# Chapter 10 Straight Lines Ex – 10.2

In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

Question 1.

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

Solution:

We know that the ordinate of each point on the x-axis is 0.
If P(x, y) is any point on the x-axis, then y = 0.
∴ Equation of x-axis is y = 0.
Also, we know that the abscissa of each point on the y-axis is 0. If P(x, y) is any point on the y-axis, then x = 0.
∴ Equation of y-axis is x = 0.

Question 2.

Passing through the point (-4,3) with slope $\frac { 1 }{ 2 }$.

Solution:

We know that the equation of a line with slope m and passing through the point (x0, y0) is given by (y – y0) = m (x – x0).

Question 3.

Passing through (0, 0) with slope m.

Solution:

We know that the equation of a line with slope m and passing through the point (x0, y0) is given by (y – y0) = m(x – x0)
Here, slope = m, x0 = 0, y0 = 0 Required equation is (y – 0) = m(x – 0)
⇒ y = mx.

Question 4.

Passing through (2,2^3) and inclined with the x-axis at an angle of 75°.

Solution:

We know that the equation of a line with slope m and passing through the point (X0, y0) is given by (y – y0) = m(x – x0)

Question 5.

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

Solution:

We know that the equation of a line with slope m and passing through the point
(x0, y0) is given by (y – y0) = m(x – x0).
Here, m = – 2, x0 = – 3, y0 = 0
y-0 = -2(x + 3) ⇒ 2x + y + 6 = 0

Question 6.

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

Solution:

We know that the equation of line with slope m and passing through the point (x0, y0) is given by (y – y0) = m(x – x0)

Question 7.

Passing through the points (-1,1) and (2, -4).

Solution:

Let the given points be A(-1, 1) and B(2, -4).

We know that the equation of a line passing through the given points (xx, y,) and (x2, y2) is given by

Question 8.

The perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.

Solution:

Here, we are given that p = 5 and ⍵ = 30°.

Question 9.

The vertices of ∆PQR are P(2, 1), Q(-2, 3) and ff(4, 5). Find equation of the median through the vertex R.

Solution:

The vertices of ∆PQR are P( 2, 1), Q(-2, 3) and R(4, 5).

Let S be the midpoint of PQ.

Question 10.

Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2,5) and (-3,6).

Solution:

Let M(2, 5) and N(-3, 6) be the end points of the given line segment.

Question 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:

Let A(1, 0) and B( 2, 3) be the given points and D divides the line segment in the ratio 1 : n.

Question 12.

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

Solution:

Let the required line make intercepts a on the x-axis and y-axis.
Then its equation is $\frac { x }{ a } +\frac { y }{ b } =1$
⇒ x + y = a … (i)
Since (i) passes through the point (2, 3), we have
2 + 3 = a ⇒ a = 5
So, required equation of the line is $\frac { x }{ 5 } +\frac { y }{ 5 } =1$ ⇒ x + y = 5.

Question 13.

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

Solution:

Let the intercepts made by the line on the x-axis and y-axis be o and 9 – a respectively.
Then its equation is $\frac { x }{ a } +\frac { y }{ 9-a } =1$
Since it passes through point (2, 2), we have $\frac { 2 }{ a } +\frac { 2 }{ 9-a } =1$
⇒ 2(9 – a) + 2a = a(9 – a)
⇒ 18 – 2a + 2a = 9a – 9a2
⇒ 18 = 9a – a2 v a2 – 9a + 18 = 0
⇒ a2 – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3 (a – 6) = 0 ⇒ a = 3, 6
Now, if a = 3 ⇒ b = 9 – 3 = 6 and if a = 6 ⇒ b = 9 – 6 = 3
So, required equation is $\frac { x }{ 3 } +\frac { y }{ 6 } =1\quad or\quad \frac { x }{ 6 } +\frac { y }{ 3 } =1$
i.e., 2x + y – 6 = 0 or x + 2y – 6 = 0.

Question 14.

Find equation of the line through the point (0, 2) making an angle $\frac { 2\pi }{ 3 }$ with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Solution:

Here, m = tan $\frac { 2\pi }{ 3 }$ = $-\sqrt { 3 }$
The equation of the line passing through point (0, 2) is y -2 = $-\sqrt { 3 }$(x – 0)
⇒ $\sqrt { 3 } x$ + y – 2 = 0
The slope of line parallel to $\sqrt { 3 } x$ + y – 2 = 0 is $-\sqrt { 3 }$.
Since, it passes through (0, -2).
So, the equation of line is
y + 2= $-\sqrt { 3 }$(x – 0)
⇒ $\sqrt { 3 } x$ + y + 2 = 0.

Question 15.

The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.

Solution:

Question 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 C.

Solution:

Assuming L along x-axis and C along y-axis, we have two points (124.942, 20) and (125.134, 110). By two point form, the point (L, C) satisfies the equation

Question 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 L (litres) along x-axis and R(rupees) along y-axis, we have two points (980,14) and (1220,16).

By two point form, the point (L, R) satisfies the equation.

Question 18.

P(a, b) is the mid-point of a lone segment between axes. Show that equation of the line is $\frac { x }{ a } +\frac { y }{ b } =2$.

Solution:

Let the line AB makes intercepts c and d on the x-axis and y-axis respectively.

Question 19.

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

Solution:

Let AB be the given line segment making intercepts a and b on the x-axis & y-axis respectively.

Then, the equation of line AB is $\frac { x }{ a } +\frac { y }{ b } =2$

So, these points are A(a, 0) and B(0, b).
Now, R(h,k) divides the line segment Ab in the ratio 1 : 2.

Question 20.

By using the concept of equation of a line, prove that the three points (3, 0), (- 2, – 2) and (8, 2) are collinear.

Solution:

Let the given points be A(3, 0), B(-2, -2) and C(8, 2). Then the equation of the line passing through A and B is

Clearly the point C(8, 2) satisfy the equation 2x – 5y – 6 = 0.
(∵ 2(8) – 5(2) – 6 = 16 – 10 – 6 = 0)
Hence, the given points lie on the same straight line whose equation is 2x – 5y – 6 = 0.