**NCERT Solutions for Class 11 Maths Chapter 9 ****Straight Lines**** Ex 9.2**

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

**Ex 9.2 Class 11 Maths Question 1.**

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

**Solution:**

We know that the y-coordinate 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 x-coordinate 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.

**Ex 9.2 Class 11 Maths Question 2.**

### Passing through the point (-4, 3) with slope ½.

**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}).

**Ex 9.2 Class 11 Maths 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.

**Ex 9.2 Class 11 Maths 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})

**Ex 9.2 Class 11 Maths 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

Required equation is y – 0 = -2(x + 3)

⇒ 2x + y + 6 = 0

**Ex 9.2 Class 11 Maths 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}).

**Ex 9.2 Class 11 Maths 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 two given points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by:

**Ex 9.2 Class 11 Maths Question 8.**

### The vertices of ∆PQR are P(2, 1), Q(-2, 3) and R(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 mid-point of PQ.

**Ex 9.2 Class 11 Maths Question 9.**

### 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.

**Ex 9.2 Class 11 Maths Question 10.**

### 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 the point D divides the line segment in the ratio 1 : n.

**Ex 9.2 Class 11 Maths Question 11.**

### 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/a + 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:

x/5 + y/5 = 1 ⇒ x + y = 5.

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**Ex 9.2 Class 11 Maths Question 12.**

### Find 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 *a* and 9 – *a* respectively.

Then, its equation is

x/a + y/(9−a) = 1

Since it passes through point (2, 2), we have 2/a + 2/(9−a) = 1

⇒ 2(9 – a) + 2a = a(9 – a)

⇒ 18 – 2a + 2a = 9a – a^{2}

⇒ 18 = 9a – a^{2}

⇒ a^{2} – 9a + 18 = 0

⇒ a^{2} – 6a – 3a + 18 = 0

⇒ a(a – 6) – 3 (a – 6) = 0

⇒ (a – 3) (a – 6) = 0

⇒ a = 3, 6

Now, if a = 3 ⇒ b = 9 – 3 = 6 and if a = 6 ⇒ b = 9 – 6 = 3

So, the required equation is:

x/3 + y/6 = 1 or x/6 + y/3 = 1

i.e., 2x + y – 6 = 0 or x + 2y – 6 = 0.

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**Ex 9.2 Class 11 Maths Question 13.**

### Find equation of the line through the point (0, 2) making an angle 2Ï€/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 2Ï€/3 = −√3

The equation of the line passing through point (0, 2) is y – 2 = −√3(x – 0)

⇒ √3x + y – 2 = 0

The slope of line parallel to √3x + y – 2 = 0 is −√3.

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

So, the equation of line is y + 2= −√3(x – 0)

⇒ √3x + y + 2 = 0.

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**Ex 9.2 Class 11 Maths Question 14.**

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

**Solution:**

**Ex 9.2 Class 11 Maths Question 15.**

### 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

**Ex 9.2 Class 11 Maths Question 16.**

### 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.

**Ex 9.2 Class 11 Maths Question 17.**

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

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**Ex 9.2 Class 11 Maths Question 18.**

### 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 and y-axis, respectively.

Then, the equation of the line AB is x/a + y/b = 2.

Now, R(h, k) divides the line segment AB in the ratio 1 : 2.

**Ex 9.2 Class 11 Maths Question 19.**

### 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:

**Related Links:**

**NCERT Solutions for Maths Class 9**