Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 10 Straight Lines Ex 10.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 10 Straight Lines Ex 10.2

In questions 1 to 7, find the equations of the line, which satisfy the given conditions:

1. Write the equations for x and y-axis.

2. passing through (- 4, 3) with slope \(\frac{1}{2}\).

3. Passing through (0, 0) with slope m.

4. Passing through (2, 2\(\sqrt{3}\)) and inclined with the x-axis at an angle of 75Â°.

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

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

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

Solutions to questions 1 to 7:

1. Equation of the x-axis is y = 0.

Equation of the y-axis is x = 0.

2. Equation of a line passing through (x_{1}, y_{1}) with slope m is (y – y_{1}) = m(x – x_{1}).

Here, x_{1} = – 4, y_{1} = 3, m = \(\frac{1}{2}\).

âˆ´ Equation of the required line is

(y – 3) = \(\frac{1}{2}\) (x + 4)

or 2y – 6 = x + 4 or x – 2y + 10 = 0

3. x_{1} = 0, y_{1} = 0, Slope = m.

Equation of the required line is

(y – 0) = m(x – 0)

or y = mx.

4. Here, x_{1} = 2, y_{1} = 2\(\sqrt{3}\)

m = tan 75Â° = tan (45Â° + 30Â°)

= \(\frac{tan 45Â° + tan 30Â°}{1 – tan 45Â° tan 30Â°}\) [âˆµ tan(A + B) = \(\frac{tan A + tan B}{1 – tan A tan B}\)

Equation of the required line is

y – 2\(\sqrt{3}\) = \(\frac{\sqrt{3}+1}{\sqrt{3}-1}\) (x – 2).

(\(\sqrt{3}\) – 1)y – 2\(\sqrt{3}\)(\(\sqrt{3}\) – 1) = (\(\sqrt{3}\) + 1)x – 2(\(\sqrt{3}\) + 1)

or (\(\sqrt{3}\) – 1)y – 6 + 2\(\sqrt{3}\) = (\(\sqrt{3}\) + 1)x – 2\(\sqrt{3}\) – 2.

or (\(\sqrt{3}\) + 1)x – (\(\sqrt{3}\) – 1)y = – 6 + 2\(\sqrt{3}\) + 2\(\sqrt{3}\) + 2.

= 4\(\sqrt{3}\) – 4 = 4(\(\sqrt{3}\) – 1).

âˆ´ Then, equation of required line is

(\(\sqrt{3}\) + 1)x – (\(\sqrt{3}\) – 1)y = 4(\(\sqrt{3}\) – 1).

5. The line AB meets x-axis at (- 3, 0) and slope of the line = – 2.

âˆ´ Equation of the line is y – 0 = – 2(x + 3)

or 2x + y + 6 = 0.

6. The line PQ intersects y-axis at (0, 2).

Slope of PQ = tan 30Â° = \(\frac{1}{\sqrt{3}}\)

Equation of PQ is

y – 2 = \(\frac{1}{\sqrt{3}}\)(x – 0).

or \(\sqrt{3y}\) – 2\(\sqrt{3}\) = x

or x – \(\sqrt{3}\)y + 2\(\sqrt{3}\) = 0.

7. The line passes through the points A(- 1, 1) and B(2, – 4). Equation of the line passing through (x_{1}, y_{1}) and (x_{2}, y_{2}) is

y – y_{1} = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x – x_{1})

Here, x_{1} = – 1, y_{1} = 1, x_{2} = 2, y_{2} = – 4.

âˆ´ Equation of AB is

y – 1 = \(\frac{-4-1}{2+1}\)(x + 1) = \(\frac{-5}{3}\)(x + 1).

or 3(y – 1) = – 5(x + 1).

or 3y – 3 = – 5x – 5.

or 5x + 3y – 3 + 5 = 0

or 5x + 3y + 2 = 0.

Question 8.

Find the equation of a line whose perpendicular distance from the origin is 5 units and the angle made by the perpendicular with positive x-axis is 30Â°.

Solution:

If a line is at a distance p from the origin and perpendicular makes an angle to with positive direction of x-axis, its equation is

xcos Ï‰ + ysin Ï‰ – p.

Here, p = 5, Ï‰ = 30Â°.

âˆ´ Equation of the required line is

x cos 30Â° + y sin 30Â° = 5.

or \(\frac{\sqrt{3}}{2}\)x + \(\frac{1}{2}\)y = 5

or \(\sqrt{3x}\) + y = 10.

Question 9.

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

Solution:

The vertices P and Q are (2,1) and (- 2, 3).

The middle point of PQ is S (\(\frac{2-2}{0}\), \(\frac{1+3}{2}\)) or S(0, 2).

âˆ´ Equation of the median RS, where R is (4, 5) and S is the point (0, 2) is

y – 5 = \(\frac{2-5}{0-4}\)(x – 4).

[Apply formula y – y_{1} = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x – x_{1})

or y – 5 = \(\frac{-3}{-4}\)(x – 4).

or 4(y – 5) = 3(x – 4) or 4y – 20 = 3x – 12.

i.e; 3x – 4y = – 20 + 12 – 8

âˆ´ Equation of median RS is 3x – 4y + 8 = 0.

Question 10.

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

Solution:

Slope of the line joining A(2, 5) and B(- 3, 6) = \(\frac{6-5}{-3-2}\) = \(\frac{1}{-5}\).

âˆ´ m = slope of any line âŠ¥ to AB = 5 [m_{1} = \(\frac{-1}{m_{2}}\).

The equation of the line passing through (- 3, 5) and perpendicular to AB is

y – 5 = 5(x + 3)

â‡’ 5x – y + 20 = 0.

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:

Slope of the line joining the points A(1, 0) and B(2, 3)

= \(\frac{3-0}{2-1}\) = \(\frac{3}{1}\) = 3.

âˆ´ Slope of the CD perpendicular to AB = – \(\frac{1}{3}\) [The lines are âŠ¥, so m_{1}m_{2} = – 1]

The point P divides AB in the ratio 1 : n.

âˆ´ Coordinates of P are

(\(\frac{1Ã—2+1Ã—n}{1+n}\), \(\frac{1Ã—3+0Ã—n}{1+n}\)) or (\(\frac{n+2}{n+1}\), \(\frac{3}{n+1}\)).

âˆ´ Equation of the line CD which is âŠ¥ to AB and passes through P is

y – \(\frac{3}{n+1}\) = – \(\frac{1}{3}\)(x – \(\frac{n+2}{n+1}\))

[Using the formula y – y_{1} = m(x – x_{1})]

or 3(n + 1)y – 9 = – (n + 1) (x – \(\frac{n+2}{n+1}\))

= – (n + 1)x + (n + 2)

or (n + 1)x + 3(n + 1)y = n + 2 + 9

or (n + 1)x + 3(n + 1)y = r + 11.

Question 12.

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

Solution:

The line making equal intercepts has the slope either 1 or – 1.

âˆ´ tan 45Â° – 1

or tan 135Â° = – 1.

Now(i) slope = 1, P(2, 3)

âˆ´ Equatirn of the line PR of slope 1 and passing through P(2, 3) is

y – 3 – 1(x – 2)

[Using the formula y – y_{1} = m(x – x_{1})]

or x – y + 1 = 0.

(ii) When the slope = – 1.

Equation of the lire with slope – 1 and passing through P(2, 3) is

y – 3 = – 1 Ã— (x – 2).

or x + y – 3 – 2 = 0,

or x + y – 5 = 0.

Question 13.

Find the equations of the lines passing through the point (2, 2), such that the sum of their intercepts on the axes is 9.

Solution:

Let the intercepts along x-axis and y-axis be a and b.

âˆ´ a + b = 9.

or b = 9 – a …………….. (1)

âˆ´ Equation of the line in intercepts form is

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 ………………….. (2)

â‡’ \(\frac{x}{a}\) + \(\frac{y}{9-a}\) = 1 [Using (1)]

Since point A(2, 2) lies on it, therefore

\(\frac{2}{a}\) + \(\frac{2}{9-a}\) = 1.

â‡’ \(\frac{2(9-a)+2a}{a(9-a)}\) = 1.

â‡’ 18 – 2a + 2a = a(9 – a)

â‡’ 18 = 9a – a^{2}

â‡’ a^{2} – 9a + 18 = 0 â‡’ (a – 3)(a – 6) = 0.

Hence, the required equations of straight lines are 2x + y = 6 and x + 2y = 6.

Question 14.

Find the equation of the line through the point (0, 2) making an angle \(\frac{2Ï€}{3}\) 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:

Slope of the line = tan \(\frac{2Ï€}{3}\)

= tan (Ï€ – \(\frac{Ï€}{3}\))

= – \(\frac{Ï€}{3}\) = – \(\sqrt{3}\).

Equation of the line AC passing through (0, 2) and having the slope – \(\sqrt{3}\) is

y – 2 = – \(\sqrt{3}\) (x – 0).

[Using the formula y – y_{1} = m(x – x_{1})]

â‡’ \(\sqrt{3}\)x + y – 2 = 8.

Another line BD is parallel to AC.

âˆ´ Slope of BD = Slope of AC = – \(\sqrt{3}\).

BD is passing through the point B(0, – 2).

âˆ´ Equation of BD is

y + 2 = – \(\sqrt{3}\)(x – 0).

or \(\sqrt{3x}\) + y + 2 = 0.

Thus, equations of AC and BD are

\(\sqrt{3x}\) + y – 2 = 0 and \(\sqrt{3x}\) + y + 2 = 0.

Question 15.

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

Solution:

Let ON be the perpendicular to AB.

The point N is (- 2, 9).

âˆ´ Slope of ON = \(\frac{9-0}{-2-0}\) = \(\frac{- 9}{2}\) [Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)]

Slope of AB which is âŠ¥ ON = \(\frac{2}{9}\) [âˆµ Lines are âŠ¥ if m_{1}m_{2} = – 1]

Now AB passes through (- 2, 9) and has the slope \(\frac{2}{9}\)

âˆ´ Equation of AB is y – 9 = \(\frac{2}{9}\)(x + 2) [y – y_{1} = m(x – x_{1})]

or 9y – 81 = 2x + 4

or 2x – 9y + 85 = 0.

Question 16.

The length L (in centimetres) 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 of C.

Solution:

L is a linear function of C.

âˆ´ L = a + bC.

For L = 124.942, C = 20.

âˆ´ 124.942 = a + 20b …………….. (1)

For L = 125.134, C = 110.

âˆ´ 125.134 = a + 110b …………….. (2)

Subtracting (1) from (2), we get

0.192 = 90b.

âˆ´ b = \(\frac{0.192}{90}\) = 0.00213.

From (1), 124.942 = a + 20 Ã— 0.00213

= a + 0.0426.

âˆ´ a = 124.942 – 0.0426

= 124.8994.

Now, L in terms of C is

L = a + bC

â‡’ L = 124.8994 + 0.00213C

Question 17.

The owner of a milk store finds that he can sell 980 litres of milk each week at â‚¹ 14/litre and 1220 litres of milk each week at â‚¹ 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at â‚¹ 17/litre.

Solution:

Let y litres milk is sold at â‚¹ x/litre.

x and y have linear relationship, i.e.,

y = a + bx, i.e., it is a straight line.

Now, y_{1} = 980 litres, x_{1} = â‚¹ 14/litre.

y_{2} = 1220 litres, x_{2} = â‚¹ 16/litre.

Slope of the line = \(\frac{1220 – 980}{16 – 14}\) = \(\frac{240}{2}\) = 120.

âˆ´ Equation of the line is

y – 980 = 120(x – 14).

When x = 17, y = 980 + 120(17 – 14)

= 980 + 120 Ã— 3

= 980 + 360 = 1340.

Hence, 1340 litres milk may be sold at â‚¹ 17/litre.

Question 18.

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

Solution:

Let the line AB makes intercepts p and q on the axes.

âˆ´ A is (p, 0) and B is (0, q).

Now, P(a, b) is the mid-point of AB.

âˆ´ \(\frac{p+0}{2}\) = a, \(\frac{0+q}{2}\) = b.

âˆ´ p = 2a, q = 2b.

Intercepts form of the line AB

\(\frac{x}{p}\) + \(\frac{x}{q}\) = 1.

or \(\frac{x}{2a}\) + \(\frac{y}{2b}\) = 1 or \(\frac{x}{a}\) + \(\frac{y}{b}\) = 2.

Question 19.

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

Solution:

Let the line AB makes intercepts a and b on the axes. The point R(h, k) divides line segment AB in the ratio 1 : 2.

So, h = \(\frac{1Ã—0+2Ã—a}{1+2}\)

= \(\frac{2a}{3}\)

âˆ´ a = \(\frac{3}{2}\)h

and k = \(\frac{1Ã—b+2Ã—0}{1+2}\) = \(\frac{b}{3}\).

âˆ´ b = 3k.

Intercepts form of the line AB is

\(\frac{x}{a}\) + \(\frac{y}{b}\) = 1.

or \(\frac{x}{3/2h}\) + \(\frac{y}{3k}\) = 1 or \(\frac{2x}{3h}\) + \(\frac{y}{3k}\) = 1.

or 2kx + hy = 3kh.

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:

Equation of the line passing through (3, 0) and (- 2, – 2) is

y – 0 = \(\frac{-2-0}{-2-3}\)(x – 3)

[Apply y – y_{1} = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x – x_{1})

= \(\frac{2}{5}\)(x – 3)

or 5y = 2x – 6 or 2x – 5y = 6.

If the point (8, 2) lies on it, (8, 2) will satisfy the equation of the line.

âˆ´ 2 Ã— 8 – 5 Ã— 2

= 16 – 10 = 6.

Hence, (8, 2) lies on it.

Thus, (3, 0), (- 2, – 2) and (8, 2) are collinear.