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

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 10 Straight Lines Miscellaneous Exercise

Question 1.

Find the value of k for which the line (k – 3)x – (4 – k^{2})y + h^{2} – 7k + 6 = 0

(a) is parallel to x-axis

(b) is parallel to y-axis

(c) passes through the origin.

Solution:

(a) Any line parallel to x-axis is of the form y = p, i.e., coeff. of x = 0.

∴ In equation (k – 3)x – (4 – k^{2})y + k^{2} – 7k + 6 = 0,

coefficient of x = k – 3 = 0.

∴ k = 3.

(b) Any line parallel to y-axis is of the form x = q.

∴ coefficient of y = 0.

In (k – 3)x – (4 – k^{2})y + k^{2} – 7x + 6 = 0,

coefficient of y = 4 – k^{2} = 0.

∴ k = ±2.

(c) If the line passes through the origin, then

x = 0, y = 0 satisfy the equation.

∴ Putting x = 0, y = 0 in

(k – 3)x – (4 – k^{2})y + k^{2} – 7k + 6 = 0,

we get k^{2} – 7k + 6 = 0.

or (k – 6)(k – 1) = 0.

∴ k = 6, 1.

Question 2.

Find the values of θ and p, if the equation xcos θ + ysin θ = p is the normal form of the line \(\sqrt{3}\)x + y + 2 = 0.

Solution:

The given line is \(\sqrt{3x}\) + y = – 2.

Dividing by \(\sqrt{(\sqrt{3})^{2}+1^{2}}\) = 2, we get

Question 3.

Find the equations of the lines which cut off intercepts on the axes whose sum and product are 1 and – 6 respectively.

Solution:

Let a and b be the intercepts, the lines makes on the axes

Sum of intercepts = a+ b = 1 …………………. (1)

Product of intercepts = ab = – 6 ……………………. (2)

From (1) and (2), a(1 – a) = – 6

or a – a^{2} = – 6 or a^{2} – a – 6 = 0.

⇒ (a – 3)(a + 2) = 0

∴ a = 3, – 2

∴ b = – 2, 3

∴ Required lines are

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

Question 4.

What are the points on y-axis whose distances from \(\frac{x}{3}\) + \(\frac{y}{4}\) = 1 are 4 units.

Solution:

Any point on y-axis is (0, y_{1})

∴ Perpendicular ditance of (0, y_{1}) to \(\frac{x}{3}\) + \(\frac{y}{4}\) = 1

or 4x + 3y – 12 = 0 is

∴ \(\frac{0+3 y_{1}-12}{\sqrt{4^{2}+3^{2}}}\) = ± 4 or \(\frac{3 y_{1}-12}{5}\) = ± 4.

∴ 3y_{1} – 12 = ± 20.

For +ve sign, 3y_{1} – 12 = 20 ⇒ y_{1} = \(\frac{32}{3}\).

For -ve sign, 3y_{1} – 12 = – 20 ⇒ y_{1} = \(\frac{- 8}{3}\).

∴ The points on the y-axis are (0, \(\frac{32}{3}\)) and (0, \(\frac{- 8}{3}\)).

Question 5.

Find perpendicular distance of the line joining the points (cos θ, sin θ) and (cos ∅, sin ∅) from the origin.

Solution:

Equation of the line joining (cos θ, sin θ) and (cos ∅, sin ∅) is

∴ The line passing through (cos θ, sin θ) and (cos ∅, sin ∅) is

x cos \(\frac{θ+∅}{2}\) + ysin \(\frac{θ+∅}{2}\) = cos \(\frac{θ-∅}{2}\).

∴ Perpendicular distance from the origin

Question 6.

Find the equation of the line parallel to y-axis and drawn through the point of intersection of lines x – 7y + 5 = 0 and 3x + y = 0.

Solution:

The point of intersection of the lines

x – 7y + 5 = 0 and 3x + y = 0 is obtained by solving these equations.

Putting y = – 3x in x – 7y + 5 = 0, we get

x – 7(- 3x) + 5 = 0 or x + 21x + 5 = 0.

or 22x + 5 = 0 or x = \(\frac{- 5}{22}\).

Any line parallel to y-axis is x = x_{1}.

Here, x_{1} = \(\frac{- 5}{22}\).

∴ The equation of the line parellel toy-axis and passing through the intersection of the given lines is x = \(\frac{- 5}{22}\) or 22x + 5 = 0.

Question 7.

Find the equation of a line drawn perpendicular to the line \(\frac{x}{4}\) + \(\frac{y}{6}\) = 1, through the point where it meets the y-axis.

Solution:

Putting x = 0, we get y = 6. The line \(\frac{x}{4}\) + \(\frac{y}{6}\) = 1 meets the y-axis.

Solution:

Putting x = 0, we get y = 6. The line \(\frac{x}{4}\) + \(\frac{y}{6}\) = 1 meets the y-axis at (0, 6).

The slope of the given line \(\frac{x}{4}\) + \(\frac{y}{6}\) = 1

or 6x + 4y = 24

is – \(\frac{6}{4}\) = – \(\frac{3}{2}\).

∴ Slope of line BC perpendicular to

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

∴ Equation of BC is y – 6 = \(\frac{2}{3}\)(x – 0).

or 3y – 18 = 2x

or 2x – 3y + 18 = 0.

Question 8.

Find the area of traingle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Solution:

The given lines are

y – x = 0 ………………….. (1)

x + y = 0 ………………….. (2)

x – k = 0 ………………….. (3)

From (1) and (2), x = 0, y = 0.

From (2) and (3), x = k, k + y = 0 or y = – k.

From (3) and (1), x = k, y – k = 0 or y = k.

∴ The vertices of the traiangle are O(0, 0), B(k, – k) and A(k, k).

∴ Base of the ∆ OAB = 2k and perpendicular distance from O to AB = k.

∴ Area of ∆ OAB = \(\frac{1}{2}\) × h × 2k.

= k^{2} sq.units.

Question 9.

Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Solution:

Consider the lines

3x + y = 2 ……………………. (1)

2x – y = 3 ………………….. (2)

Adding 5x = 5 or x = 1.

Putting x = 1 in (1), 3 + y = 2.

∴ y = – 1.

∴ Lines (1) and (2) intersect at (1, – 1).

This point will lie on the line px + 2y – 3 = 0. if

p × 1 + 2(- 1) – 3 = 0 or p = 5.

∴ For p = 5, the given lines intersect at (1, – 1).

Question 10.

If three lines whose equations are y = m_{1}x + c_{1}, y = m_{2}x + c_{2} and y = m_{3}x + c_{3} are concurrent, then show that m_{1}(c_{2} – c_{3}) + m_{2}(c_{3} – c_{1}) + m_{3}(c_{1} – c_{2}) = 0.

Solution:

Consider the lines

y = m_{1}x + c_{1} ………………….. (1)

and y = m_{2}x + c_{2} ………………… (2)

Subtracting, we get

0 = (m_{1} – m_{2})x + c_{1} – c_{2}.

Putting the values of x and y in the third equation

y = m_{3}x + c_{3}, we get

⇒ – m_{1}(c_{1} – c_{2}) + c_{1}(m_{1} – m_{2}) + m_{3}(c_{1} – c_{2}) – c_{3}(m_{1} – m_{2}) = 0

⇒ m_{1}(- c_{1} + c_{2} + c_{1} – c_{3}) + m_{2}(- c_{1} + c_{3}) + m_{3}(c_{1} – c_{2}) = 0

i.e; m_{1}(c_{2} – c_{3}) + m_{2}(c_{3} – c_{1}) + m_{3}(c_{1} – c_{2}) = 0.

Question 11.

Find the equation of the lines passing through the points (3, 2) and making an angle of 45° with the line x – 2y = 3.

Solution:

Le the line AB be x – 2y = 3.

Its slope = \(\frac{1}{2}\).

Let m be the slope of the line PA passing through the point P(3, 2).

Angle between PA and AB is 45°.

or m = 3.

Equation of the line PA, passing through P(3, 2) and making an angle of 45° with AB is

y – 2 = 3(x – 3) = 3x – 9

or 3x – y – 9 + 2 = 0 or 3x – 7 = 7.

For – ve sign,

\(\frac{2m-1}{m+2}\) = – 1 or 2m – 1 = – m – 2.

or 3m = – 1 ⇒ m = – \(\frac{1}{3}\).

∴ Equation of PB is

∴ y – 2 = – \(\frac{1}{3}\)(x – 3)

3y – 6 = – x + 3 or x + 3y – 6 – 3 = 0

or x + 3y – 9 = 0.

Question 12.

Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 and that has equal intercepts on the axis?

Solution:

The given lines are

2x – 3y = – 1 ……………… (1)

4x + 7y = 3 ……………. (2)

Multiplying eq. (1) by 2, we get

4x – 6y = – 2 ………………. (3)

Substracting (3) from (2), we get

13y = 5.

∴ y = \(\frac{5}{13}\).

Putting the value of y in (1), 2x – \(\frac{3×5}{13}\) = – 1

Let this point be denoted by P.

PA and PB are the lines that make equal intercepts on the axes.

They make angles of 135° and 45° with positive direction of x-axis.

∴ Their slopes are tan 135° and tan 45° i.e., – 1 and 1 respectively.

∴ Equation of PA is

y – \(\frac{5}{13}\) = – (x – \(\frac{1}{13}\))

or 13y – 5 = – 13x + 1

13x + 13y – 6 = 0.

Equation of PB is

y – \(\frac{5}{13}\) = 1. (x – \(\frac{1}{13}\)).

or 13y – 5 = 13x – 1

or 13x – 13y + 5 – 1 = 0

or 13x – 13y + 4 = 0.

Question 13.

Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is \(\frac{y}{x}\) = \(\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\).

Solution:

Let PQ be the line y = mx + c, whose slope is in.

The line PO makes an angle θ with PQ. Let its slope be m_{1}.

∴ Equation of the line OP is

y = m_{1}x.

or y = \(\frac{y}{x}\) = m

∴ \(\frac{y}{x}\) = \(\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\).

Question 14.

In what ratio, the line segment joining (- 1, 1) and (5, 7) is divided by the line x + y = 4?

Solution:

The line segment joining the points A(- 1, 1) and B(5, 7) is divided by P(x_{1}, y_{1}) in some ratio. Let this ratio be k : 1.

∴ Point P is (\(\frac{5k-1}{k+1}\), \(\frac{7k+1}{k+1}\)).

The point P lies on the line x + y = 4.

∴ \(\frac{5k-1}{k+1}\) + \(\frac{7k+1}{k+1}\) = 4.

or 5k – 1 + 7k + 1 = 4k = 4.

or 8k = 4.

∴ k = \(\frac{1}{2}\).

∴ P divides AB in the ratio \(\frac{1}{2}\) : 1, i.e; 1 : 2.

Question 15.

Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Solution:

The equation of line AB is 4x + 7y + 5 = 0 ………………… (1)

The equation of PQ is 2x – y = 0 ………………… (2)

The point P is (1, 2).

From eq. (2) is y = 2x.

Putting it in eq. (1), we get

4x + 7 × 2x + 5 = 0.

⇒ 18x + 5 = 0 ⇒ x = \(\frac{- 5}{18}\).

∴ y = 2x = – 2 × \(\frac{5}{18}\) = – \(\frac{5}{9}\).

∴ The point Q, where the given lines meet is (\(\frac{- 5}{18}\), \(\frac{- 5}{9}\)).

The point P is (1, 2).

Question 16.

Find the direction in which a straight line must be drawn through the point (- 1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

Solution:

A line passing through P(- 1, 2) is

y – 2 = m(x + 1),

where m is its slope.

⇒ y = mx + m + 2

Putting this value of y in x + y = 4, we get

x + mx + m + 2 = 4.

or (1 + m)x = 4 – 2 – m

∴ x = \(\frac{2-m}{1+m}\).

∴ The co-ordinates of Q the point of intersection Q of AB and PQ is (\(\frac{2-m}{1+m}\), \(\frac{2+5m}{1+m}\)).

The point P is (- 1, 2) and PQ = 3 (Given).

⇒ 1 + m^{2} = (1 + m^{2})^{2} = 1 + 2m + m^{2}

⇒ m = 0.

∴ Slope of PQ is zero i.e; it is parallel to x-axis.

Question 17.

The hypotenuse of a right traiangle has its ends at the points (1, 3) and (- 4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Solution:

Let ABC be the right triangle such that ∠C = 90°,

There are infinitely many such lines.

Let m be the slope of AC. Then, the slope of BC = – \(\frac{1}{m}\).

∴ Equation of AC is y – 3 = m(x – 1).

or x – 1 = \(\frac{1}{m}\)(y – 3).

Equation of BC is x + 4 = – m(y – 1).

⇒ y – 1 = – \(\frac{1}{m}\)(x + 4).

For a given value of m, we can find these equations.

For m = 0, these lines are x + 4 = 0, y – 3 = 0.

For m = ∞, the lines are x – 1 = 0, y – 1 = 0.

Question 18.

Find the image of the point (3, 8) with respect to a line x + 3y = 7, assuming the line to be a plane mirror.

Solution:

Let AB be the line x + 3y = 7 and let the image of P(3, 8) be Q(x_{1}, y_{1}).

Middle point M(\(\frac{x_{1}+3}{2}, \frac{y_{1}+8}{2}\)) lies on AB.

∴ (\(\frac{x_{1}+3}{2}\)) + 3(\(\frac{y_{1}+8}{2}\)) = 7

or x_{1} + 3 + 3y_{1} + 24 = 14

or x_{1} + 3y_{1} + 13 = 0 ………………… (1)

Now, slope of AB = – \(\frac{1}{3}\)

or y_{1} – 8 = 3(x_{1} – 3) = 3x_{1} – 9

or y_{1} = 3x_{1} – 1 ………………………. (2)

Putting the value of y_{1} in (1), we get

x_{1} + 3(3x_{1} – 1) + 13 = 0

10x_{1} + 10 = 0 or x_{1} = – 1

Putting x_{1} = – 1 in eq. (2), we get

y_{1} = – 3 – 1 = – 4.

∴ The image Q of P is (- 1, – 4).

Question 19.

1f the line y = 3x + 1 and 2y = x +3 are equally inclined to the line y = mx + 4, find the value of m.

Solution:

Slope of the line PQ, i.e., y = 3x + 4 is 3.

Slope of QR = m.

The angle θ between the lines is given by

tan θ = |\(\frac{m-3}{1+3m}\)|.

Slope of the line PR, i.e.,

2y = x + 3 is \(\frac{1}{2}\).

Slope of QR is m.

The angle between PR and QR is given by

For +ve sign, \(\frac{m-3}{3m+1}\) = \(\frac{2m-1}{m+2}\).

or (3m + 1)(2m – 1) = (m – 3)(m + 2)

or 6m^{2} – m – 1 = m^{2} – m – 6

5m^{2} = – 5 or m^{2} = – 1 (Not admissible)

For -ve sign, \(\frac{m-3}{3m+1}\) = – \(\frac{2m-1}{m+2}\).

or (3m + 1)(2m – 1) = – (m – 3)(m + 2).

or 6m^{2} – m – 1 = – m^{2} + m + 6.

or 7m^{2} – 2m – 7 = 0.

Question 20.

If the sum of perpendicular distances of a variable point P(x, y) from the lines x + y – 5 = 0 and – 2y + 7 = 0 is always 10, show that P must move on a line.

Solution:

Perpendicular distance p_{1} from P(x, y) to the line x + y – 5 = 0 is given by

Perpendicular distance p_{2} from P(x, y) to the line 3x – 2y + 7 = 0 is given by

which is the equation of a straight line. Hence, P moves on a line.

Question 21.

Find the equation of the line, which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Solution:

The given parallel lines are

9x + 6y – 7 = 0 …………….. (1)

and 3x + 2y + 6 = 0

Multiplying by 3, we get

9x + 6y + 18 = 0 …………………. (2)

Let the third line parallel to the lines (1) and (2) and then which is equidistant from them be

9x + 6y + c = 0 ………………… (3)

Distance between (1) and (3) = \(\frac{|-7-c|}{\sqrt{9^{2}+6^{2}}}\).

Distance between (2) and (3) = \frac{|c-18|}{\sqrt{9^{2}+6^{2}}}.

∴ The third line being equidistant from the given two lines, we get

\(\frac{|7+c|}{\sqrt{117}}\) = \(\frac{|c-18|}{\sqrt{117}}\)

or 7 + c = -(c – 18) or 2c = 11.

∴ c = \(\frac{11}{2}\).

∴ The line mid-way between the given lines is

9x + 6y + \(\frac{11}{2}\) = 0 or 18x + 12y + 11 = 0.

Question 22.

A ray of light passes through the point (1, 2) reflects on the x-axis at a point A and the reflected ray passes through the point (5, 3). Find the co-ordinates of A.

Solution:

Angle of incidence = Angle of reflection.

If AN ⊥ OX, then

∠PAN = ∠QAN,

where P is (1, 2) and Q is (5, 3).

⇒ ∠QAX = ∠PAO.

∴ If ∠QAX = θ, then ∠PAO = θ.

∴ ∠XAP = 180° – θ.

∴ Slope of AQ = tan θ = \(\frac{3-0}{5-k}\).

where the point A is (k, 0).

Slope of AP = tan(180° – θ) = \(\frac{2-0}{1-k}\) = – tan θ.

∴ \(\frac{3}{5-k}\) = – \(\frac{2}{1-k}\)

or 3 – 3k = – 10 + 2k

or 5k = 13.

∴ k = \(\frac{13}{5}\).

∴ The point A is (\(\frac{13}{5}\), 0).

Question 23.

Prove that the product of the lengths of perpendiculars drawn from the point (\(\sqrt{a^{2}-b^{2}}\), 0) and (- \(\sqrt{a^{2}-b^{2}}\), 0) to the line \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 is b^{2}.

Solution:

The perpendicular distances p_{1} from (\(\sqrt{a^{2}-b^{2}}\), 0) and (- \(\sqrt{a^{2}-b^{2}}\), 0) to the line cos θ + sin θ = 1 are

Question 24.

A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y = 4 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow.

Solution:

Equation of the two paths OA and OB are

2x – 3y – 4 = 0 ……………… (1)

3x + 4y – 5 = 0 ……………… (2)

Solving these equations, we get

\(\frac{x}{15+16}\) = \(\frac{y}{-12+10}\) = \(\frac{1}{8+9}\)

∴ x = \(\frac{31}{17}\), y = \(\frac{- 2}{17}\).

∴ These paths meet at (\(\frac{31}{17}\), \(\frac{- 2}{17}\)).

The third path is AB whose equation is 6x – 7y + 8 = 0.

The shortest path from O to the path AB is the perpendicular path from O to AB.

∴ Slope of AB = \(\frac{6}{7}\)

∴ Slope of ⊥ path = \(\frac{-7}{6}\)

∴ Equation of perpendicular path OM is

y + \(\frac{2}{17}\) = – \(\frac{7}{6}\)(x – \(\frac{31}{17}\))

⇒ y + \(\frac{2}{17}\) = – \(\frac{7}{6}\)x + \(\frac{217}{102}\).

Multiplying by 102, we get

102y + 12 = – 119x+ 217

or 119x + 102y – 205 = 0.

∴ Equation of the shortest path is

119x + 102y – 205 = 0.

Hence, the time taken to reach AB will be minimum.