â€“1) and C (1, 2), find the equation and length of altitude from vertex A.

Solution:

Let AD be the altitude of triangle ABC from vertex A.

So, AD is perpendicular to BC.

Given:

Vertices A (2, 3), B (4, â€“1) and C (1, 2)

Let the slope of the line BC = m1

m1Â = (- 1 â€“ 2)/(4 â€“ 1)

m1Â = -1

Let the slope of the line AD be m2

m1Â Ã— m2Â = -1

-1 Ã— m2Â = -1

m2Â = 1

The equation of the line passing through the point (2, 3) and having a slope of 1 is

y â€“ 3 = 1 Ã— (x â€“ 2)

y â€“ 3 = x â€“ 2

y â€“ x = 1

Equation of the altitude from vertex A = y â€“ x = 1

Length of AD = Length of the perpendicular from A (2, 3) to BC

The equation of BC is

y + 1 = -1 Ã— (x â€“ 4)

y + 1 = -x + 4

x + y â€“ 3 = 0 â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

Now, compare equation (1) to the general equation of the line, i.e., Ax + By + C = 0; we get

[where, A = 1, B = 1 and C = -3]

âˆ´ The equation and the length of the altitude from vertex A are y â€“ x = 1 and

âˆš2 units, respectively.

18. If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p2 = 1/a2 + 1/b2Â

Solution:

The equation of a line whose intercepts on the axes are a and b is x/a + y/b = 1

bx + ay = ab

bx + ay â€“ ab = 0 â€¦â€¦â€¦â€¦â€¦â€¦..(1)

Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

Now, square on both sides; we get

âˆ´ 1/p2 = 1/a2 + 1/b2

Hence, proved.

Miscellaneous EXERCISE PAGE NO: 233

1. Find the values of k for which the line (k â€“ 3) x â€“ (4 â€“ k2) y + k2 â€“ 7k + 6 = 0 is

(a) Parallel to the x-axis

(b) Parallel to the y-axis

(c) Passing through the origin

Solution:

It is given that

(kÂ â€“ 3)Â xÂ â€“ (4 â€“Â k2)Â yÂ +Â k2Â â€“ 7kÂ + 6 = 0 â€¦ (1)

(a) Here, if the line is parallel to the x-axis

Slope of the line = Slope of the x-axis

It can be written as

(4 â€“Â k2)Â yÂ = (kÂ â€“ 3)Â xÂ +Â k2Â â€“ 7kÂ + 6 = 0

We get

By further calculation,

k â€“ 3 = 0

k = 3

Hence, if the given line is parallel to the x-axis, then the value of k is 3.

(b) Here, if the line is parallel to the y-axis, it is vertical, and the slope will be undefined.

So, the slope of the given line

k2 = 4

k = Â± 2

Hence, if the given line is parallel to the y-axis, then the value of k is Â± 2.

(c) Here, if the line is passing through (0, 0), which is the origin satisfies the given equation of the line.

(k â€“ 3) (0) â€“ (4 â€“ k2) (0) + k2 â€“ 7k + 6 = 0

By further calculation,

k2 â€“ 7k + 6 = 0

Separating the terms,

k2 â€“ 6k â€“ k + 6 = 0

We get

(k â€“ 6) (k â€“ 1) = 0

k = 1 or 6

Hence, if the given line is passing through the origin, then the value of k is either 1 or 6.

2. Find the values of Î¸ and p, if the equation x cos Î¸ + y sin Î¸ = p is the normal form of the line âˆš3x + y + 2 = 0.

Solution:

3. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and â€“6, respectively.

Solution:

Consider the intercepts cut by the given lines on the a and b axes.

a + b = 1 â€¦â€¦ (1)

ab = â€“ 6 â€¦â€¦.. (2)

By solving both equations, we get

a = 3 and b = -2 or a = â€“ 2 and b = 3

We know that the equation of the line whose intercepts on the a and b axes is

Case I â€“ a = 3 and b = â€“ 2

So, the equation of the line is â€“ 2x + 3y + 6 = 0, i.e. 2x â€“ 3y = 6

Case II â€“ a = -2 and b = 3

So, the equation of the line is 3x â€“ 2y + 6 = 0, i.e. -3x + 2y = 6

Hence, the required equation of the lines are 2x â€“ 3y = 6 and -3x + 2y = 6

4. What are the points on theÂ y-axis whose distance from the line x/3 + y/4 = 1 is 4 units?

Solution:

Consider (0, b) as the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4 units.

It can be written as 4x + 3y â€“ 12 = 0 â€¦â€¦. (1)

By comparing equation (1) to the general equation of line Ax + By + C = 0, we get

A = 4, B = 3 and C = â€“ 12

We know that the perpendicular distance (d) of a line Ax + By + C = 0 from (x1, y1) is written as

By cross multiplication,

20 = |3b â€“ 12|

We get

20 = Â± (3b â€“ 12)

Here, 20 = (3b â€“ 12) or 20 = â€“ (3b â€“ 12)

It can be written as

3b = 20 + 12 or 3b = -20 + 12

So, we get

b = 32/3 or b = -8/3

Hence, the required points are (0, 32/3) and (0, -8/3).

5. Find the perpendicular distance from the origin to the line joining the pointsÂ

Solution:

6. Find the equation of the line parallel to the y-axis and draw through the point of intersection of the lines xÂ â€“ 7yÂ + 5 = 0 and 3xÂ +Â yÂ = 0.

Solution:

Here, the equation of any line parallel to the y-axis is of the form

x = a â€¦â€¦. (1)

Two given lines are

x â€“ 7y + 5 = 0 â€¦â€¦ (2)

3x + y = 0 â€¦â€¦ (3)

By solving equations (2) and (3), we get

x = -5/22 and y = 15/22

(-5/ 22, 15/22) is the point of intersection of lines (2) and (3)

If the line x = a passes through point (-5/22, 15/22), we get a = -5/22

Hence, the required equation of the line is x = -5/22

7. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point where it meets the y-axis.

Solution:

It is given that

x/4 + y/6 = 1

We can write it as

3x + 2y â€“ 12 = 0

So, we get

y = -3/2 x + 6, which is of the form y = mx + c

Here, the slope of the given line = -3/2

So, the slope of line perpendicular to the given line = -1/ (-3/2) = 2/3

Consider the given line intersects, the y-axis at (0, y)

By substituting x as zero in the equation of the given line,

y/6 = 1

y = 6

Hence, the given line intersects the y-axis at (0, 6).

We know that the equation of the line that has a slope of 2/3 and passes through the point (0, 6) is

(y â€“ 6) = 2/3 (x â€“ 0)

By further calculation,

3y â€“ 18 = 2x

So, we get

2x â€“ 3y + 18 = 0

Hence, the required equation of the line is 2x â€“ 3y + 18 = 0

8. Find the area of the triangle formed by the linesÂ yÂ â€“Â xÂ = 0,Â xÂ +Â yÂ = 0 andÂ xÂ â€“Â kÂ = 0.

Solution:

It is given that

y â€“ x = 0 â€¦â€¦ (1)

x + y = 0 â€¦â€¦ (2)

x â€“ k = 0 â€¦â€¦. (3)

Here, the point of intersection of

Lines (1) and (2) is

x = 0 and y = 0

Lines (2) and (3) is

x = k and y = â€“ k

Lines (3) and (1) is

x = k and y = k

So, the vertices of the triangle formed by the three given lines are (0, 0), (k, -k) and (k, k).

Here, the area of triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is

Â½ |x1 (y2 â€“ y3) + x2 (y3 â€“ y1) + x3 (y1 â€“ y2)|

So, the area of the triangle formed by the three given lines is

= Â½ |0 (-k â€“ k) + k (k â€“ 0) + k (0 + k)| square units

By further calculation,

= Â½ |k2 + k2| square units

So, we get

= Â½ |2k2|

= k2 square units

9. Find the value ofÂ pÂ so that the three lines 3xÂ +Â yÂ â€“ 2 = 0,Â pxÂ + 2yÂ â€“ 3 = 0 and 2xÂ â€“Â yÂ â€“ 3 = 0 may intersect at one point.

Solution:

It is given that

3x + y â€“ 2 = 0 â€¦â€¦ (1)

px + 2y â€“ 3 = 0 â€¦.. (2)

2x â€“ y â€“ 3 = 0 â€¦â€¦ (3)

By solving equations (1) and (3), we get

x = 1 and y = -1

Here, the three lines intersect at one point, and the point of intersection of lines (1) and (3) will also satisfy line (2)

p (1) + 2 (-1) â€“ 3 = 0

By further calculation,

p â€“ 2 â€“ 3 = 0

So we get

p = 5

Hence, the required value of p is 5.

10. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 â€“ c3) + m2 (c3 â€“ c1) + m3 (c1 â€“ c2) = 0.

Solution:

It is given that

y = m1x + c1 â€¦.. (1)

y = m2x + c2 â€¦.. (2)

y = m3x + c3 â€¦.. (3)

By subtracting equation (1) from (2), we get

0 = (m2 â€“ m1) x + (c2 â€“ c1)

(m1 â€“ m2) x = c2 â€“ c1

So we get

Taking out the common terms,

m1 (c2 â€“ c3) + m2 (c3 â€“ c1) + m3 (c1 â€“ c2) = 0

Therefore, m1 (c2 â€“ c3) + m2 (c3 â€“ c1) + m3 (c1 â€“ c2) = 0

11. Find the equation of the lines through the point (3, 2), which makes an angle of 45Â° with the line xÂ â€“2yÂ = 3.

Solution:

Consider m1 as the slope of the required line

It can be written as

y = 1/2 x â€“ 3/2 which is of the form y = mx + c

So, the slope of the given line m2 = 1/2

We know that the angle between the required line and line x â€“ 2y = 3 is 45o

If Î¸ is the acute angle between lines l1 and l2 with slopes m1 and m2,

It can be written as

2 + m1 = 1 â€“ 2m1 or 2 + m1 = â€“ 1 + 2m1

m1 = â€“ 1/3 or m1 = 3

Case I â€“ m1 = 3

Here, the equation of the line passing through (3, 2) and having a slope 3 is

y â€“ 2 = 3 (x â€“ 3)

By further calculation,

y â€“ 2 = 3x â€“ 9

So, we get

3x â€“ y = 7

Case II â€“ m1 = -1/3

Here, the equation of the line passing through (3, 2) and having a slope -1/3 is

y â€“ 2 = â€“ 1/3 (x â€“ 3)

By further calculation,

3y â€“ 6 = â€“ x + 3

So, we get

x + 3y = 9

Hence, the equations of the lines are 3x â€“ y = 7 and x + 3y = 9

12. Find the equation of the line passing through the point of intersection of the lines 4xÂ + 7yÂ â€“ 3 = 0 and 2xÂ â€“ 3yÂ + 1 = 0 that has equal intercepts on the axes.

Solution:

Consider the equation of the line having equal intercepts on the axes as

x/a + y/a = 1

It can be written as

x + y = a â€¦.. (1)

By solving equations 4x + 7y â€“ 3 = 0 and 2x â€“ 3y + 1 = 0, we get

x = 1/13 and y = 5/13

(1/13, 5/13) is the point of intersection of two given lines.

We know that equation (1) passes through the point (1/13, 5/13).

1/13 + 5/13 = a

a = 6/13

So, equation (1) passes through (1/13, 5/13).

1/13 + 5/13 = a

We get

a = 6/13

Her, equation (1) becomes

x + y = 6/13

13x + 13y = 6

Hence, the required equation of the line is 13x + 13y = 6

13. Show that the equation of the line passing through the origin and making an angleÂ Î¸ with the line y = mx + c is .

Solution:

Consider y = m1x as the equation of the line passing through the origin

14. In what ratio, the line joining (â€“1, 1) and (5, 7) is divided by the line xÂ +Â yÂ = 4?

Solution:

By cross multiplication,

â€“ k + 5 = 1 + k

We get

2k = 4

k = 2

Hence, the line joining the points (-1, 1) and (5, 7) is divided by the line x + y = 4 in the ratio 1: 2.

15. Find the distance of the line 4xÂ + 7yÂ + 5 = 0 from the point (1, 2) along the line 2xÂ â€“Â yÂ = 0.

Solution:

It is given that

2x â€“ y = 0 â€¦.. (1)

4x + 7y + 5 = 0 â€¦â€¦ (2)

Here, A (1, 2) is a point on the line (1).

Consider B as the point of intersection of lines (1) and (2).

By solving equations (1) and (2), we get x = -5/18 and y = â€“ 5/9

So, the coordinates of point B are (-5/18, -5/9).

From the distance formula, the distance between A and B

Hence, the required distance is
.

16. Find the direction in which a straight line must be drawn through the point (â€“1, 2) so that its point of intersection with the lineÂ xÂ +Â yÂ = 4 may be at a distance of 3 units from this point.

Solution:

Consider y = mx + c as the line passing through the point (-1, 2).

So, we get

2 = m (-1) + c

By further calculation,

2 = -m + c

c = m + 2

Substituting the value of c

y = mx + m + 2 â€¦â€¦ (1)

So the given line is

x + y = 4 â€¦â€¦. (2)

By solving both equations, we get

By cross multiplication,

1 + m2 = m2 + 1 + 2m

So, we get

2m = 0

m = 0

Hence, the slope of the required line must be zero, i.e., the line must be parallel to the x-axis.

17. The hypotenuse of a right-angled triangle has its ends at points (1, 3) and (âˆ’4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Solution:

Consider ABC as the right angles triangle where âˆ C = 90o

Here, infinity such lines are present.

m is the slope of AC

So, the slope of BC = -1/m

Equation of AC â€“

y â€“ 3 = m (x â€“ 1)

By cross multiplication,

x â€“ 1 = 1/m (y â€“ 3)

Equation of BC â€“

y â€“ 1 = â€“ 1/m (x + 4)

By cross multiplication,

x + 4 = â€“ m (y â€“ 1)

By considering the values of m, we get

If m = 0,

So, we get

y â€“ 3 = 0, x + 4 = 0

If m = âˆž,

So, we get

x â€“ 1 = 0, y â€“ 1 = 0 we get x = 1, y = 1

18. Find the image of the point (3, 8) with respect to the lineÂ xÂ + 3y = 7, assuming the line to be a plane mirror.

Solution:

It is given that

x + 3y = 7 â€¦.. (1)

Consider B (a, b) as the image of point A (3, 8).

So line (1) is the perpendicular bisector of AB.

On further simplification,

a + 3b = â€“ 13 â€¦.. (3)

By solving equations (2) and (3), we get

a = â€“ 1 and b = â€“ 4

Hence, the image of the given point with respect to the given line is (-1, -4).

19. If the linesÂ yÂ = 3xÂ + 1 and 2yÂ =Â xÂ + 3 are equally inclined to the lineÂ yÂ =Â mxÂ + 4, find the value ofÂ m.

Solution:

It is given that

y = 3x + 1 â€¦â€¦ (1)

2y = x + 3 â€¦â€¦ (2)

y = mx + 4 â€¦â€¦ (3)

Here, the slopes of

Line (1), m1 = 3

Line (2), m2 = Â½

Line (3), m3 = m

We know that lines (1) and (2) are equally inclined to line (3), which means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).

On further calculation,

â€“ m2 + m + 6 = 1 + m â€“ 6m2

So, we get

5m2 + 5 = 0

Dividing the equation by 5,

m2 + 1 = 0

m = âˆš-1, which is not real.

Therefore, this case is not possible.

If

20. If the sum of the perpendicular distances of a variable point P (x,Â y) from the linesÂ xÂ +Â yÂ â€“ 5 = 0 and 3xÂ â€“ 2yÂ + 7 = 0 is always 10. Show that P must move on a line.

Solution:

In the same way, we can find the equation of the line for any signs of (x + y â€“ 5) and (3x â€“ 2y + 7)

Hence, point P must move on a line.

21. Find the equation of the line which is equidistant from parallel lines 9xÂ + 6yÂ â€“ 7 = 0 and 3xÂ + 2yÂ + 6 = 0.

Solution:

Here,

9h + 6k â€“ 7 = 3 (3h + 2k + 6) or 9h + 6k â€“ 7 = â€“ 3 (3h + 2k + 6)

9h + 6k â€“ 7 = 3 (3h + 2k + 6) is not possible as

9h + 6k â€“ 7 = 3 (3h + 2k + 6)

By further calculation,

â€“ 7 = 18 (which is wrong)

We know that

9h + 6k â€“ 7 = -3 (3h + 2k + 6)

By multiplication,

9h + 6k â€“ 7 = -9h â€“ 6k â€“ 18

We get

18h + 12k + 11 = 0

Hence, the required equation of the line is 18x + 12y + 11 = 0

22. A ray of light passing through the point (1, 2) reflects on theÂ x-axis at point A, and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Solution:

Consider the coordinates of point A as (a, 0).

Construct a line (AL) which is perpendicular to the x-axis.

Here, the angle of incidence is equal to the angle of reflection

âˆ BAL = âˆ CAL =Â Î¦

âˆ CAX =Â Î¸

It can be written as

âˆ OAB = 180Â° â€“ (Î¸Â + 2Î¦) = 180Â° â€“ [Î¸Â + 2(90Â° â€“Â Î¸)]

On further calculation,

= 180Â° â€“Â Î¸Â â€“ 180Â° + 2Î¸

=Â Î¸

So, we get

âˆ BAX = 180Â° â€“Â Î¸

By cross multiplication,

3a â€“ 3 = 10 â€“ 2a

We get

a = 13/5

Hence, the coordinates of point A are (13/5, 0).

23. Prove that the product of the lengths of the perpendiculars drawn from points to the line.

Solution:

It is given that

We can write it as

bx cos Î¸ + ay sin Î¸ â€“ ab = 0 â€¦.. (1)

24. A person standing at the junction (crossing) of two straight paths represented by the equations 2xÂ â€“ 3yÂ + 4 = 0 and 3xÂ + 4yÂ â€“ 5 = 0 wants to reach the path whose equation is 6xÂ â€“ 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.

Solution:

It is given that

2x â€“ 3y + 4 = 0 â€¦â€¦ (1)

3x + 4y â€“ 5 = 0 â€¦â€¦. (2)

6x â€“ 7y + 8 = 0 â€¦â€¦ (3)

Here, the person is standing at the junction of the paths represented by lines (1) and (2).

By solving equations (1) and (2), we get

x = â€“ 1/17 and y = 22/17

Hence, the person is standing at point (-1/17, 22/17).

We know that the person can reach path (3) in the least time if they walk along the perpendicular line to (3) from point (-1/17, 22/17)

Here, the slope of line (3) = 6/7

We get the slope of the line perpendicular to the line (3) = -1/ (6/7) = â€“ 7/6

So, the equation of the line passing through (-1/17, 22/17) and having a slope of -7/6 is written as

By further calculation,

6 (17y â€“ 22) = â€“ 7 (17x + 1)

By multiplication,

102y â€“ 132 = â€“ 119x â€“ 7

We get

1119x + 102y = 125

Therefore, the path that the person should follow is 119x + 102y = 125

## NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 10 â€“ Straight Lines includes the main topics such as

10.1 Introduction

10.2 Slope of a Line

10.2.1 Slope of a line when coordinates of any two points on the line are given

10.2.2 Conditions for parallelism and perpendicularity of lines in terms of their slopes

10.2.3 Angle between two lines

10.2.4 Collinearity of three points

10.3 Various Forms of the Equation of a Line

10.3.1 Horizontal and vertical lines

10.3.2 Point-slope form

10.3.3 Two-point form

10.3.4 Slope-intercept form

10.3.5 Intercept â€“ form

10.3.6 Normal form

10.4 General Equation of a Line

10.4.1 Different forms of Ax + By + C = 0

10.5 Distance of a Point From a Line

10.5.1 Distance between two parallel lines

Students can utilise the NCERT Solutions for Class 11 for any quick references to comprehend the above and other complex topics. Further, referring to these improves the problem-solving skills of the students.

## NCERT Solutions for Class 11 Maths Chapter 10 â€“ Straight Lines

The Straight Lines is a part of the unit Coordinate Geometry, which adds up to 10 marks of the total 80 marks. A total of 4 exercises are present in this chapter to provide them with the maximum study resources. Some of the concepts discussed in Chapter 10 of NCERT Solutions for Class 11 Maths are given below.

1. Slope (m) of a non-vertical line.
2. If a line makes an angle Ã¡ with the positive direction of the x-axis, then the slope of the line is given by m = tan Î±, Î± â‰  90Â°.
3. The slope of a horizontal line is zero, and the slope of a vertical line is undefined.
4. Two lines are parallel if and only if their slopes are equal.
5. Two lines are perpendicular if and only if the product of their slopes is â€“1.
6. Three points A, B and C are collinear if and only if the slope of AB = slope of BC.
7. The equation of the horizontal line having distance a from the x-axis is either y = a or y = â€“ a
8. The equation of the vertical line having distance b from the y-axis is either x = b or x = â€“ b
9. The point (x, y) lies on the line with slope m and through the fixed point (xo , yo ) if and only if its coordinates satisfy the equation y â€“ yo = m (x â€“ xo).
10. The point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c
11. If a line with slope m makes x-intercept d. Then, the equation of the line is y = m (x â€“ d).
12. The equation of the line having a normal distance from the origin p and angle between normal and the positive x-axis Ï‰ is given by cosÏ‰+ sinÏ‰ = pyx.
13. Any equation of the form Ax + By + C = 0, with A and B, not zero, simultaneously, is called the general linear equation or general equation of a line

Brief recall of two-dimensional geometry from earlier classes, Shifting of origin, Slope of a line and angle between two lines, Various forms of equations of a line: parallel to the axis, point-slope form, slope-intercept form, two-point form, intercept form, and normal form, General equation of a line, equation of the family of lines passing through the point of intersection of two lines, the distance of a point from a line are the topics that the students must have grasped by learning these concepts.

Disclaimer â€“Â

Dropped Topics â€“Â

10.2.4 Collinearity of Three Points (Examples 4â€“5 and Ques. 8, 13â€“14 in Exercise 10.1)

10.3.6 Normal Forms
Ques. 8 in Exercise 10.2
10.4 General Equation of a Line
Ques. 3 in Exercise 10.3
Ques. 2 (Miscellaneous Exercise)
Fourth Last Point in the Summary
10.6 Equation of Family of Lines Passing Through the Points of Intersection of Two Lines

10.7 Shifting of Origin

## Frequently Asked Questions on NCERT Solutions for Class 11 Maths Chapter 10

Q1

### What are the concepts covered in Chapter 10 Straight Lines of NCERT Solutions for Class 11 Maths?

Chapter 10 Straight Lines of NCERT Solutions for Class 11 Maths provides students with an overview of coordinate geometry, which is an important topic in the Class 11 Maths CBSE Syllabus. The graphical representation of straight lines and their algebraic forms are explained in simple language to help students to score well in the annual exam. Students will also understand the applications of various representations of a first-degree equation apart from the general equation of a line.
Q2

### Explain the equation of a line in general form as per Chapter 10 of NCERT Solutions for Class 11 Maths.

There are numerous forms of an equation in which a graphical line can be represented. The general equation is the most common form of a linear equation. For a line which is represented in two variables, x and y, of the first degree, Ax + By + C = 0 is the general equation where A and B cannot be 0. A, B and C are constants which belong to the set of real numbers, and x and y are variables which represent the coordinates of the respective axes.
Q3

### Mention the topics and subtopics covered in Chapter 10 of NCERT Solutions for Class 11 Maths.

The topics and subtopics covered in Chapter 10 of NCERT Solutions for Class 11 Maths are as follows:
10.1 Introduction
10.2 Slope of a Line
10.2.1 Slope of a line when coordinates of any two points on the line are given
10.2.2 Conditions for parallelism and perpendicularity of lines in terms of their slopes
10.2.3 Angle between two lines
10.2.4 Collinearity of three points
10.3 Various Forms of the Equation of a Line
10.3.1 Horizontal and vertical lines
10.3.2 Point-slope form
10.3.3 Two-point form
10.3.4 Slope-intercept form
10.3.5 Intercept â€“ form
10.3.6 Normal form
10.4 General Equation of a Line
10.4.1 Different forms of Ax + By + C = 0
10.5 Distance of a Point From a Line
10.5.1 Distance between two parallel lines