NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines provided here can be extremely beneficial for the students in their Maths 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 for the questions given in the textbook. Experienced teachers of Maths develop these solutions, students can practise and improve their understanding of essential topics.

We know that Chapter 10 structures to be an indispensable part of the Class 11 Maths syllabus. The important topics covered in the chapter are slope of a line, Horizontal and vertical Lines, Point-slope form, Two-point form, Slope-intercept form, Intercept â€“ form as well as Normal form. Thus, to assist students in mastering all these topics, our experts have made sure that they solve all the questions using simple methods and formulas.Â

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

### Access Answers to NCERT Class 11 Maths Chapter 10- Straight Lines

Â

Exercise 10.1 Solutions 14 Questions

Exercise 10.2 Solutions 20 Questions

Exercise 10.3 Solutions 18 Questions

Miscellaneous Exercise On Chapter 10 Solutions 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, AD which gives 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 (x_{1},y_{1}) , (x_{2}, y_{2}) and (x_{3},y_{3}) is

Are of âˆ†Â ABC = Â½ [x_{1} (y_{2} â€“ y_{3}) + x_{2} (y_{3} â€“ y_{1}) + x_{3} (y_{1} â€“ y_{2})]

= Â½ [-4 (7 + 5) + 0 (-5 â€“ 5) + 5 (5 â€“ 7)] unit^{2}

= Â½ [-4 (12) + 5 (-2)] unit^{2}

= Â½ (58) unit^{2}

= 29 unit^{2}

Are of âˆ†Â ACD = Â½ [x_{1} (y_{2} â€“ y_{3}) + x_{2} (y_{3} â€“ y_{1}) + x_{3} (y_{1} â€“ y_{2})]

= Â½ [-4 (-5 + 2) + 5 (-2 â€“ 5) + (-4) (5 â€“ (-5))] unit^{2}

= Â½ [-4 (-3) + 5 (-7) â€“ 4 (10)] unit^{2}

= Â½ (-63) unit^{2}

= -63/2 unit^{2}

Since area cannot be negative area âˆ† ACD = 63/2 unit^{2}

Area (ABCD) = 29 + 63/2

= 121/2 unit^{2}

**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 vertices of the triangle.**

**Solution:**

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

Where, AB = BC = AC = 2a

In the above figure, by 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 co-ordinates of point C are (0, a) and that of B are (0,-a)

Since 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 Pythagoras theorem

(AC)^{2}Â = OA^{2}Â + OC^{2}

(2a)^{2}= a^{2}Â +Â OC^{2}

4a^{2}Â â€“ a^{2}Â =Â OC^{2}

3a^{2 }=Â OC^{2}

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 (x _{1}, y_{1}) and Q (x_{2}, y_{2}) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.**

**Solution:**

Given:

Points P (x_{1}, y_{1}) and Q(x_{2}, y_{2})

**(i)** When PQ is parallelÂ to y axis then x_{1}Â = x_{2}

So, the distance between P and Q isÂ given by

= |y_{2} â€“ y_{1}|

**(ii)** When PQ is parallel to the x-axis then y_{1}Â = y_{2}

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

=Â

= |x_{2} â€“ x_{1}|

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

**Solution:**

Let us consider (a, 0) 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 the sides we get,

a^{2} â€“ 14a + 85 = a^{2} 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 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 (x_{1}, y_{1}) and

(x_{2}, y_{2}) is given by m = (y_{2} â€“ y_{1})/(x_{2} â€“ x_{1}) where, x â‰ x_{1}

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 the Pythagoras 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 (x_{1}, y_{1}) and

(x_{2}, y_{2}) is given by m = (y_{2} â€“ y_{1})/(x_{2} â€“ x_{1}) where, x â‰ x_{1}

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

the slope of the line BC (m_{2}) = (-1-5)/(-1-3) = -6/-4 = 3/2

the slope of the line CA (m_{3}) = (4+1)/(4+1) = 5/5 = 1

It is observed that, m_{1}.m_{3}Â = -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 y-axis measured anticlockwise.**

**Solution:**

We know that, if a line makes an angle of 30Â° with the positive direction of y-axis measured anti-clock-wise , then the angle made by the line with the positive direction of x- axis measure 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€“**âˆš**3

**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 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 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, 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 CD = (2+1)/(-3+2) = 3/-1 = -3

Hence, slope of BC = Slope of CD

âˆ´Â BCÂ âˆ¥Â CD

Thus the pair of opposite sides are quadrilateral areÂ 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 = (y_{2} â€“ y_{1})/(x_{2} â€“ x_{1}) where, x â‰ x_{1}

m = (-2 â€“(-1))/(4-3)

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

= -1/1

= -1

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

**Solution:**

Let us consider â€˜m_{1}â€™Â and â€˜mâ€™ be the slope of the two given lines such that m_{1 }= 2m

We know that if Î¸ is the angle between the lines l1 and l2 with slope m_{1}Â and m_{2}, then

1+2m^{2}Â = -3m

2m^{2}Â +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:

2m^{2}Â â€“ 3m + 1 = 0

2m^{2}Â 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 (x _{1}, y_{1}) and (h, k). If slope of the line is m, show that k â€“ y_{1}Â = m (h â€“ x_{1}).**

**Solution:**

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

The slope of the line passing through (x_{1}, y_{1}) and (h, k) is (k â€“ y_{1})/(h â€“ x_{1})

So,

(k â€“ y_{1})/(h â€“ x_{1})Â = m

(k â€“ y_{1}) = m (h â€“ x_{1})

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, slope of AB = slope of BC

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

let us simplify we get,

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

-ab = ka- kh â€“ab +bh

ka +bh = kh

Divide both the 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, the 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 the 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 satisfy the given conditions:**

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

**Solution:**

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

âˆ´Â Equation of x-axis is y = 0.

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

âˆ´Â Equation of 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 (x_{0}, y_{0}), if and only if, its coordinates satisfy the equation y â€“ y_{0}Â = m (x â€“ x_{0})

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 (x_{0}, y_{0}), if and only if, its coordinates satisfy the equation y â€“ y_{0}Â = m (x â€“ x_{0})

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 75 ^{o}.**

**Solution:**

Given: point (2, 2âˆš3) and Î¸ = 75Â°

Equation of line: (y â€“ y_{1}) = m (x â€“ x_{1})

where, m = slope of line = tanÂ Î¸ and (x_{1}, y_{1}) 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 (x_{1}, y1), if and only if, its coordinates satisfy the equation y â€“ y_{1}Â = m (x â€“ x_{1})

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 aÂ line L with slope m makes x-intercept d, then equation of L is

y = m(x âˆ’ d).

If the distance is 3 units to the left of 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 30 ^{o}Â with positive direction of the x-axis.**

**Solution:**

Given: Î¸ = 30Â°

We know that slope, m = tan Î¸

m = tan30Â°Â = (1/âˆš3)

We know thatÂ 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.

If 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 (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 30 ^{o}.**

**Solution:**

Given: p = 5 and Ï‰ = 30Â°

We know thatÂ the equation of the line having normal distance p from the origin and angle Ï‰ which the normal makes with the positive direction of 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 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 (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 = (y_{2} â€“ y_{1})/(x_{2} â€“ x_{1})

= (6 â€“ 5)/(-3 â€“ 2)

= 1/-5 = -1/5

We know thatÂ two non-vertical lines are perpendicular to each other if and 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 (x_{0}, y_{0}), if and only if, its coordinates satisfy the equation y â€“ y_{0}Â = m (x â€“ x_{0})

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 (x_{1}, y_{1}) and (x_{2}, y_{2}) internally in the ratio m: n are

We know that slope, m = (y_{2} â€“ y_{1})/(x_{2} â€“ x_{1})

= (3 â€“ 0)/(2 â€“ 1)

= 3/1

= 3

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

Then, m = (-1/m) =Â -1/3

_{0}, y_{0}), if and only if, its coordinates satisfy the equation y â€“ y_{0}Â = m (x â€“ x_{0})

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 thatÂ equation of the line intercepts a and b on x-and 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 equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.**

**Solution:**

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

Given: sum of intercepts = 9

a + b = 9

b = 9 â€“ a

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

x/a + y/(9 â€“ a) = 1

Given: the line passes through the 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) = 118/a(9 â€“ a) = 1

18 = a (9Â â€“Â a)

18 = 9aÂ â€“Â a^{2}

a^{2}Â â€“ 9a + 18 = 0

Upon factorizing, we get

a^{2}Â 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 equation of the line through the point (0, 2) making an angleÂ 2Ï€/3Â with the positive x-axis. Also, find the equation of 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

_{0}, y_{0}), if and only if, its coordinates satisfy the equation y â€“ y_{0}Â = m (x â€“ x_{0})

yÂ â€“Â 2 = -âˆš3 (xÂ â€“Â 0)

yÂ â€“Â 2 = -âˆš3 x

âˆš3 x + y â€“ 2 = 0

Given, equation of line parallel to 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 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 = (y_{2} â€“ y_{1})/(x_{2} â€“ x_{1})

= (9 â€“ 0)/(-2-0)

= -9/2

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

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

_{0}, y_{0}), if and only if, its coordinates satisfy the equation y â€“ y_{0}Â = m (x â€“ x_{0})

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 line is 2x â€“ 9y + 85 = 0.

**16. The length L (in centimetre) 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 (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 selling price and demand, how many litres could he sell weekly at Rs. 17/litre?**

**Solution:**

Assuming the relationship between selling price and demand is linear.

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

_{1}, y_{1}) and (x_{2}, y_{2}) 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 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 the 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 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 thatÂ the coordinates of a point dividing the line segment joins the points (x_{1}, y_{1}) and (x_{2}, y_{2}) 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)

_{1}, y_{1}) and (x_{2}, y_{2}) is given by

3h(y â€“ 3k) = -6kx

3hy â€“ 9hk = -6kx

6kx + 3hy = 9hk

Let us divide both the sides by 9hk, 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 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 (x_{1}, y_{1}) and (x_{2}, y_{2}) 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 the 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 â€“ intercept form and find their slopes and the y â€“ 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

Slope â€“ 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

Slope â€“ 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

Slope â€“ 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

Equation of line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepts on x axis and y â€“ 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

Intercept on x â€“ axis is 4

Intercept on y â€“ axis is 6

**(ii) **4x â€“ 3y = 6

Given:

The equation is 4x â€“ 3y = 6

Equation of line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepts on x axis and y â€“ 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

Intercept on x â€“ axis is 3/2

Intercept on y â€“ axis is -2

**(iii) **3y + 2 = 0

Given:

The equation is 3y + 2 = 0

Equation of line in intercept form is given by x/a + y/b = 1, where â€˜aâ€™ and â€˜bâ€™ are intercepts on x axis and y â€“ 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

Intercept on x â€“ axis is 0

Intercept on y â€“ axis is -2/3

**3. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between 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

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

So now, x â€“ âˆš3y + 8 = 0

x â€“ âˆš3y = -8

Divide both the sides by âˆš(1^{2} + (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 120^{o} + y sin 120^{o} = 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 line from origin = 4

Angle between perpendicular and positive x â€“ axis = 120Â°

**(ii) **y â€“ 2 = 0

Given:

The equation is y â€“ 2 = 0

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

So now, 0 Ã— x + 1 Ã— y = 2

Divide both sides by âˆš(0^{2} + 1^{2}) = âˆš1 = 1

0 (x) + 1 (y) = 2

This is in the form of: x cos 90^{o} + y sin 90^{o} = 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 line from origin = 2

Angle between perpendicular and positive x â€“ axis = 90Â°

**(iii) **x â€“ y = 4

Given:

The equation is x â€“Â y + 4 = 0

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

So now, x â€“ y = 4

Divide both the sides by âˆš(1^{2} + 1^{2}) = 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 315^{o} + y sin 315^{o} = 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 line from origin = 2âˆš2

Angle between perpendicular and positive x â€“ 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 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 (x_{1}, y_{1}) is given by

âˆ´ The distance is 5units.

**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 line isÂ x/3 + y/4 = 1

4x + 3y = 12

4x + 3y â€“ 12 = 0 â€¦. (1)

Now, compare equation (1) with 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 (x_{1}, y_{1}) 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´Â The required points on the x â€“ 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 + C_{1}Â = 0 and Ax + By + C_{2}Â = 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 + C_{1}Â = 0 and Ax + By + C_{2}Â = 0 is given by

âˆ´Â The distance between parallel lines is |p+r|/l**âˆš**2

**7. Find 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 line have same slope.

âˆ´Â Slope of other line = m = 3/4

Equation of line having slope m and passing through (x_{1}, y_{1}) is given by

y â€“ y_{1} = m (x â€“ x_{1})

âˆ´Â Equation of 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.]

Slope of the given line is 1/7

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

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

So, the equation of line with slope -7 and 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)

Slope of line (1) is m_{1}Â = -âˆš3, while the slope of line (2) is m_{2}Â = -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 right angle. Find the value of h.**

**Solution:**

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

Then, m_{1} = (1-3)/(4-h) = -2/(4-h)

Let the slope of line 7x â€“ 9y â€“ 19 = 0 be m_{2}

7x â€“ 9y â€“ 19 = 0

So, y = 7/9x â€“ 19/9

m_{2}Â = 7/9

Since, the given lines are perpendicular

m_{1}Â Ã— m_{2}Â = -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 (x _{1}, y_{1}) and parallel to the line Ax + By + C = 0 is A (x â€“ x_{1}) + B (y â€“ y_{1}) = 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 (x_{1}, y_{1}) and having slope m = -A/B is

y â€“ y_{1} = m (x â€“ x_{1})

y â€“ y_{1}= -A/B (x â€“ x_{1})

B (y â€“ y_{1}) = -A (x â€“ x_{1})

âˆ´Â A(x â€“ x_{1}) + B(y â€“ y_{1}) = 0

So, the line through point (x_{1}, y_{1}) and parallel to the line Ax + By + C = 0 is A (x â€“ x_{1}) + B (y â€“ y_{1}) = 0

Hence proved.

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

**Solution:**

Given: m_{1}Â = 2

Let the slope of the first line be m_{1}

And let the slope of the other line be m_{2}.

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 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 m_{1}

m_{1}Â = (2 â€“ 4)/(-1 â€“ 3)

= -2/(-4)

= 1/2

And let the slope of the line perpendicular to AB be m_{2}

m_{2} = -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 perpendicular from the point (â€“1, 3) to the line 3x â€“ 4y â€“ 16 = 0.**

**Solution:**

Let us consider the co-ordinates 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 m_{1}

m_{1 }= (b-3)/(a+1)

And let the slope of the line 3x â€“ 4y â€“ 16 = 0 be m_{2}

y = 3/4x â€“ 4

m_{2} = 3/4

Since these two lines are perpendicular, m_{1}Â Ã— m_{2}Â = -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 co-ordinates of the foot of perpendicular is (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 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

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 p ^{2}Â + 4q^{2}Â = k^{2}**

**Solution:**

Given:

The equations of 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 (x_{1}, y_{1}) is given by

q = k cos Î¸ sin Î¸

Multiply both sides by 2, we get

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

2q = k sin 2Î¸

Squaring both sides, we get

4q^{2}Â = k^{2}Â sin^{2}2Î¸ â€¦â€¦â€¦â€¦â€¦â€¦â€¦(4)

Now add (3) and (4) we get

p^{2}Â + 4q^{2}Â = k^{2}Â cos^{2}Â 2Î¸ + k^{2}Â sin^{2}Â 2Î¸

p^{2}Â + 4q^{2}Â = k^{2}Â (cos^{2}Â 2Î¸ + sin^{2}Â 2Î¸) [Since, cos^{2}Â 2Î¸ + sin^{2}Â 2Î¸ = 1]

âˆ´Â p^{2}Â + 4q^{2}Â = k^{2}

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 the 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 slope of line BC = m_{1}

m_{1}Â = (- 1 â€“ 2)/(4 â€“ 1)

m_{1}Â = -1

Let slope of line AD be m_{2}

AD is perpendicular to BC

m_{1}Â Ã— m_{2}Â = -1

-1 Ã— m_{2}Â = -1

m_{2}Â = 1

The equation of the line passing through 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

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 (x_{1}, y_{1}) is given by

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

Length of AD =

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 perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p ^{2} = 1/a^{2} + 1/b^{2}Â **

**Solution:**

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 (x_{1}, y_{1}) is given by

Now square on both the sides we get

âˆ´ 1/p^{2} = 1/a^{2} + 1/b^{2}

Hence proved.

Miscellaneous EXERCISE PAGE NO: 233

**1. Find the values of k for which the line (k â€“ 3) x â€“ (4 â€“ k ^{2}) y + k^{2} 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

(*k*Â â€“ 3)Â *x*Â â€“ (4 â€“Â *k*^{2})Â *y*Â +Â *k*^{2}Â â€“ 7*k*Â + 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 *k*^{2})Â *y*Â = (*k*Â â€“ 3)Â *x*Â +Â *k*^{2}Â â€“ 7*k*Â + 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

k^{2} = 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 line.

(k â€“ 3) (0) â€“ (4 â€“ k^{2}) (0) + k^{2} â€“ 7k + 6 = 0

By further calculation

k^{2} â€“ 7k + 6 = 0

Separating the terms

k^{2} 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 a and b axes.

a + b = 1 â€¦â€¦ (1)

ab = â€“ 6 â€¦â€¦.. (2)

By solving both the 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 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 lineÂ 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 (x_{1}, y_{1}) 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 toÂ y-axis and drawn through the point of intersection of the linesÂ 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 intersect 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 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 (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is

Â½ |x_{1} (y_{2} â€“ y_{3}) + x_{2} (y_{3} â€“ y_{1}) + x_{3} (y_{1} â€“ y_{2})|

So the area of triangle formed by the three given lines

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

By further calculation

= Â½ |k^{2} + k^{2}| square units

So we get

= Â½ |2k^{2}|

= k^{2} 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 = m _{1}x + c_{1}, y = m_{2}x + c_{2} and y = m_{3}x + c_{3} are concurrent, then show that m_{1} (c_{2} â€“ c_{3}) + m_{2} (c_{3} â€“ c_{1}) + m_{3} (c_{1} â€“ c_{2}) = 0.**

**Solution:**

It is given that

y = m_{1}x + c_{1} â€¦.. (1)

y = m_{2}x + c_{2} â€¦.. (2)

y = m_{3}x + c_{3} â€¦.. (3)

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

0 = (m_{2} â€“ m_{1}) x + (c_{2} â€“ c_{1})

(m_{1} â€“ m_{2}) x = c_{2} â€“ c_{1}

So we get

Taking out the common terms

m_{1} (c_{2} â€“ c_{3}) + m_{2} (c_{3} â€“ c_{1}) + m_{3} (c_{1} â€“ c_{2}) = 0

Therefore, m_{1} (c_{2} â€“ c_{3}) + m_{2} (c_{3} â€“ c_{1}) + m_{3} (c_{1} â€“ c_{2}) = 0.

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

**Solution:**

Consider m_{1} 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 m_{2} = 1/2

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

If Î¸ is the acute angle between lines l_{1} and l_{2} with slopes m_{1} and m_{2}

It can be written as

2 + m_{1} = 1 â€“ 2m_{1} or 2 + m_{1} = â€“ 1 + 2m_{1}

m_{1} = â€“ 1/3 or m_{1} = 3

Case I â€“ m_{1} = 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€“ m_{1} = -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 4 xÂ + 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 point (1/13, 5/13)

1/13 + 5/13 = a

a = 6/13

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

1/13 + 5/13 = a

We get

a = 6/13

Here the 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 = m_{1}x 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 4 xÂ + 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 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 the equations we get

By cross multiplication

1 + m^{2} = m^{2} + 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 the 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 = 90^{o}

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 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 + 3yÂ = 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 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), m_{1} = 3

Line (2), m_{2} = Â½

Line (3), m_{3} = m

We know that the 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€“ m^{2} + m + 6 = 1 + m â€“ 6m^{2}

So we get

5m^{2} + 5 = 0

Dividing the equation by 5

m^{2} + 1 = 0

m = âˆš-1, which is not real.

Therefore, this case is not possible.

If

**20. If 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 line for any signs of (x + y â€“ 5) and (3x â€“ 2y + 7)

Hence, point P must move on a line.

**21. Find equation of the line which is equidistant from parallel lines 9 xÂ + 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 angle of reflection

âˆ BAL = âˆ CAL =Â *Î¦*

âˆ CAX =Â *Î¸*

It can be written as

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

On further calculation

= 180Â° â€“Â *Î¸*Â â€“ 180Â° + 2*Î¸*

=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 the 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 2 xÂ â€“ 3yÂ + 4 = 0 and 3xÂ + 4yÂ â€“ 5 = 0 wants to reach the path whose equation is 6xÂ â€“ 7yÂ + 8 = 0 in the least time. Find 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 he walks along the perpendicular line to (3) from point (-1/17, 22/17)

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

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

So the equation of 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

Chapter 10 Straight Lines of NCERT Solutions for Class 11 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

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

The Straight Lines is a part of the unit Coordinate Geometry, that 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 as given below:

- Slope (m) of a non-vertical line.
- If a line makes an angle Ã¡ with the positive direction of x-axis, then the slope of the line is given by m = tan Î±, Î± â‰ 90Â°.
- Slope of a horizontal line is zero and slope of vertical line is undefined.
- Two lines are parallel if and only if their slopes are equal.
- Two lines are perpendicular if and only if the product of their slopes is â€“1.
- Three points A, B and C are collinear if and only if slope of AB = slope of BC.
- Equation of the horizontal line having distance a from the x-axis is either y = a or y = â€“ a.
- Equation of the vertical line having distance b from the y-axis is either x = b or x = â€“ b.
- The point (x, y) lies on the line with slope m and through the fixed point (x
_{o}, y_{o}), if and only if its coordinates satisfy the equation y â€“ y_{o}= m (x â€“ x_{o}). - 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.
- If a line with slope m makes x-intercept d. Then equation of the line is y = m (x â€“ d).
- The equation of the line having normal distance from origin p and angle between normal and the positive x-axis Ï‰ is given by cosÏ‰+ sinÏ‰ = pyx .
- Any equation of the form Ax + By + C = 0, with A and B are 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 axis, point-slope form, slope-intercept form, two-point form, intercept form, and normal form, General equation of a line, equation of family of lines passing through the point of intersection of two lines, distance of a point from a line are the topics that the students must have grasped by learning these concepts.