ML Aggarwal Solution Class 10 Chapter 12 Equation of a Straight Line Exercise 12.1

 Exercise 12.1

Question 1

Find the slope of a line whose inclination is

(i) 45°

(ii) 30°

Sol :

(i) tan 45° = 1

(ii) tan 30°$=\frac{1}{\sqrt{3}}$

Question 2

Find the inclination of a line whose gradient is

(i) 1

(ii) √3

(iii) $\frac{1}{\sqrt{3}}$

Sol :

(i) tan θ = 1 ⇒ θ = 45°

(ii) tan θ = √3 ⇒ θ = 60°

(iii) $\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}$

Question 3

Find the equation of a straight line parallel 1 to x-axis which is at a distance

(i) 2 units above it

(ii) 3 units below it.

Sol :

(i) A line which is parallel to x-axis is y = a

⇒ y = 2

⇒ y – 2 = 0

(ii) A line which is parallel to x-axis is y = a

⇒ y = -3

⇒ y + 3 = 0

Question 4

Find the equation of a straight line parallel to y-axis which is at a distance of:

(i) 3 units to the right

(ii) 2 units to the left.

Sol :

(i) The equation of line parallel to y-axis is at a distance of 3 units to the right is x = 3 ⇒ x – 3 = 0

(ii) The equation of line parallel to y-axis at a distance of 2 units to the left is x = -2 ⇒ x + 2 = 0

Question 5

Find the equation of a straight line parallel to y-axis and passing through the point ( – 3, 5).

Sol :

The equation of the line parallel to y-axis passing through ( – 3, 5) to x = -3

⇒ x + 3 = 0

Question 6

Find the equation of the a line whose

(i) slope = 3, y-intercept = – 5

(ii) slope $=-\frac{2}{7}, y$ -intercept =3

(iii) gradient $=\sqrt{3}, y$ -intercept $=-\frac{4}{3}$

(iv) inclination = 30°,y-intercept = 2

Sol :

Equation of a line whose slope and y-intercept is given is

y = mx + c

where m is the slope and c is the y-intercept

(i) y = mx + c

⇒ y = 3x + (-5)

⇒ y = 3x – 5

(ii) $y=m x+c \Rightarrow y=\frac{-2}{7} x+3$

$\Rightarrow 7 y=-2 x+21$

$\Rightarrow 2 x+7 y-21=0$

(iii) y=mx+c

$\Rightarrow y=\sqrt{3} x+\left(-\frac{4}{3}\right)$

$\Rightarrow y=\sqrt{3} x-\frac{4}{3}$

$\Rightarrow 3 y=3 \sqrt{3} x-4$

$\Rightarrow 3 \sqrt{3} x-3 y-4=0$

(iv) Inclination $=30^{\circ}$

$\therefore$ slope $=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

$\therefore$ Equation $y=m x+c \Rightarrow y=\frac{1}{\sqrt{3}} x+2$

$\Rightarrow \sqrt{3} y=x+2 \sqrt{3}$

$\Rightarrow  x-\sqrt{3} y+2 \sqrt{3}=0$

Question 7

Find the slope and y-intercept of the following lines:

(i) x – 2y – 1 = 0

(ii) 4x – 5y – 9 = – 0

(iii) 3x +5y + 7 = 0

(iv) $\frac{x}{3}+\frac{y}{4}=1$

(v) y – 3 = 0

(vi) x – 3 = 0

Sol :

We know that in the equation

y=mx+c, m is the slope and c is the y-intercept.

Now using this, we find,

(i) x-2 y-1=0 

$\Rightarrow x-1=2 y$

$\Rightarrow 2 y=x-1$

$\Rightarrow y=\frac{1}{2} x-\frac{1}{2}$

Here slope $=\frac{1}{2}$ and y-intercept $=-\frac{1}{2}$

(ii) $4 x-5 y-9=0 \Rightarrow 4 x-9=5 y$

$\Rightarrow 5 y=4 x-9$

$\Rightarrow y=\frac{4}{5} x-\frac{9}{5}$

Here slope $=\frac{4}{5}$ and intercept $=\frac{-9}{5}$

(iii) 3x+5y+7=0 

$\Rightarrow 5 y=-3 x-7$

$\Rightarrow y=\frac{-3}{5} x-\frac{7}{5}$

Here slope $=\frac{-3}{5}$ and y-intercept $=\frac{-7}{5}$

(iv) $\frac{x}{3}+\frac{y}{4}=1$

$\Rightarrow 4 x+3 y=12$

$\Rightarrow 3 y=-4 x+12$

$\Rightarrow y=\frac{-4}{3} x+\frac{12}{3}$

$\Rightarrow y=\frac{-4}{3} x+4$

Here, $=\frac{-4}{3}$ and y-intercept =4

(v) y-3=0 

$\Rightarrow y=3 \Rightarrow y=0, x+3$

Here slope =0 and y-intercept =3

(vi) x-3=0

Here in this equation, slope cannot be defined and does not meet y-axis.

Question 8

The equation of the line PQ is 3y – 3x + 7 = 0

(i) Write down the slope of the line PQ.

(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.

Sol :

Equation of line PQ is 3y–3x+7=0

Writing in form of y=mx+c

3y=3x-7 

$\Rightarrow y=\frac{3 x}{3}-\frac{7}{3}$

$\Rightarrow  y=x-\frac{7}{3}$

(i) Here slope=1

(ii) $\therefore$ Angle which makes PQ with x-axis is Q

But $\tan . \theta=1 \therefore \theta=45^{\circ}$

Question 9

The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.

Sol :

Slope of the line y = x + 1 after comparing

it with y = mx + c, m = 1

$\Rightarrow  \tan \theta=1$

$\Rightarrow \theta=45^{\circ}$

and slope of line $y=\sqrt{3} x-1$

$m=\sqrt{3} \Rightarrow \tan \theta=\sqrt{3}$

$\Rightarrow \theta=60^{\circ}$

Now in Δ formed by the given two lines and x-axis.

Ext. angle=Sum of interior opposite angle.

$\Rightarrow 60^{\circ}=\theta+45^{\circ}$

$\Rightarrow \theta=60^{\circ}-45^{\circ}=15^{\circ}$

Question 10

Find the value of p, given that the line $\frac{y}{2}=x-p$ passes through the point ( – 4, 4) (1992).

Sol :

Equation of line is $\frac{y}{2}=x-p$

It passes through the points (-4, 4)

It will satisfy the equation

$\Rightarrow \frac{4}{2}=-4-p$

$\Rightarrow 2=-4-p$

$\Rightarrow \quad p=-4-2$

$\Rightarrow p=-6$

Hence, p=-6

Question 11

Given that (a, 2a) lies on the line $\frac{y}{2}=3 x-6$ find the value of a

Sol :

∵ Point (a, 2a) lies on the line

$\frac{y}{2}=3 x-6$

$\frac{y}{2}=3 x-6$

$\therefore \frac{2 a}{2}=3(a)-6$

$\Rightarrow a=3 a+6$

-3a+a=-6

$\Rightarrow-2 a=-6$

$\Rightarrow a=\frac{-6}{-2}$

∴a=3

Question 12

The graph of the equation y = mx + c passes through the points (1, 4) and ( – 2, – 5). Determine the values of m and c.

Sol :

Equation of the line is y = mx + c

∴ it passes through the points (1, 4)

∴ 4 = m x 1 + c

⇒ 4 = m + c

⇒ m + c = 4 … (i)

Again it passes through the point (-2, -5)

∴ 5 = m (-2) + c

⇒ 5 = -2 m + c

⇒ 2m – c = 5 …(ii)

Adding (i) and (ii)

3m = 9

⇒ m = 3

Substituting the value of m in (i)

3 + c = 4

⇒ c = 4 – 3 = 1

Hence m = 3, c = 1

Question 13

Find the equation of the line passing through the point (2, – 5) and making an intercept of – 3 on the y-axis.

Sol :

∴ The line intersects y-axis making an intercept of -3

∴ the co-ordinates of point of intersection will be (0, -3)

Now the slope of line $(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{-3+5}{0-2}=\frac{2}{-2}=-1$

∴ Equation of the line will be,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-(-5)=-1(x-2)$

$\Rightarrow y+5=-x+2$

$\Rightarrow x+y+5-2=0$

$\Rightarrow x+y+3=0$

Question 14

Find the equation of a straight line passing through ( – 1, 2) and whose slope is $\frac{2}{5}$

Sol :

Equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right)$

$y-2=\frac{2}{5}(x+1)$

⇒ 5y – 10 = 2x + 2

⇒ 2x – 5y + 2 + 10 = 0

⇒ 2x – 5y + 12 = 0

Question 15

Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).

Sol :

The equation of line whose slope is wand passes through a given point is

$y-y_{1}=m\left(x-x_{1}\right)$

Here m = tan 60° = √3 and point is (0, -3)

∴ y + 3 = √3 (x – 0)

⇒ y + 3 = √3x

⇒ √3x – y – 3 = 0

Question 16

Find the gradient of a line passing through the following pairs of points.

(i) (0, – 2), (3, 4)

(ii) (3, – 7), ( – 1, 8)

Sol :

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Given

(i) (0, -2), (3, 4)

(ii) (3, -7), (-1, 8)

(i) $m=\frac{4+2}{3-0}=\frac{6}{3}=2$ ∴gradient=2

(ii) $m=\frac{8+7}{-1-3}=\frac{15}{-4}$ ∴gradient$=-\frac{15}{4}$

Question 17

The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find :

(i) The gradient of EF

(ii) The equation of EF

(iii) The coordinates of the point where the line EF intersects the x-axis.

Sol :

Co-ordinates of points E (0, 4) and F (3, 7) are given, then

(i) The gradient of EF

$(i) \therefore$ gradient $(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-4}{3-0}=\frac{3}{3}=1$

(ii) Equation of line EF,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-7=1(x-3)$

$\Rightarrow y-7=x-3$

$\Rightarrow x-y-3+7=0$

$\Rightarrow x-y+4=0$

(iii) Co-ordinates of point of intersection of EF and the x-axis will be y=0

Substitutes the value y in the above equation

x-y+4=0 

$\Rightarrow x-0+4=0(\because y=0)$

$\Rightarrow x=-4$

Hence co-ordinates are (-4,0)

Question 18

Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.

Sol :

Putting y = 0, we will get the intercept made on x-axis,

2x – 3y + 12 = 0

⇒ 2x – 3 × 0 + 12 = 0

⇒ 2x – 0 + 2 = 0

⇒ 2x = -12

⇒ x = -6

and putting x = 0, we get the intercepts made on y-axis,

2x – 3y + 12 = 0

⇒ 2 × 0 – 3y + 12 = 0

⇒ -3y = -12

$\Rightarrow y=\frac{-12}{-3}=4$

Question 19

Find the equation of the line passing through the points P (5, 1) and Q (1, – 1). Hence, show that the points P, Q and R (11, 4) are collinear.

Sol :

The two given points are P (5, 1), Q(1, -1).

∴ Slope of the line (m)

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-1}{1-5}=\frac{-2}{-4}=\frac{1}{2}$

Equation of the line,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y+1=\frac{1}{2}(x-1)$

$\Rightarrow 2 y+2=x-1$

$\Rightarrow x-2 y-1-2=0$...(i)

If point R(11,4) be on it, then it will satisfy it.

Now substituting the value of x and y in

(i) $11-2 \times 4-3=11-8-3=11-11=0$

$\therefore$ R satisfies it

Hence P, Q and R are collinear.

Question 20

Find the value of ‘a’ for which the following points A (a, 3), B (2,1) and C (5, a) are collinear. Hence find the equation of the line.

Sol :

Given That

A(a, 3), B (2, 1) and C (5, a) are collinear.

Slope of AB = Slope of BC

$=\frac{1-3}{2-a}=\frac{a-1}{5-2}$

$=\frac{-2}{2-a}=\frac{a-1}{3}$

-6=(a-1)(2-a) (Cross-multiplication)

$-6=2 a-a^{2}-2+a$

$-6=3 a-a^{2}-2$

$a^{2}-3 a+2-6=0$

$a^{2}-3 a-4=0$

$a^{2}-4 a+a-4=0$

a(a-4)+1(a-4)=0

(a+1)(a-4)=0

a=-1, or a=4

a=-1 (∵does not satisfy the equation)

∴a=4

Slope of $\mathrm{BC}=\frac{a-1}{5-2}=\frac{4-1}{3}=\frac{3}{3}=1 \mathrm{~m}$

Equation of BC ;(y-1)=1(x-2)

$y-1=x-2 \Rightarrow x-y=-1+2$

x-y=1

Question 21

Use a graph paper for this question. The graph of a linear equation in x and y, passes through A ( – 1, – 1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and ($\frac{1}{2}$, k). (2005)

Sol :

Points (h, 4) and ($\frac{1}{2}$, k) lie on the line passing

through A(-1, -1) and B(2, 5)

Figure to be added

From the graph, we see that $h=\left(\frac{3}{2}\right)$

and k=2

Question 22

ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, – 4). Find

(i) the coordinates of A

(ii) the equation of the diagonal BD.

Sol :

Given that

ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4)

Coordinates of $\mathrm{O}=\left(\frac{5+2}{2}, \frac{8-4}{2}\right)=(3.5,2)$

For the line AC

$3.5=\frac{x+4}{2}$

$\Rightarrow x+4=7$

$\Rightarrow x=7-4=3$

x=3, y=-3

$2=\frac{y+7}{2}$

$\Rightarrow y+7=4$

$\Rightarrow y=4-7=-3$

Thus, the coordinates of A are (3,-3)

(ii) Equation of diagonal BD is given by

$y-8=\frac{-4-8}{2-5}(x-5)$

$\Rightarrow y-8=\frac{-12}{-3}(x-5)$

$\Rightarrow y-8=4 x-20$

$\Rightarrow 4 x-y-12=0$

Question 23

In ∆ABC, A (3, 5), B (7, 8) and C (1, – 10). Find the equation of the median through A.

Sol :

AD is median

⇒ D is mid point of BC

$\therefore \text{D is }\left(\frac{7+1}{2}, \frac{8-10}{2}\right)$

i.e (4, -1)

slope of AD

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5+1}{3-4}=\frac{6}{-1}=-6$

$\therefore$ Equation of $\mathrm{AD}$

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y+1=-6(x-4)$

$\Rightarrow y+1=-6 x+24$

$\Rightarrow y+6 x=-1+24$

$\Rightarrow 6 x+y=23$

Question 24

Find the equation of a line passing through the point ( – 2, 3) and having x-intercept 4 units. (2002)

Sol :

x-intercept = 4

∴ Co-ordinates of the point will be (4, 0)

Now slope of the line passing through the points (-2, 3) and (4, 0)

$(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-3}{4+2}=\frac{-3}{6}=-\frac{1}{2}$

$\therefore$ Equation of the line will be

$y-{y}_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-0=-\frac{1}{2}(x-4)$

$\Rightarrow 2 y=-x+4$

$\Rightarrow x+2 y=4$ or x+2y-4=0

Question 25

Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.

Sol :

x-intercept = 6

∴ The line will pass through the point (6, 0)

y -intercept = -4 ⇒ c = -4

$\therefore$ The line will pass through the point (0,-4)

Now $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-4-0}{0-6}=\frac{-4}{-6}=\frac{2}{3}$

$\therefore$ Equation of line will be

y=m x+c

$\Rightarrow y=\frac{2}{3} x+(-4)=\frac{2}{3} x-4$

$\Rightarrow 3 y=2 x-12 \Rightarrow 2 x-3 y=12$

Question 26

Write down the equation of the line whose gradient is $\frac{1}{2}$ and which passes through P where P divides the line segment joining A ( – 2, 6) and B (3, – 4) in the ratio 2 : 3. (2001)

Sol :

P divides the line segment joining the points

A (-2, 6) and (3, -4) in the ratio 2 : 3

∴ Co-ordinates of P will be

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{2 \times 3+3 \times(-2)}{2+3}$

$=\frac{6-6}{5}=\frac{0}{5}=0$

$y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{2 \times(-4)+3(6)}{2+3}$

$=\frac{-8+18}{5}=\frac{10}{5}=2$

$\therefore$ Co-ordinates are (0,2)

Now slope (m) of the line passing through $(0,2)=\frac{3}{2}$

$\therefore$ Equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-2=\frac{3}{2}(x-0)$

$\Rightarrow 2 y-4=3 x$

$\Rightarrow 3 x-2 y+4=0$

Question 27

Find the equation of the line passing through the point (1, 4) and intersecting the line x – 2y – 11 = 0 on the y-axis.

Sol :

line x – 2y – 11 = 0 passes through y-axis

x = 0,

Now substituting the value of x in the equation x – 2y – 11 = 0

$\therefore-2 y-11=0 \Rightarrow-2 y=11 \Rightarrow y^{\prime}=-\frac{11}{2}$

$\therefore$ Co-ordinates of point will be $\left(0,-\frac{11}{2}\right)$

Now slope of the line joining the points

(1,4) and $\left(0,-\frac{11}{2}\right)$

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-\frac{11}{2}-4}{0-1}=\frac{-\frac{19}{2}}{-1}=\frac{19}{2}$

and equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y+\frac{11}{2}=\frac{19}{2}(x-0)$

$\Rightarrow 2 y+11=19 x$

$\Rightarrow 19 x-2 y-11=0$

Question 28

Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.

Sol :

Let the line containing the point P (3, 2)

passes through x-axis at A (x, 0) and y-axis at B (0, y)

OA = OB given

∴ x = y

Now slope of the line $(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{0-y}{x-0}=\frac{-x}{x}=-1(\because x=y)$

$\therefore$ Equation of the line will be $y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-2=-1(x-3)$

$\Rightarrow y-2=-x+3$

$\Rightarrow  x+y-2-3=0$

$\Rightarrow  x+y-5=0$

$\Rightarrow x+y=5$

Question 29

Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:

(i) the coordinates of the fourth vertex D.

(ii) length of diagonal BD.

(iii) equation of side AB of the parallelogram ABCD. (2015)

Sol :

Three vertices of a parallelogram ABCD taken in order are

A (3, 6), B (5, 10) and C (3, 2)

(i) We need to find the co-ordinates of D

We know that the diagonals of a parallelogram bisect each other

Let (x, y) be the co-ordinates of D

$\therefore$ Mid-point of diagonal $\mathrm{AC}=\left(\frac{3+3}{2}, \frac{6+2}{2}\right)$

=(3,4)

And, mid-point of diagonal $\mathrm{BD}=\left(\frac{5+x}{2}, \frac{10+y}{2}\right)$

Thus, we have

$\frac{5+x}{2}=3$ and $\frac{10+y}{2}=4$

$\Rightarrow 5+x=6$ and 10+y=8

$\Rightarrow x=1$ and y=-2

$\therefore$ Coordinate of D=(1,-2)

(ii) Length of diagonal BD

$=\sqrt{(1-5)^{2}+(-2-10)^{2}}=\sqrt{(4)^{2}+(-12)^{2}}$

$=\sqrt{16+144}=\sqrt{160}$ units

(iii) Equation of the side joining A(3,6) and D(1,-2) is given by

$\frac{x-3}{3-1}=\frac{y-6}{6+2}$

$\Rightarrow \frac{x-3}{2}=\frac{y-6}{8}$

$\Rightarrow 4(x-3)=y-6$

$\Rightarrow 4 x-12=y-6$

$\Rightarrow 4 x-y=6$

Thus, the equation of the side joining A(3,6) and D(1,-2) is 4 x-y=6

Question 30

A and B arc two points on the x-axis and y-axis respectively. P (2, – 3) is the mid point of AB. Find the

(i) the co-ordinates of A and B.

(ii) the slope of the line AB.

(iii) the equation of the line AB. (2010)

Sol :

Points A and B are on x-axis and y-axis respectively

Let co-ordinates of A be (X, O) and of B be (O, Y)

P (2, -3) is the midpoint of AB

Then, $2=\frac{X+O}{2}$ and $-3=\frac{O+Y}{2}$

$\Rightarrow x=4, y=-6$

(i) Hence co-ordinates of A are (4,0) and of B are (0,-6)

(ii) Slope of $AB=\frac{y_{2}-y_{1}}{x_{2}-y_{1}}$

$=\frac{-6-0}{0-4}=\frac{-6}{-4}=\frac{3}{2}$

(iii) Equation of AB will be $y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y=(-3)=\frac{3}{2}(x-2)$ $(\because P$ lies on it $)$

$\Rightarrow y+3=\frac{3}{2}(x-2)$

$\Rightarrow 2 y+6=3 x-6$

$\Rightarrow 3 x-2 y=6+6$

$\Rightarrow 3 x-2 y=12$

Question 31

Find the equations of the diagonals of a rectangle whose sides are x = – 1, x = 2 , y = – 2 and y = 6.

The equations of sides of a rectangle whose equations are

$x_{1}=-1, x_{2}=2, y_{1}=-2, y_{2}=6$

These lines form a rectangle when they intersect at A, B, C, D respectively

Co-ordinates of A, B, C and D will be

(-1, -2), (2, -2), (2, 6) and (-1, 6) respectively.

AC and BD are its diagonals

(i) Slope of the diagonal AC

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{6+2}{2+1}=\frac{8}{3}$

$\therefore$ Equation of AC will be 

$y-y_{1}=m\left(x-x_{1}\right)$

$=y+2=\frac{8}{3}(x+1)$

$\Rightarrow 3 y+6=8 x+8$

8x-3y+8-6=0

$\Rightarrow 8 x-3 y+2=0$

(ii) Slope of $\mathrm{BD}=\frac{y_{2}-y_{1}}{x_{2}-x_{2}}=\frac{6+2}{-1-2}=\frac{8}{-3}$

$\therefore$ Equation BD will be $y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y+2=-\frac{8}{3}(x-2)$

3y+6=-8x+16

$\Rightarrow 8 x+3 y+6-16=0$

$\Rightarrow 8 x+3 y-10=0$

Question 32

Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x + 1y – 3 and 2x – 3y = 7

Sol :

5x + 7y = 3 …(i)

2x – 3 y = 7 …(ii)

Multiply (i) by 3 and (ii) by 7,

15x + 21y = 9

14x – 21y = 49

Adding we get,

$29 x=58 \Rightarrow x=\frac{58}{29}=2$

Substituting the value of x in (i)

$5 \times 2+7 y=3 \Rightarrow 10+7 y=3$

$\Rightarrow 7 y=3-10 \Rightarrow 7 y=-7 \Rightarrow y=-1$

$\therefore$ Point of intersection of lines is (2,-1)

Now slope of the line joining the points

(2,-1) and the origin (0,0)

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0+1}{0-2}=-\frac{1}{2}$

Equation of line will be

$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-0=-\frac{1}{2}(x-0)$

$\Rightarrow 2 y=-x$

$\Rightarrow x+2 y=0$

Question 33

Point A (3, – 2) on reflection in the x-axis is mapped as A’ and point B on reflection in the y-axis is mapped onto B’ ( – 4, 3).

(i) Write down the co-ordinates of A’ and B.

(ii) Find the slope of the line A’B, hence find its inclination.

Sol :

A’ is the image of A (3, -2) on reflection in the x-axis.

∴ Co-ordinates of A’ will be (3, 2)

Again B’ (- 4, 3) in the image of A’, when reflected in the y-axis

∴ Co-ordinates of B will be (4, 3)

(ii) Slope of the line joining, the points A’ (3, 2) and B (4, 3)

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-3}{3-4}=\frac{-1}{-1}=1$

Now tan $\theta=1$

$\therefore \theta=45^{\circ}$

Hence angle of inclination =45°