# NCERT Solutions For Class 11 Maths Chapter 10

## NCERT Solutions Class 11 Maths Straight Lines

NCERT solutions class 11 maths chapter 10 straight lines is one of the ideal tools to prepare straight lines for class 11 mathematics exam. The NCERT solutions for class 11 maths chapter 10 straight lines is provided here so that students can prepare for their exam more effectively. The NCERT solutions for class 11 maths chapter 10 is prepared in such a way that it covers all the key areas and concepts of straight line. Students must practice the solutions regularly to score good marks in the exam. Check the NCERT solutions for class 11 maths chapter 10 given below.

### NCERT Solutions Class 11 Maths Chapter 10 Exercises

Exercise – 10.1

Q-1. Construct a quadrilateral in the Cartesian plane with vertices (-2, 5), (0, 6), (4, -4) and (-3, -1). Also, find the area of the quadrilateral.

Solution.

Let, MNOP be the given quadrilateral having vertices M (-2, 5), N (0, 6), O (4, -4) and P (-3, -1).

Now, plot M, N, O and P on the Cartesian plane and join MN, NO, OP and PM.

The quadrilateral, thus, formed will be

As we need to find the area of the quadrilateral MNOP, so draw a diagonal, say, MO.

Now,

Area of (MNOP) = area of (ΔMNO$\Delta MNO$) + area of ( ΔMOP$\Delta MOP$ )

Area of a triangle with vertices (a1, b1), (a2, b2), (a3, b3) and (a4, b4) is given by:

12$\frac{ 1 }{ 2 }$ | a1 (b2 – b3 ) + a2 (b3 – b1 ) + a3 (b1 – b2 ) |

Now,

Vertices of ΔMNO$\Delta MNO$ is M (-2, 5), N (0, 6) and O (4, -4)

Area of ΔMNO$\Delta MNO$ = 12|[2(6(4))+0(4(5))+4(5(6))]|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -6 – \left ( -4 \right ) \right ) + 0\left ( -4 – \left ( 5\right ) \right ) + 4\left ( 5 – \left ( 6 \right ) \right ) \right ] \right |$

= 12|[2(6+4)+0(45)+4(56)]|=12|[20+04]|=12|24|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -6 + 4 \right ) + 0\left ( -4 – 5 \right ) + 4\left ( 5 – 6 \right ) \right ] \right | \\ = \frac{ 1 }{ 2 }\left | \left [ -20 + 0 – 4 \right ] \right | \\ = \frac{ 1 }{ 2 }\left | -24 \right |$

= 12×24$\frac{ 1 }{ 2 } \times 24$

= 12 unit2

Vertices of ΔMOP$\Delta MOP$ is M (-2, 5), O (4, -4) and P (-3, -1).

Area of ΔMOP$\Delta MOP$ = 12|[2(4(1))+4(1(5))3(5(4))]|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -4 – \left ( -1 \right ) \right ) + 4\left ( -1 – \left ( 5\right ) \right ) – 3\left ( 5 – \left ( -4 \right ) \right ) \right ] \right |$

= 12|[2(4+1)+4(15)3(5+4)]|=12|[62427]|=12|45|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -4 + 1 \right ) + 4\left ( -1 – 5 \right ) – 3\left ( 5 + 4 \right ) \right ] \right | \\ = \frac{ 1 }{ 2 }\left | \left [ 6 – 24 – 27 \right ] \right | \\ = \frac{ 1 }{ 2 }\left | -45 \right |$

= 12×45$\frac{ 1 }{ 2 } \times 45$

= 22.5 unit2

Area of MNOP = Area of ΔMNO$\Delta MNO$ + Area of ΔMOP$\Delta MOP$

= 12 + 22.5 = 34.5 22.5 unit2

Q-2. Consider an equilateral triangle, each of whose sides are 2b which lies on the y- axis in such a manner that, the mid- point of each of its base is at the origin. Obtain all the vertices of the equilateral triangle.

Solution.

Let, XYZ be the given equilateral triangle each of whose side is 2b.

Since, XYZ is an equilateral triangle,

So, XY = YZ = ZX = 2b

Now, consider that the base YZ lies along y-axis in such a way that it’s mid- point is at the origin.

So, YA = AZ = b, where A is the origin.

Thus, by our observation, the coordinates of the point Y are (0, b), while the coordinates of the point Z is (0, -a).

We know that, the line joining the vertex of an equilateral triangle having the mid- point of its opposite sides is perpendicular.

Thus, the vertex X lies on the y- axis.

By applying the Pythagoras theorem to ΔXAZ$\Delta XAZ$, we will get

( XZ )2 = ( AX )2 + ( AZ )2

(2b)2=(AX)2+(b)2$\Rightarrow \left ( 2b \right )^{ 2 } = \left ( AX \right )^{ 2 } + \left ( b \right )^{ 2 }$ (AX)2=(4b2b2)$\Rightarrow \left ( AX \right )^{ 2 } = \left ( 4b^{ 2 } – b^{ 2 } \right )$ (AX)2=3b2$\Rightarrow \left ( AX \right )^{ 2 } = 3b^{ 2 }$ AX=±3b$\Rightarrow AX = \pm \sqrt{ 3 }b$

The coordinate of the point X = ( ±3b$\pm \sqrt{ 3 }b$ , 0 )

Hence, the vertices of the equilateral triangle XYZ are X( 3b$\sqrt{ 3 }b$ , 0 ), Y( 0, b ) and Z( 0, -b ) or X( 3b$-\sqrt{ 3 }b$ , 0 ), Y( 0, b ) and Z( 0, -b )

Q-3. What is the distance between X (a1, b1) and Y (a2, b2) if:

(a) XY is parallel to x- axis

(b) XY is parallel to y- axis

Solution.

Given points are X (a1, b1) and Y (a2, b2)

(a) The given condition is:

XY is parallel to the x- axis, i.e., b1 = b2

Therefore, the distance hence measured between X and Y in this case = (a2a1)2+(b2b1)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= (a2a1)2+(b2b2)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 2 } \right )^{ 2 } }$

= (a2a1)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } }$

= | a2 – a1 |

(b) The given condition is:

XY is parallel to the y- axis, i.e., a1 = a2

Therefore, the distance hence measured between X and Y in this case = (a2a1)2+(b2b1)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= (a2a2)2+(b2b1)2$\sqrt{ \left ( a_{ 2 } – a_{ 2 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= (b2b1)2$\sqrt{ \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= | b2 – b1 |

Q-4. Consider two points (6, 5) and (4, 2). Get a point on the y- axis which is equivalent from the given two points.

Solution.

Let us assume that, (0, y) be the given point on y- axis which is equivalent to the two given points (6, 5) and (4, 2).

Now,

(60)2+(5a)2=(40)2+(2a)2$\sqrt{ \left ( 6 – 0 \right )^{ 2 } + \left ( 5 – a \right )^{ 2 }} = \sqrt{ \left ( 4 – 0 \right )^{ 2 } + \left ( 2 – a \right )^{ 2 }}$ 36+2510a+a2=16+44a+a2$\Rightarrow \sqrt{ 36 + 25 – 10a + a^{ 2 }} = \Rightarrow \sqrt{ 16 + 4 – 4a + a^{ 2 }}$ 41=6a$\Rightarrow \sqrt{ 41 } = \sqrt{ 6a }$

Squaring both side, we will get

6a = 41

a = 416$\frac{ 41 }{ 6 }$

a = 6.8 unit

Therefore, (0, 416$\frac{ 41 }{ 6 }$ ) is the point on Y – axis which is equivalent to the given points (6, 5) and (4, 2).

Q-5. What is the slope of a line passing through the origin and, the mid- point of the line- segment joining the two points O (0, -5) and A (9, 0)?

Solution.

The two points of the line- segment is O (0, -5) and A (9, 0).

Then, the co-ordinates of the mid-point of line-segment OA are-

(0+92,5+02)$\left( \frac{ 0 + 9 }{ 2 } , \frac{ -5 + 0 }{ 2 } \right )$ = ( 4.5, -2.5 )

We know that the slope (say, m) of the non- vertical line which passes through the points (a1, b1) and (a2, b2) is:

m = b1b2a1a2$\frac{ b_{1} – b_{2} }{ a_{1} – a_{2} }$, a2 ≠ a1

Hence, the slope of the line- segment passing through (0, 0) and (4.5, -2.5) is given by:

2.504.50=2.54.5=2545=59$\frac{ -2.5 – 0 }{ 4.5 – 0 } = \frac{ -2.5 }{ 4.5 } = \frac{ -25 }{ 45 } = \frac{ -5 }{ 9 }$

Therefore, the required slope of the line-segment = 59$\frac{ -5 }{ 9 }$

Q-6. Prove that the points (5, 5), (4, 6) and (-2, -2) are the vertices of the right- angled triangle, without using Pythagoras theorem.

Solution.

There are three points given in the question. Let, those points be the vertices of the triangle XYZ, namely, X (5, 5), Y (4, 6) and Z (-2, -2).

We know that, the slope (say, m) of the non- vertical line which passes through the points (a1, b1) and (a2, b2) is:

m = b1b2a1a2$\frac{ b_{1} – b_{2} }{ a_{1} – a_{2} }$, a2 ≠ a1

The slope of XY( m1 ) = 6545$\frac{ 6 – 5 }{ 4 – 5 }$ = -1.

The slope of YZ( m2 ) = 2624=86=43$\frac{ -2 – 6 }{ -2 – 4} = \frac{ -8 }{ -6 } = \frac{ 4 }{ 3 }$

The slope of ZX( m3 ) = 5+25+2=77=1$\frac{ 5 + 2 }{ 5 + 2 } = \frac{ 7 }{ 7 } = 1$

Here, we can see that, m1. m3 = -1

This proves that, the line- segments XY and ZX is perpendicular to each other, i.e., the triangle XYZ is right- angled at X (5, 5).

Hence, proved that the given points, i.e., (5, 5), (4, 6) and (-2, -2) are the vertices of the right- angled triangle.

Q-7. What is the slope of the line which makes an angle 60$60^{\circ}$ along the positive direction of the Y- axis which is measured in anticlockwise sequence.

Solution.

By mathematical calculation rule,

If a line makes an angle of 60$60^{\circ}$ with any of the positive direction of the axis which is measured in an anticlockwise sequence, then the angle made by the line with positive direction of the corresponding axis is also measured in anticlockwise sequence is given by,

90+x$90^{\circ} + x^{\circ}$.

Then, if the angle measured by the line in the x- axis measured in positive direction of the x- axis, measured in anticlockwise direction is-

90+60=150$90^{\circ} + 60^{\circ} = 150^{\circ}$

Therefore, the slope of the given line is tan 150$150^{\circ}$ = tan( 18030)$180^{\circ} – 30^{\circ} )$ = – tan 30$30^{\circ}$ = 13$\frac{ 1 }{ \sqrt{ 3 }}$.

Q-8. What will be the value of a so that the points (a, -2), (3, 2) and (5, 6) get collinear to each other?

Solution.

Let us assume that, the points X (a, -2), Y (3, 2) and Z (5, 6) are collinear.

Now,

Slope of XY = Slope of YZ

2(2)3a=6253$\Rightarrow \frac{2 – \left ( -2 \right )}{ 3 – a } = \frac{ 6 – 2 }{ 5 – 3 }$ 2+23a=42$\Rightarrow \frac{2 + 2 }{ 3 – a } = \frac{ 4 }{ 2 }$ 43a=2$\Rightarrow \frac{ 4 }{ 3 – a } = 2$

4 = 6 – 2a

2a = 2

a = 1

Therefore, the value of a = 1.

Q-9. Prove that the points (-3, -2), (5, 0), (4, 4) and (-4, 2) are the vertices of a parallelogram without using the distance formula.

Solution.

Let us denote the given points as P (-3, -2), Q (5, 0), R (4, 4) and S (-4, 2).

Now,

Slope of the side PQ = 0+25+3=28=14$\frac{ 0 + 2 }{ 5 + 3 } = \frac{ 2 }{ 8 } = \frac{ 1 }{ 4 }$

Slope of the side RS = 2444=28=14$\frac{ 2 – 4 }{ -4 – 4 } = \frac{ -2 }{ -8 } = \frac{ 1 }{ 4 }$

Thus,

Slope of the side PQ = Slope of the side RS

Hence,

PQ and RS are parallel to each other.

Now,

Slope of the side QR = 4045=41=4$\frac{ 4 – 0 }{ 4 – 5 } = \frac{ 4 }{ -1 } = -4$

Slope of the side PS = 2+24+3=41=4$\frac{ 2 + 2 }{ -4 + 3 } = \frac{ 4 }{ -1 } = -4$

Thus,

Slope of the side QR = Slope of the side PS

Hence,

QR and PS are parallel to each other.

Hence, both of the opposite side‘s pairs of the quadrilateral PQRS are parallel. Thus, PQRS is a parallelogram.

Therefore, the given points P (-3, -2), Q (5, 0), R (4, 4) and S (-4, 2) are the vertices of the parallelogram.

Q-10. Consider two points (4, -2) and (5, -3). What is the angle between the x- axis and the line joining the given two points?

Solution.

The slope of the line joining the given two points, namely, X (4, -2) and Y (5, -3) is given by:

M = 3(2)54$\frac{ -3 – \left( -2 \right ) }{ 5 – 4 }$ = -3 + 2 = -1

Thus,

The angle of inclination, say Θ$\Theta$, of the line joining the given points X (4, -2) and Y (5, -3) is:

tanΘ$\Theta$ = m = -1

Since, Θ$\Theta$ is (9045$90^{\circ} – 45^{\circ}$ ) = 135$135^{\circ}$

Therefore, between the x- axis and the line joining the given two points, the angle of inclination is 135$135^{\circ}$.

Q-11. Assume that the slope of a line AB is twice the slope of the other line PQ. Get all the slopes of the lines when the tangent of the angle between them is 13$\frac{ 1 }{ 3 }$.

Solution.

Let us assume that, the slope of the given two lines be m1 and m2, so that

m2 = 2 m1

Now,

It is known that if Θ$\Theta$ is the angle between the two given lines, namely AB and PQ, whose slopes are m1 and m2, then

tanΘ$\Theta$ = m2m11+m1m2$\left | \frac{ m_{ 2 } – m_{ 1 } }{ 1 + m_{ 1 } m_{ 2 }} \right |$

According to the question, the tangent of the angle between the two given lines is 13$\frac{ 1 }{ 3 }$.

13$\frac{ 1 }{ 3 }$ = 2m1m11+m1×2m1$\left | \frac{ 2m_{ 1 } – m_{ 1 } }{ 1 + m_{ 1 } \times 2m_{ 1 }} \right |$

13=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \left | \frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }} \right |$ 13=m11+(2m1)2or13=(m11+(2m1)2)=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }} \; or \; \frac{ 1 }{ 3 } = -\left (\frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }} \right ) = \frac{ m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }}$

CASE- I:

13=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }}$ 1+(2m1)2=3m1$\Rightarrow 1 + \left ( 2m_{ 1 } \right )^{ 2 } = -3 m_{ 1 }$ 1+3m1+(2m1)2=0$\Rightarrow 1 + 3 m_{ 1 } + \left ( 2m_{ 1 } \right )^{ 2 } = 0$ (2m1)2+2m1+m1+1=0$\Rightarrow \left ( 2m_{ 1 } \right )^{ 2 } + 2 m_{ 1 } + m_{ 1 } + 1 = 0$ 2m1(m1+1)+1(m1+1)=0$\Rightarrow 2m_{ 1 } \left ( m_{ 1 } + 1 \right ) + 1\left ( m_{ 1 } + 1 \right ) = 0$ (m1+1)(2m1+1)=0$\Rightarrow \left ( m_{ 1 } + 1 \right ) \left ( 2m_{ 1 } + 1 \right ) = 0$ m1=1orm=12$\Rightarrow m_{ 1 } = -1 \; or \; m = -\frac{ 1 }{ 2 }$

Let, m1 = -1, then the slopes of the lines are -1 and -2.

Let, m=12$m = -\frac{ 1 }{ 2 }$, then the slopes of the given lines, namely AB and PQ are 12$-\frac{ 1 }{ 2 }$ and -1.

CASE- II:

13=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \frac{ m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }}$ 1+(2m1)2=3m1$\Rightarrow 1 + \left ( 2m_{ 1 } \right )^{ 2 } = 3 m_{ 1 }$ 13m1+(2m1)2=0$\Rightarrow 1 – 3 m_{ 1 } + \left ( 2m_{ 1 } \right )^{ 2 } = 0$ (2m1)22m1m1+1=0$\Rightarrow \left ( 2m_{ 1 } \right )^{ 2 } – 2 m_{ 1 } – m_{ 1 } + 1 = 0$ 2m1(m11)1(m11)=0$\Rightarrow 2m_{ 1 } \left ( m_{ 1 } – 1 \right ) – 1\left ( m_{ 1 } – 1 \right ) = 0$ (m11)(2m11)=0$\Rightarrow \left ( m_{ 1 } – 1 \right ) \left ( 2m_{ 1 } – 1 \right ) = 0$ m1=1orm=12$\Rightarrow m_{ 1 } = 1 \; or \; m = \frac{ 1 }{ 2 }$

Let, m1 = 1, then the slopes of the lines are 1 and 2.

Let, m=12$m = \frac{ 1 }{ 2 }$, then the slopes of the given lines namely, AB and PQ are 12$\frac{ 1 }{ 2 }$ and 1.

Therefore, the slopes of the lines AB and PQ are -1 and -2 or,  12$-\frac{ 1 }{ 2 }$ and -1 or,  12$\frac{ 1 }{ 2 }$ and 1 or,  1 and 2.

Q-12. Consider a line passing through two points (a1, b1) and (j, k). Assume that the slope of the line passing through these points is m. Prove that:

k – b1 = m (j – a1).

Solution.

As we know that:

The slope of the line, say AB, passing through the two given points (a1, b1) and (j, k) is given by:

kb1ja1$\frac{ k – b_{ 1 } }{ j – a_{ 1 } }$

kb1ja1$\frac{ k – b_{ 1 } }{ j – a_{ 1 } }$ = m

kb1=m(ja1)$\Rightarrow k – b_{ 1 } = m \left (j – a_{ 1 } \right )$

Therefore, kb1=m(ja1)$\Rightarrow k – b_{ 1 } = m \left (j – a_{ 1 } \right )$.

Hence, proved.

Q-13. Assume that the three points (a, 0), (p, r) and (0, h) lies on the same line. Prove that:

pa+rh=1$\frac{ p }{ a } + \frac{ r }{ h } = 1$.

Solution.

As per the given condition in the question,

Points X( a, 0 ), Y( p, r ) and Z( 0, h ) lies on the same line, then

The slope of XY = The slope of YZ

r0pa=hr0p$\Rightarrow \frac{ r – 0 }{ p – a } = \frac{ h – r }{ 0 – p }$ rpa=hrp$\Rightarrow \frac{ r }{ p – a } = \frac{ h – r }{ – p }$

-pr = (h – r) (p – a)

-pr = hp – ha – rp + ra

hp + ra = ha

Dividing both side by ha, we will get

hpha+raha=haha$\frac{ hp }{ ha } + \frac{ ra }{ ha } = \frac{ ha }{ ha }$ pa+rh=1$\Rightarrow \frac{ p }{ a } + \frac{ r }{ h } = 1$

Hence, proved.

Q-14. Take the records of population and year graph given below. What will be the slope of the line XY? By using this, find the population in the year 2005.

Solution.

Here we can observe that the line XY passes through points X (1985, 90) and Y (1995, 95).

Then, the slope of the line XY is given by

959019951985=510=12$\frac{ 95 – 90 }{ 1995 – 1985 } = \frac{ 5 }{ 10 } = \frac{ 1 }{ 2 }\\$

Let, a be the population of the year 2005.

Then, as per the data in the graph, the line XY will exactly pass through the point Z( 2005, y ).

Slope of XY = Slope of YZ

12=y9520051995$\Rightarrow \frac{ 1 }{ 2 } = \frac{ y – 95 }{ 2005 – 1995 }$ 12=y9510$\Rightarrow \frac{ 1 }{ 2 } = \frac{ y – 95 }{ 10 }$ 102=y95$\Rightarrow \frac{ 10 }{ 2 } = y – 95\\$

So, y – 95 = 5

So, y = 100

Hence, the slope of the line XY is 12$\frac{ 1 }{ 2 }$, whereas, in the year 2005, the expected population will be about 100 crores.

Exercise – 10.2

Q-1. Write all the possible equations for the two axis, which is x- axis and y- axis.

Solution.

As we know that,

The x- coordinates at all the points on the y- axis is 0.

So, the equations on the y – axis is y = 0.

The y- coordinates at all the points on the x- axis is 0.

So, the equations on the x – axis is x = 0.

Q-2. Write the equation for the line which passes through the point (-5, 4) with having slope 14$\frac{ 1 }{ 4 }$.

Solution.

It is known that the equation for the line passing through any point, say, (x0, y0), with having slope m, is given by-

(y – y0) = m (x – x0)

Therefore, the equation for the line passing through (-5, 4) with having slope 14$\frac{ 1 }{ 4 }$, is

(y – 4) = 14$\frac{ 1 }{ 4 }$( x – ( -5 ))

(y – 4) = 14$\frac{ 1 }{ 4 }$( x + 5 )

4(y – 4) = (x + 5)

4y – 16 = x + 5

i.e., x – 4y + 11 = 0

Hence, the equation of the line which passes through (-5, 4) is given by-

x – 4y + 11 = 0

Q-3. Write the equation for the line which passes through the point (0, 0) with having slope S.

Solution.

It is known that the equation for the line passing through any point, say, (x0, y0), with having slope m, is –

(y – y0) = m (x – x0)

Therefore, the equation for the line passing through (0, 0) with having slope S, is

(y – 0) = S (x – 0)

i.e., y = Sx

Hence, the equation of the line which passes through (0, 0) is given by-

y = Sx

Q-4. Write the equation for the line which passes through ( 3, 33$\sqrt{ 3}$ ) which is inclined on x- axis at an angle 75$75^{\circ}$.

Solution.

The slope of the line inclined on the x- axis at an angle 75$75^{\circ}$ is m = tan 75$75^{\circ}$

m=tan(45+30)$\Rightarrow m = tan\left ( 45^{\circ} + 30^{\circ} \right )$ m=tan45+tan301tan45.tan30$\Rightarrow m = \frac{ tan 45^{\circ} + tan 30^{\circ} }{ 1 – tan 45^{\circ} . tan 30^{\circ} }$ m=1+1311.13=3+13313$\Rightarrow m = \frac{ 1 + \frac{ 1 }{ \sqrt{ 3 } } }{ 1 – 1.\frac{ 1 }{ \sqrt{ 3 } }} = \frac{\frac{ \sqrt{ 3 } + 1 }{ \sqrt{ 3 } }}{ \frac{ \sqrt{ 3 } – 1 }{ \sqrt{ 3 } } }$ m=3+131$\Rightarrow m = \frac{ \sqrt{ 3 } + 1 }{ \sqrt{ 3 } – 1 }$

It is known that the equation for the line passing through any point, say, ( x0, y0 ), with having slope m, is –

(y – y0) = m (x – x0)

Therefore, the equation for the line passing through ( 3, 33$\sqrt{ 3 }$ ) inclined at an angle of 75$75^{\circ}$ on the x- axis is given by:

( y – 33$\sqrt{ 3 }$ ) = 3+131$\frac{ \sqrt{ 3 } + 1 }{ \sqrt{ 3 } – 1 }$ ( x – 3 )

( y – 33$\sqrt{ 3 }$ ) ( 3$\sqrt{ 3 }$ – 1 ) = (3$\sqrt{ 3 }$ + 1 )( x – 3 )

y (3$\sqrt{ 3 }$ – 1 ) – 33$\sqrt{ 3 }$( 3$\sqrt{ 3 }$ – 1 ) = x ( 3$\sqrt{ 3 }$ + 1 ) – 3( 3$\sqrt{ 3 }$ + 1 )

( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 33$\sqrt{ 3 }$ + 3 – 9 + 33$\sqrt{ 3 }$

( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 63$\sqrt{ 3 }$ – 6

i.e., ( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 6( 3$\sqrt{ 3 }$ – 1 )

Hence, the equation of the line which passes through ( 3, 33$\sqrt{ 3 }$ ) is given by-

( 3$\sqrt{ 3 }$ + 1 )x – ( 3$\sqrt{ 3 }$ – 1 )y = 6( 3$\sqrt{ 3 }$ – 1 )

Q-5. Write the equation for the line which intersects x – axis at a distance 4 units away from the left side of the origin having slope -3.

Solution.

We know that,

If a line having slope m intersects x- axis at a distance d, then the equation of such lines is given by:

y = m(x – d)

Thus, The equation of the line intersecting x- axis at a distance 4 units from the left side of the origin, i.e., d = -4, and having slope m = -3, is given by:

y = -3[x – (- 4)]

y = -3x – 12

Therefore, the equation of the line required is given by:

3x + y + 12 = 0

Q-6. Write the equation for the line which intersects the y – axis at a distance 3 units above the origin which makes an angle 60$60^{\circ}$ along the positive direction of the corresponding axis, i.e., x- axis.

Solution.

We know that,

If a line having slope m makes a y- intercept c, then the desired equation for the line is given by:

y = mx + c

Since, c = 3 units and makes an angle of 60$60^{\circ}$ along the positive direction of the x- axis, i.e.,

m = tan60$60^{\circ}$ = 3$\sqrt{ 3 }$

Therefore, the equation of the line required is:

y = 3$\sqrt{ 3 }$x + 3

i.e., 3$\sqrt{ 3 }$x – y + 3 = 0

Q-7. Write the equation of the line passing through the two points (-2, 2) and (3, -5).

Solution.

We know that,

m=y2y1x2x1$m = \frac{ y_{ 2 } – y_{ 1 } }{ x_{ 2 } – x_{ 1 } }$

Also, the equation for the line passing through the two given points (x1, y1) and (x2, y2) is given by:

y – y1 = y2y1x2x1$\frac{ y_{ 2 } – y_{ 1 } }{ x_{ 2 } – x_{ 1 } }$ ( x – x1 )

Hence, the required equation for the line which passes through (-2, 2) and (3, -5) is

y – 2  = 523(2)$\frac{ -5 – 2 }{ 3 – \left( -2 \right ) }$ ( x + 2 )

y – 2 = 75$\frac{ -7 }{ 5 }$( x + 2 )

5(y – 2) = -7 (x + 2)

5y – 10 = -7x – 14

7x + 5y + 4 = 0

Therefore, the required equation is 7x + 5y + 4 = 0

Q-8. Write the equation for the line which is at a perpendicular distance of 7 units from the point (0, 0) and the perpendicular makes an angle of 45$45^{\circ}$ along the positive x- axis.

Solution.

Let, k be the length of the normal from the origin to a line and Θ$\Theta$ be the angle of the normal to a line in the positive direction of the x- axis, then

The equation of the line is:

acosΘ+bsinΘ=k$a cos\Theta + b sin\Theta = k$

Since, k = 5 units and Θ=45$\Theta = 45^{\circ}$

Therefore,

The equation of the line satisfying the conditions is given by:

acos45+bsin45=7$a cos 45^{\circ} + b sin 45^{\circ} = 7$ a12+b12=7$a \frac{ 1 }{ \sqrt{ 2 } } + b \frac{ 1 }{ \sqrt{ 2 } } = 7$

i.e., a+b=72$a + b = 7\sqrt{ 2 }$

i.e., a+b72=0$a + b – 7\sqrt{ 2 } = 0$

Q-9. Consider a triangle ABC whose vertices are A (3, 2), B (-3, 4) and C (5, 6). Write the equation for the median through one of the vertex of the triangle say, C.

Solution.

As per the data given in the question,

The vertices of the Δ$\Delta$ABC are A( 3, 1 ), B( -3, 4 ) and C( 5, 6 ).

Let us assume that, CM be the median of the triangle through vertex C.

It means, M is the mid- point of the side AB.

Now, by mid- point theorem, the required coordinates of the point M is given by:

(332,2+42)$\left ( \frac{ 3 – 3 }{ 2 }, \frac{ 2 + 4 }{ 2 } \right )$ = ( 0, 3 )

We know that, the equation for the line passing through the two given points (a1, b1) and (a2, b2) is given by:

b – b1 = b2b1a2a1$\frac{ b_{ 2 } – b_{ 1 } }{ a_{ 2 } – a_{ 1 } }$ ( a – a1 )

Here, (a1, b1) = (5, 6) and (a2, b2) = (0, 3)

So, by substitution these values in the equation for the line CM passing through two given points is:

(b – 6) = 3605$\frac{ 3 – 6 }{ 0 – 5 }$( a – 5 )

(b – 6) = 35$\frac{ 3 }{ 5 }$( a – 5 )

5 (b – 6) = 3 (a – 5)

5b – 30 = 3a – 15

3a – 5b + 15 = 0

Therefore, the equation of the median through vertex R is 3a – 5b + 15 = 0

Q-10. Get the equation for the line which is passing through (-4, 5) and which is perpendicular to the line passing through two points (3, 6) and (-4, 7).

Solution.

We know that,

m=y2y1x2x1$m = \frac{ y_{ 2 } – y_{ 1 } }{ x_{ 2 } – x_{ 1 } }$

So, the slope of the line joining the two given points, namely A (3, 6) and B (-4, 7), i.e.,

m = 7643=17$\frac{ 7 – 6 }{ -4 – 3 } = \frac{ 1 }{ -7 }$

Now,

It is known that the two non- vertical lines are perpendicular to each other if and only if each of their slopes is a negative reciprocal of each other.

Hence, the slope of the line which is perpendicular to the line through the given points (3, 6) and (-4, 7) is:

1m=117=7$-\frac{ 1 }{ m } = -\frac{ 1 }{ \frac{ -1 }{ 7 } } = 7$

Hence, the equation for the line passing through the point (-4, 6), having slope 7 is given by:

(y – 6) = 7 (x + 4)

y – 5 = 7x + 28

i.e., 7x – y + 33 = 0

Therefore, the required equation for the line is 7x – y + 33 = 0

Q-11. Consider a line which is perpendicular to the line segment joining the two points (2, 0) and (3, 4) dividing it in 1 : p ratio. Find the equation for such a line.

Solution.

By having the section formula, we know that

The coordinates of the point dividing the line segment joining the two points (2, 0) and (3, 4) in the ratio 1 : p which is given by:

(p(2)+2(3)1+p,p(0)+2(4)1+p)=(2p+6p+1,8p+1)$\left ( \frac{ p \left ( 2 \right ) + 2 \left ( 3 \right )}{ 1 + p }\; , \; \frac{ p \left ( 0 \right ) + 2 \left ( 4 \right )}{ 1 + p } \right ) = \left ( \frac{ 2p + 6 }{ p + 1 } , \frac{ 8 }{ p + 1 }\right )$

Now, the slope for the line joining the two given points (2, 0) and (3, 4) is given by:

m = 4032=41$\frac{ 4 – 0 }{ 3 – 2 } = \frac{ 4 }{ 1 }$ = 4

It is known that the two non- vertical lines are perpendicular to each other if and only if each of their slopes is a negative reciprocal of each other.

Hence, The slope of the line which is perpendicular to the line through the given points (2, 0) and (3, 4 ) is:

1m=14$-\frac{ 1 }{ m } = -\frac{ 1 }{ 4 }$

Hence, the equation for the line passing through the point (2p+6p+1,8p+1)$\left ( \frac{ 2p + 6 }{ p + 1 } , \frac{ 8 }{ p + 1 }\right )$, having slope 14$-\frac{ 1 }{ 4 }$ is given by:

$\Rightarrow$   (y8p+1)=14(x2p+6p+1)$\left (y\; -\; \frac{ 8 }{ p + 1 } \right ) = -\frac{ 1 }{ 4 } \left ( x \;-\; \frac{ 2p + 6 }{ p + 1 } \right )\\$

$\\\Rightarrow$   4(y8p+1)=(x2p+6p+1)$4\left (y\; – \;\frac{ 8 }{ p + 1 } \right ) = -\left ( x\; – \;\frac{ 2p + 6 }{ p + 1 } \right )$

$\Rightarrow$   4(p+1)y32=x(p+1)+(2p+6)$4\left (p + 1 \right )y – 32 = -x\left ( p + 1 \right ) + \left ( 2p + 6 \right )$

$\Rightarrow$   x(p+1)4(p+1)y=(2p+632)$x\left (p + 1 \right) – 4\left (p + 1 \right)y = -\left( 2p + 6 – 32 \right )$

$\Rightarrow$   x(p+1)4(p+1)y=262p$x\left (p + 1 \right) – 4\left (p + 1 \right)y = 26 – 2p$

Therefore, the equation is x(p+1)4(p+1)y=262p$\Rightarrow x\left (p + 1 \right) – 4\left (p + 1 \right)y = 26 – 2p$.

Q-12. What will be the equation for the line which cuts off intercepts equally on the co-ordinate axes and which passes through the point (3, 4).

Solution.

It is known that, the equation of any line in the intercept form is:

xm+yn=1$\frac{ x }{ m } + \frac{ y }{ n } = 1$                    …………(i)

Where, m and n are intercepts on the x- axis and y- axis, respectively.

From the data given in the question it is known that, the line cuts off an equal intercepts on the co-ordinate axes, which clearly means that m = n.

So, from equation (i), we have

xm+ym=1$\frac{ x }{ m } + \frac{ y }{ m } = 1$

x+y=m$\Rightarrow x + y = m$                                     …………..(ii)

Thus, the line passes through the point (3, 4), equation ( ii ) reduces to

3 + 4 = m

So, m = 7

On substituting the value of m in equation (ii), we will get

x + y = 7

Hence, the equation of the line required is x + y = 7.

Q-13. A line is passing through (3, 3) and making intercepts on the coordinate axes. The sum of the two intercepts is 12. Find the equation of the line.

Solution.

It is known that, the equation of any line in the intercept form is:

xm+yn=1$\frac{ x }{ m } + \frac{ y }{ n } = 1$                    …………(i)

Where, m and n are intercepts on the x- axis and y- axis, respectively.

From the data given in the question it is known that,

m + n = 12

So, m = 12 – n

Substituting the value of m in equation (i), we will get

x12n+yn=1$\frac{ x }{ 12 – n } + \frac{ y }{ n } = 1$                    …………(ii)

Since, the line is passing through the point (3, 3). Hence, from equation (ii), we will get

312n+3n=1$\frac{ 3 }{ 12 – n } + \frac{ 3 }{ n } = 1\\$

$\Rightarrow$   3(n)+3(12n)n(12n)=1$\frac{ 3( n ) + 3 ( 12 – n )}{ n ( 12 – n )} = 1$

$\\\Rightarrow$   3n+363n12nn2=1$\frac{ 3n + 36 – 3n }{ 12n – n^{ 2 }} = 1$

$\Rightarrow$   3612nn2=1$\frac{ 36 }{ 12n – n^{ 2 }} = 1$

$\Rightarrow$   36=12nn2$36 = 12n – n^{ 2 }$

n2$\Rightarrow n^{2}$   12n+36=0$– 12n + 36 = 0$

$\Rightarrow$   n26n6n+36=0$n^{ 2 } – 6n – 6n + 36 = 0$

$\Rightarrow$   n(n6)6(n6)$n\left( n – 6 \right ) – 6\left( n – 6 \right )$

(n – 6) (n – 6) = 0

So, n = 6 or n = 6

Now,

As n = 6, so m = 6

Hence, the required equation of the line is given by:

x6+y6=1$\frac{ x }{ 6 } + \frac{ y }{ 6 } = 1\\$

Therefore, Equation is x + y = 6.

Q-14. Write the equation for the line which is passing through (0, 3) and makes an angle

π6$\frac{ \pi }{ 6 }$ along the positive direction of x- axis. Also, find the equation for the line which is parallel to it and which crosses the y-axis at the distance of 3 units to the negative direction of y- axis from (0, 0).

Solution.

As the line is making an angle π6$\frac{ \pi }{ 6 }$ in the positive direction of x- axis,

So the slope of the line is m = tanπ6$\frac{ \pi }{ 6 }$ = tan30$30^{\circ}$ = 13$\frac{ 1 }{ \sqrt{ 3 }}$

Thus, the equation of the line which is passing through (0, 3) and having slope 13$\frac{ 1 }{ \sqrt{ 3 }}$ is

(y – 3) = 13$\frac{ 1 }{ \sqrt{ 3 }}$ ( x –  0 )

3$\sqrt{ 3 }$( y – 3 ) = x

i.e., x – 3$\sqrt{ 3 }$y + 33$\sqrt{ 3 }$ = 0

Hence, the slope of the line parallel to x – 3$\sqrt{ 3 }$y + 33$\sqrt{ 3 }$ = 0 is 13$\frac{ 1 }{ \sqrt{ 3 }}$ to the line.

In the question, it is given that the line parallel to x – 3$\sqrt{ 3 }$y + 33$\sqrt{ 3 }$ = 0 crosses y- axis at a distance 3 units to the negative direction of y- axis from the origin, i.e., it passes through point ( 0, -3 ).

Therefore, the equation of the line which is passing through ( 0, -3 ) and having slope 13$\frac{ 1 }{ \sqrt{ 3 }}$ is:

y – ( -3 ) = 13$\frac{ 1 }{ \sqrt{ 3 }}$ ( x – 0 )

y + 3 = 13$\frac{ 1 }{ \sqrt{ 3 }}$ ( x – 0 )

3$\sqrt{ 3 }$( y + 3 ) = x

i.e., x – 3$\sqrt{ 3 }$y – 33$\sqrt{ 3 }$ = 0

Hence, the required equation of the line = x – 3$\sqrt{ 3 }$y – 33$\sqrt{ 3 }$ = 0.

Q-15. What will be the equation of the line perpendicular to the line from the origin meets at       (-3, 10)?

Solution.

The slope of the line which joins (0, 0) and the point (-3, 10) is

m  = 10030=103$\frac{ 10 – 0 }{ -3 – 0 } = -\frac{ 10 }{ 3 }$

So, accordingly, the slope of the line which is perpendicular to the line joining (0, 0) and the point (-3, 10) is-

m1 = –1m=1103=310$\frac{ 1 }{ m } = -\frac{ 1 }{ \frac{ -10 }{ 3 }} = { 3 } { 10 }$

Thus, the equation of the line which passes from (-3, 10) and having slope m1 is given by

(y – 10) = 310${ 3 } { 10 }$( x + 3 )

10( y – 10 ) = 3 ( x + 3 )

10y – 100 = 3x + 9

i.e., 30x – 10 y + 91 = 0

Hence, the required equation of the line is 30x – 10 y + 91 = 0

Q-16. Consider a copper rod having length L (in centimeters) which is a linear function of the Celsius temperature which is C. Express L in terms of C, for an experiment. Assume that, L = 127.952 when C = 24, and L = 128.987 when C = 114.

Solution.

As per the data given in the question, we have

If C = 24, then the value of L = 127.952, whereas if C = 114, then the value of L = 128.987.

So, the points are (24, 127.952) and (114, 128.987) which satisfies the linear relation between L and C.

Consider L along the y- axis and C along the x- axis and also, we have two points i.e., (24, 127.952) and (114, 128.987) in XY plane.

Hence, the linear relation of the line between C and L which is passing through (24, 127.952) and (114, 128.987) is given by:

(L – 127.952) = 128.987127.95211424$\frac{ 128.987 – 127.952 }{ 114 – 24 }$( C – 24 )

L – 127.952 = 1.03590$\frac{ 1.035 }{ 90 }$(C – 24)

i.e., L = 1.03590$\frac{ 1.035 }{ 90 }$( C – 24 ) + 127.952

Hence, the linear relation between C and L required = L = 1.03590$\frac{ 1.035 }{ 90 }$( C – 24 ) + 127.952

Q-17. A milk store owner observed that, he can sell at least 1000 liters of milk every week at a price of Rs. 15/liter and also, 1240 liters of milk every week at a price of Rs. 17/liter. Consider a linear relation between the demand and estimated selling price, find how many liters would he will sell every week at a cost of Rs.18/liters?

Solution.

It is given in the question that there is a linear relation between the selling price and the demand.

Consider demand along the y- axis and the estimated selling price per liter along the x- axis. (15, 1000) and (17, 1240) are the two given points along the XY plane which satisfies the linear relation between the selling price and the demand.

Hence, the linear relation of the line between S (Selling price) and D (demand) which is passing through (15, 1000) and (17, 1240) is given by:

y – 1000 = 124010001715=2402=120$\frac{ 1240 – 1000 }{ 17 – 15 } = \frac{ 240 }{ 2 } = 120$( x – 15 )

y – 1000 = 2402$\frac{ 240 }{ 2 }$( x – 15 )

y – 1000 = 120 (x – 15)

i.e., y = 120 (x – 15) + 1000

If, x = Rs. 18/liters

y = 120(18 – 15) + 1000

y = 120 (3) + 1000

y = 1360

Hence, the milk store owner will sell 1360 liters of milk weekly at the rate of Rs. 18/liter.

Q-18. S (m, n) be the mid-point of the line- segment between the axes. Prove that the equation of such line is-

xm+yn=2$\frac{ x }{ m } + \frac{ y }{ n } = 2$

Solution.

Let, PR be the line segment between the axes and let, S (m, n) be its mid-point.

Thus, the co-ordinates of P and R be (0, y) and (x, 0), respectively.

As, point S (m, n) is the mid-point of the line PR.

So, (0+x2,y+02)=(m,n)$\left( \frac{ 0 + x }{ 2 },\; \frac{ y + 0 }{ 2 } \right ) = \left( m, n \right )$

(x2,y2)=(m,n)$\Rightarrow \left( \frac{ x }{ 2 },\; \frac{ y }{ 2 } \right ) = \left( m, n \right )$ x2=mandy2=n$\Rightarrow \frac{ x }{ 2 } = m \; and \; \frac{ y }{ 2 } = n$

x = 2m and y = 2n

Hence, the coordinates of P and R are (0, 2n) and (2m, 0).

Therefore, the equation of the line which is passing through the points (0, 2n) and (2m, 0) is given by:

(y – 2n) = 02n2m0$\frac{ 0 – 2n }{ 2m – 0 }$ ( x – 0 )

(y – 2n) = 2n2m$\frac{ -2n }{ 2m }$( x – 0 )

(y – 2n) = nm$\frac{ -n }{ m }$( x )

m (y – 2n) = -n( x )

my – 2mn = -nx

i.e., nx + my – 2mn = 0

By dividing both side with mn, we will get

nxmn+mymn2mnmn=0$\frac{ nx }{ mn } + \frac{ my }{ mn } – \frac{ 2mn }{ mn } = 0$ xm+yn2=0$\frac{ x }{ m } + \frac{ y }{ n } – 2 = 0$ xm+yn=2$\frac{ x }{ m } + \frac{ y }{ n } = 2$

Hence, the required equation of the line is xm+yn=2$\frac{ x }{ m } + \frac{ y }{ n } = 2$.

Q-19. The point R (m, n) divides the line segment PQ in the ratio of 2: 3 between both the axes. What will be the equation of the line?

Solution.

As per the data given in the question, we know that PQ is the line segment which is divided in the ratio of 2 : 3 by the point R (m, n).

Let, the coordinates of P and Q is (x, 0) and (0, y), respectively.

As we know, PQ is the line segment which is divided in the ratio of 2 : 3 by the point R( m, n ).

By using the section formula,

( m, n ) = 2×0+3×x2+3,2×y+3×02+3$\frac{ 2 \times 0 + 3 \times x }{ 2 + 3 },\; \frac{ 2 \times y + 3 \times 0 }{ 2 + 3 }$

(m,n)=(3x5,2y5)$( m,\;n ) = \left(\frac{3x}{5}, \; \frac{2y}{5}\right)$

$\\\Rightarrow$ m=3x5andn=2y5$m = \frac{ 3x }{ 5 } \; and \; n = \frac{ 2y }{ 5 }$

$\Rightarrow$    x=5m3andy=5n2$x = \frac{ 5m }{ 3 } \; and \; y = \frac{ 5n }{ 2 }$

Thus, the coordinates of P and Q are ( 5m3$\frac{ 5m }{ 3 }$, 0 ) and ( 0, 5n2$\frac{ 5n }{ 2 }$ ).

Then,

The equation for the line PQ which is passing through points (5m3,0)and(0,5n2)$\left ( \frac{ 5m }{ 3 }, 0 \right ) \;and\; \left ( 0,\frac{ 5n }{ 2 } \right )$ is given by:

(y – 0) = (5n2005m3)(x5m3)$\left( \frac{ \frac{ 5n }{ 2 } – 0 }{ 0 – \frac{ 5m }{ 3 } } \right ) ( x – \frac{ 5m }{ 3 })$

y = (5n×35n×2)(x5m3)$\left( \frac{ 5n \times 3 }{ -5n \times 2 } \right ) \left( x – \frac{ 5m }{ 3 } \right )$

y = 3n2m(x5m3)$\frac{ -3n }{ 2m } \left( x – \frac{ 5m }{ 3 } \right )$

$\Rightarrow$    2m×y=3n×3x5m3$2m \times y = -3n \times \frac{ 3x – 5m }{ 3 }$

$\Rightarrow$    2m×y×3=3n×3x+3n×5m$2m \times y \times 3 = -3n \times 3x + 3n \times 5m$

6my = -9nx + 15mn

9nx + 6my – 15mn = 0

3(3nx + 2my – 5 mn) = 0

i.e., 3nx + 2my – 5mn = 0

Hence, the equation of the line PQ required is given by 3nx + 2my – 5mn = 0.

Q-20. Show that the three points (4, 1), (-3, -3) and (11, 5) are collinear by using the concept of the equation of a line.

Solution.

If we will prove that the line which passes through the points (4, 1) and (-3, -3) passes through the point (11, 5) also, then we can prove that the three points (4, 1), (-3, -3) and (11, 5) are collinear.

Now, the required equation of the line passing through (4, 1) and (-3, -3) will be given by:

(y – 1) = 3134$\frac{ -3 – 1 }{ -3 – 4 }$( x – 4 )

y – 1 = 47$\frac{ -4 }{ -7 }$( x – 4 )

7y – 7 = 4 (x – 4)

7y = 4x – 16 + 7

4x – 7y = 9

Since, the line passes through the point (11, 5), so, x = 11 and y = 5

Now,

LHS = 4 × 11 – 7 × 5 = 44 – 35 = 9 = RHS

Hence, the line which is passing through the points (4, 1) and (-3, -3) also passes through the other point (11, 5). Hence, (4, 1), (-3, -3) and (11, 5) are collinear.

Exercise 10.3

Q1. Reduce the following equations into slope-intercept form and find their slopes and the y intercepts.

(i) x + 6y = 0

(ii) 6x + 3y – 6 = 0

(iii) y = 1

Sol:

(i) x + 6y = 0

It can be written as:

y=16a+0$y = -\frac{1}{6}a + 0$ ————–(1)

The given equation is in the form of y = mx + c,

Where m=17$m = -\frac{1}{7}$ and c = 0

Hence, equation (1) is in the slope – intercept form, where the slope and the y – intercept are 16$-\frac{1}{6}$ and 0 respectively.

(ii) 6x + 3y – 6 = 0

It can be written as:

y=13(6x+6)$y = \frac{1}{3}\left ( -6x + 6 \right )$

y=2x+63$y = -2x + \frac{6}{3}$ —————— (2)

The given equation is in the form of y = mx + c,

Where m = – 2 and c = 2

Hence, equation (2) is in the slope – intercept form, where the slope and the y – intercept are -2 and 3 respectively.

(iii) y = 1

It can be written as:

y = 0.x + 1 —————— (3)

The given equation is in the form of y = mx + c,

Where m = 0 and c = 1

Hence, equation (3) is in the slope – intercept form, where the slope and the y – intercept are 0 and 1 respectively.

Q2. Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y – 14 = 0

(ii) 4x – 3y = 6

(iii) 3y + 2 = 0

Sol:

(i) 3x + 2y – 14 = 0

It can be written as:

3x + 2y = 14

3x14+2y14=1$\frac{3x}{14} + \frac{2y}{14} = 1$

3x14+y7=1$\frac{3x}{14} + \frac{y}{7} = 1$ ————— (1)

The given equation is in the form of xa+yb=1$\frac{x}{a} + \frac{y}{b} = 1$

Where a = 14 / 3 and b = 7.

Hence, equation (1) is in intercept form, where the intercepts on the x and y axes are 14 / 3 and 7 respectively.

(ii) 4x – 3y = 12

It can be written as:

4x123y12=1$\frac{4x}{12} – \frac{3y}{12} = 1$ x3y4=1