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 – k2)y + h2 – 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 – k2)y + k2 – 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 – k2)y + k2 – 7x + 6 = 0,
coefficient of y = 4 – k2 = 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 – k2)y + k2 – 7k + 6 = 0,
we get k2 – 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 – a2 = – 6 or a2 – 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, y1)
∴ Perpendicular ditance of (0, y1) 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.
∴ 3y1 – 12 = ± 20.
For +ve sign, 3y1 – 12 = 20 ⇒ y1 = \(\frac{32}{3}\).
For -ve sign, 3y1 – 12 = – 20 ⇒ y1 = \(\frac{- 8}{3}\).
∴ 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 = x1.
Here, x1 = \(\frac{- 5}{22}\).
∴ 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.
= k2 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 = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) = 0.
Solution:
Consider the lines
y = m1x + c1 ………………….. (1)
and y = m2x + c2 ………………… (2)
Subtracting, we get
0 = (m1 – m2)x + c1 – c2.
Putting the values of x and y in the third equation
y = m3x + c3, we get
⇒ – m1(c1 – c2) + c1(m1 – m2) + m3(c1 – c2) – c3(m1 – m2) = 0
⇒ m1(- c1 + c2 + c1 – c3) + m2(- c1 + c3) + m3(c1 – c2) = 0
i.e; m1(c2 – c3) + m2(c3 – c1) + m3(c1 – c2) = 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 m1.
∴ Equation of the line OP is
y = m1x.
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(x1, y1) 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 + m2 = (1 + m2)2 = 1 + 2m + m2
⇒ 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(x1, y1).
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 x1 + 3 + 3y1 + 24 = 14
or x1 + 3y1 + 13 = 0 ………………… (1)
Now, slope of AB = – \(\frac{1}{3}\)
or y1 – 8 = 3(x1 – 3) = 3x1 – 9
or y1 = 3x1 – 1 ………………………. (2)
Putting the value of y1 in (1), we get
x1 + 3(3x1 – 1) + 13 = 0
10x1 + 10 = 0 or x1 = – 1
Putting x1 = – 1 in eq. (2), we get
y1 = – 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 6m2 – m – 1 = m2 – m – 6
5m2 = – 5 or m2 = – 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 6m2 – m – 1 = – m2 + m + 6.
or 7m2 – 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 p1 from P(x, y) to the line x + y – 5 = 0 is given by
Perpendicular distance p2 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 b2.
Solution:
The perpendicular distances p1 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.