*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 .**

**Solution:**

We know that the equation of a line with slope m and passing through the point (x_{0}, y_{0}) is given by (y – y_{0}) = m (x – x_{0}).

**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 (x_{0}, y_{0}) is given by (y – y_{0}) = m(x – x_{0})

Here, slope = m, x_{0} = 0, y_{0} = 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 (X_{0}, y_{0}) is given by (y – y_{0}) = m(x – x_{0})

**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

(x_{0}, y_{0}) is given by (y – y_{0}) = m(x – x_{0}).

Here, m = – 2, x_{0} = – 3, y_{0} = 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 (x_{0}, y_{0}) is given by (y – y_{0}) = m(x – x_{0})

**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

⇒ 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

⇒ 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

Since it passes through point (2, 2), we have

⇒ 2(9 – a) + 2a = a(9 – a)

⇒ 18 – 2a + 2a = 9a – 9a^{2}

⇒ 18 = 9a – a^{2} v a^{2} – 9a + 18 = 0

⇒ a^{2} – 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

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 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 =

The equation of the line passing through point (0, 2) is y -2 = (x – 0)

⇒ + y – 2 = 0

The slope of line parallel to

+ y – 2 = 0 is .

Since, it passes through (0, -2).

So, the equation of line is

y + 2= (x – 0)

⇒ + 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 .**

**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

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.