11th NCERT/CBSE Straight Lines Miscellaneous Exercise Questions 24
Hi

Question (1)

Find the values of k for which the line
(k - 3)x - (4 - k2)y + k2 -7k + 6 = 0 is (a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

Solution

The given equation of line is (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 ---- (1)
Slope of line $l = \frac{{ - a}}{b}$
$l = \frac{{ - \left( {k - 3} \right)}}{{ - \left( {4 - k} \right)}} = \frac{{k - 3}}{{4 - k}}$
(i) line parallel x-axis
∴ slope of line = slope of x-axis
$\therefore \frac{{k - 3}}{{4 - k}} = 0$
k = 0

(ii) Line parallel y-axis
slope of line = slope of y-axis
slope of y-axis is not define
∴ Denominator = 0
4 - k = 0
k = 4

(iii) Line is passing through (0, 0)
∴ c = 0
∴ k2 - 7k + 6 = 0
(k - 6) ( k-1) = 0
k = 6 or k = 1

Question (2)

Find the values of θ and p, if the equation xcosθ + ysinθ is the normal form of the line $\sqrt 3 x + y + 2 = 0$.

Solution

The equation of the given line is $\sqrt 3 x + y + 2 = 0$.
This equation can be reduced as
$\sqrt 3 x + y + 2 = 0$.
$\Rightarrow - \sqrt 3 x - y = 2$
On dividing both sides by $\sqrt {{{\left( { - \sqrt 3 } \right)}^2} + {{\left( { - 1} \right)}^2}} = 2$, we obtain
$- \frac{{\sqrt 3 }}{2}x - \frac{1}{2}y = \frac{2}{2}$
$\Rightarrow \left( { - \frac{{\sqrt 3 }}{2}} \right)x + \left( { - \frac{1}{2}} \right)y = 1 - - - (1)$
On comparing equation (1) to xcosθ + ysinθ = p, we obtain
$\cos \theta = - \frac{{\sqrt 3 }}{2},\sin \theta = - \frac{1}{2},$ and p = 1
Since the values of sin θ and cos θ are negative, $\theta = \pi + \frac{\pi }{6} = \frac{{7\pi }}{6}$

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 intercept made by line on axis be 'a' and 'b'. Equation of line be $\frac{x}{a} + \frac{y}{b} = 1$
Given that a+b = 1, ab=-6, a = 1-b
∴ ab = -6
(1-b)b = -6
b - b2 = -6
b2 - b - 6 = 0
∴ (b-3) ( b + 2) = 0
b = 3 or b = -2
if b = 3, a = 1-b = 1-3 = -2
then equation of line will be
$\frac{x}{{ - 2}} + \frac{y}{3} = 1$
-3x + 2y = 6
3x - 2y + 6 = 0
If b = -2, a = 1-b = 1-(-2) = 3
then equation of line will be
$\frac{x}{3} + \frac{y}{{ - 2}} = 1$
2x - 3y = 6

Question (4)

What are the points on the y-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units

Solution

Let (0, b) be the point on the y-axis whose distance from line
$\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.
The given line can be written as 4x + 3y – 12 = 0 --- (1)
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = –12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
$d = \frac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$
Therefore, if (0, b) is the point on the y-axis whose distance from line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units, then:
$4 = \frac{{\left| {4\left( 0 \right) + 3\left( b \right) - 12} \right|}}{{\sqrt {{4^2} + {3^2}} }}$
$\Rightarrow 4 = \frac{{\left| {3\left( b \right) - 12} \right|}}{5}$
⇒ 20 = |3b - 12|
⇒ 20 = ±(3b -12)
⇒ 20 = (3b -12) or 3b = -(3b-12)
⇒ 3b = 20 +12 or 3b = -20 +12
$\Rightarrow b = \frac{{32}}{5} \;or \;b = - \frac{8}{3}$
Thus, the required points are $\left( {0,\frac{{32}}{3}} \right)\;and\;\left( {0. - \frac{8}{3}} \right)$

Question (5)

Find the perpendicular distance from the origin to the line joining the points (cosθ, sinθ) and (cosφ, sinφ)

Solution

The equation of the line joining the points (cosθ, sinθ) and (cosφ, sinφ) is given by
$\frac{{y - \sin \theta }}{{\sin \phi - \sin \theta }} = \frac{{x - \cos \theta }}{{\cos \phi - \sin \theta }}$
$\Rightarrow \frac{{y - \sin \theta }}{{2\cos \left( {\frac{{\phi + \theta }}{2}} \right)\sin \left( {\frac{{\phi - \theta }}{2}} \right)}} = \frac{{x - \cos \theta }}{{2\sin \left( {\frac{{\phi + \theta }}{2}} \right)\sin \left( {\frac{{\phi - \theta }}{2}} \right)}}$
$\Rightarrow \frac{{y - \sin \theta }}{{\cos \left( {\frac{{\phi + \theta }}{2}} \right)}} = \frac{{x - \cos \theta }}{{ - \sin \left( {\frac{{\phi + \theta }}{2}} \right)}}$
$\Rightarrow - y\sin \left( {\frac{{\phi + \theta }}{2}} \right) + \sin \theta \sin \left( {\frac{{\phi + \theta }}{2}} \right) = x\cos \left( {\frac{{\phi + \theta }}{2}} \right) - \cos \theta \cos \left( {\frac{{\phi + \theta }}{2}} \right)$
$x\cos \left( {\frac{{\phi + \theta }}{2}} \right) + y\sin \left( {\frac{{\phi + \theta }}{2}} \right) - \sin \theta \sin \left( {\frac{{\phi + \theta }}{2}} \right) - \cos \theta \cos \left( {\frac{{\phi + \theta }}{2}} \right) = 0$
Let 'p' be the perpendicular distance of line from (0, 0)
$p = \frac{{\left| c \right|}}{{\sqrt {{a^2} + {b^2}} }}$
$p = \frac{{\left| { - \sin \theta \sin \left( {\frac{{\phi + \theta }}{2}} \right) + \cos \theta \cos \left( {\frac{{\phi + \theta }}{2}} \right)} \right|}}{{\sqrt {{{\cos }^2}\left( {\frac{{\phi + \theta }}{2}} \right) + {{\sin }^2}\left( {\frac{{\phi + \theta }}{2}} \right)} }}$
$p = \sin \theta \sin \left( {\frac{{\theta + \phi }}{2}} \right) + \cos \theta \cos \left( {\frac{{\theta + \phi }}{2}} \right)$
Use follwoing formula cosαcosβ + sinαsinβ = cos(α - β)
$p = \cos \left( {\theta - \frac{{\phi + \theta }}{2}} \right)$
$p = \cos \left( {\frac{{2\theta - \phi - \theta }}{2}} \right)$
$p = \cos \left( {\frac{{\theta - \phi }}{2}} \right)$

Question (6)

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

Solution

The two given lines are
x – 7y + 5 = 0 ---- (1)
3x + y = 0 ---- (2)
The equation of required line passing through the point of intersection of l1 and l2 is given by
l : l1 + λl2 = 0
∴ ( x - 7y +5) + λ (3x + y) = 0 ----(1)
(1+3λ)x + (-7+λ)y + 5 = 0
slope of line
$l:\frac{{ - a}}{b} = \frac{{ - \left( {1 + 3\lambda } \right)}}{{ - 7 + \lambda }}$
line parallel to y-axis
slope of l = slope of y-axis
slope of y axis is not define
∴ slope of line is also not define
⇒ Deno. of slope = 0
∴ 7 - λ = 0
λ = 7
Replacing value of λ in equation (1) we get
x - 7y + 5 +7(3x+y) = 0
x - 7y + 5 + 21x +7y = 0
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

The equation of the given line is
${l_1}:\frac{x}{4} + \frac{y}{6} = 1$
Line meets y-axis at A
x-coordinate of A = 0
$\frac{y}{6} = 1 \Rightarrow y = 6$
∴ A(0, y) = (0, 6)
${l_1}:\frac{x}{4} + \frac{y}{6} = 1$
6x + 4y -24 = 0
3x + 2y - 12 = 0
Equation of line perpendicular to l1 will be
2x - 3y + k = 0
it passes through A(0, 6)
2(0) - 3(6) + k = 0
k = 18
∴ Equation of required line is 2x-3y + 18 = 0

Question (8)

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

Solution

The equations of the given lines are
y – x = 0 … (1)
x + y = 0 … (2)
x – k = 0 … (3)
The point of intersection of lines (1) and (2) is given by
x = 0 and y = 0
The point of intersection of lines (2) and (3) is given by
x = k and y = –k
The point of intersection of lines (3) and (1) is given by
x = k and y = k
Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and (k, k).
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is
$\frac{1}{2}\left| {{x_1}\left( {{y_2} - {y_1}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$
Therefore, area of the triangle formed by the three given lines
$= \frac{1}{2}\left| {0\left( { - k - k} \right) + k\left( {k - 0} \right) + k\left( {0 + k} \right)} \right|$ square units
$= \frac{1}{2}\left| {{k^2} + {k^2}} \right| = {k^2}$square 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

The equations of the given lines are
3x + y – 2 = 0 … (1)
px + 2y – 3 = 0 … (2)
2x – y – 3 = 0 … (3)
On solving equations (1) and (3), we obtain
x = 1 and y = –1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).
p (1) + 2 (–1) – 3 = 0
p – 2 – 3 = 0
p = 5
Thus, the required value of p is 5.

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

The equations of the given lines are
m1x - y + c1
y = m1x + c1... (1)
y = m2x + c2
y = m2x + c2... (2)
y = m3x + c3
y = m3x + c3.....(3)
lines are concurrents
$\left| {\begin{array}{*{20}{c}}{{m_1}}&{ - 1}&{{c_1}}\\{{m_2}}&{ - 1}&{{c_2}}\\{{m_3}}&{ - 1}&{{c_3}}\end{array}} \right| = 0$
⇒ m1(-c3 + c2) 1(m2c3 - m3c2) + c1(-m2+m3) = 0
⇒ m1(c2-c3) + m2c3 -m3c2 -m2c1+m3c1= 0
$rArr; m1(c2 - c3) + m2c3-m2c1 + m3c1 - m3c2= 0$rArr; m1(c2-c3)+m2(c3-c1)+m3(c1-c2) = 0

Question (11)

Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.

Solution

Let the slope of the required line be m1.
The given line can be written as $y = \frac{1}{2}x - \frac{3}{2}$, which is of the form y = mx + c
∴ Slope of the given line $= {m_2} = \frac{1}{2}$
It is given that the angle between the required line and line x – 2y = 3 is 45°.
We know that if θ is the acute angle between lines l1 and l2 with slopes m1 and m2 respectively, then $\tan \theta = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
$\therefore \;\tan 45 = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$
$\Rightarrow 1 = \left| {\frac{{\frac{1}{2} - {m_1}}}{{1 + \frac{{{m_1}}}{2}}}} \right|$
$\Rightarrow 1 = \left| {\frac{{1 - 2{m_1}}}{{2 + {m_1}}}} \right|$
$\Rightarrow 1 = \pm \left( {\frac{{1 - 2{m_1}}}{{2 + {m_1}}}} \right)$
$\Rightarrow 1 = \frac{{1 - 2{m_1}}}{{2 + {m_1}}} \;\; or\;\; 1 = - \left( {\frac{{1 - 2{m_1}}}{{2 + {m_1}}}} \right)$
⇒ 2 + m1 = 1 - 2m1 or 2 +m1 = -1 + 2m1
$\Rightarrow {m_1} = - \frac{1}{3} \;\;or\;\; {m_1} = 3$
Case I: m1 = 3
The equation of the line passing through (3, 2) and having a slope of 3 is:
y – 2 = 3 (x – 3)
y – 2 = 3x – 9
3x – y = 7
Case II: ${m_1} = - \frac{1}{3}$
The equation of the line passing through (3, 2) and having a slope of $- \frac{1}{3}$ is:
$y - 2 = - \frac{1}{3}\left( {x - 3} \right)$
3y-6 = -x + 3
x +3y = 9
Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

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 that has equal intercepts on the axes.

Solution

l1: 4x + 7y – 3 = 0 and l2: 2x – 3y + 1 = 0
The equation of required line "l" which [asses through the point of intersection of l1 and l2 is given bt
l = l1 + λl2 = 0
∴ (4x + 7y - 3) + λ (2x-3y+1) = 0 ----(1)
(4+2λ)x + (7-3λ)y-3+λ = 0
Slope of line l $= - \frac{a}{b} = - \frac{{\left( {4 + 2\lambda } \right)}}{{7 - 3\lambda }}$
The required line has equal intercepts on axis
∴ slope = - 1
$- \frac{{\left( {4 + 2\lambda } \right)}}{{7 - 3\lambda }} = - 1$
4 + 2λ = 7 -3λ
5λ = 3
$\lambda = \frac{3}{5}$
Replacea the value of $\lambda = \frac{3}{5}$ i equation(1) we get
$\left( {4x + 7y - 3} \right) + \frac{3}{5}\left( {2x - 3y + 1} \right) = 0$
20x + 35y -15 +6x -9y +3 =0
26x + 26y -12 = 0
13x + 13y -6 = 0

Question (13)

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

Solution

Let the equation of the line passing through the origin be y = m1x.
If this line makes an angle of θ with line y = mx + c, then angle θ is given by
$\therefore \tan \theta = \left| {\frac{{{m_1} - m}}{{1 + {m_1}m}}} \right|$
$\Rightarrow \tan \theta = \left| {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right|$
$\Rightarrow \tan \theta = \pm \left( {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right)$
$\Rightarrow \tan \theta = \frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}} \;or \; \tan \theta = - \left( {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right)$
Case I : $\tan \theta = \frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}$
$\Rightarrow \tan \theta + \frac{y}{x}m\tan \theta = \frac{y}{x} - m$
$\Rightarrow m + \tan \theta = \frac{y}{x}\left( {1 - m\tan \theta } \right)$
$\Rightarrow \frac{y}{x} = \frac{{m + \tan \theta }}{{1 - m\tan \theta }}$
Case II : $\tan \theta = - \left( {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right)$
$\Rightarrow \tan \theta + \frac{y}{x}m\tan \theta = - \frac{y}{x} + m$
$\Rightarrow \frac{y}{x}\left( {1 + m\tan \theta } \right) = m - \tan \theta$
$\Rightarrow \frac{y}{x} = \frac{{m - \tan \theta }}{{1 + m\tan \theta }}$
Therefore, the required line is given by $\Rightarrow \frac{y}{x} = \frac{{m \pm \tan \theta }}{{1 \pm m\tan \theta }}$

Question (14)

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

Solution

Let line x+y = 4
Let p(x+y) dividing the line segment AB where A(-1, 1) and B(5, 7) in ratio k:1
By section formula
$p\left( {x,y} \right) = \left( {\frac{{k\left( 5 \right) + 1\left( { - 1} \right)}}{{k + 1}},\frac{{7\left( k \right) + 1}}{{k + 1}}} \right)$
$p\left( {x,y} \right) = \left( {\frac{{5k - 1}}{{k + 1}},\frac{{7k + 1}}{{k + 1}}} \right)$
As line x + y = 4, divides AB in ratio p ∈ line
$\frac{{5k - 1}}{{k + 1}} + \frac{{7k + 1}}{{k + 1}} = 4$
5k - 1 +7k + 1 = 4k +4
12k = 4k + 4
8k = 4
$k = \frac{1}{2}$ ∴ line divides AB in ratio 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 given lines are
2x – y = 0 ... (1)
4x + 7y + 5 = 0 ... (2)
A (1, 2) is a point on line (1).
Let B be the point of intersection of lines (1) and (2).
On solving equations (1) and (2), we obtain $x = \frac{{ - 5}}{{18}}\; and \; y = \frac{{ - 5}}{9}$
∴ Coordinates of point B are $\left( {\frac{{ - 5}}{{18}},\frac{{ - 5}}{9}} \right)$
By using distance formula, the distance between points A and B can be obtained as
$AB = \sqrt {{{\left( {1 + \frac{5}{{18}}} \right)}^2} + {{\left( {2 + \frac{5}{9}} \right)}^2}} unit$
$AB = \sqrt {{{\left( {\frac{{23}}{{18}}} \right)}^2} + {{\left( {\frac{{23}}{9}} \right)}^2}} unit$
$AB = \sqrt {{{\left( {\frac{{23}}{{2 \times 9 }}} \right)}^2} + {{\left( {\frac{{23}}{9}} \right)}^2}} unit$
$AB = \sqrt {{{\left( {\frac{{23}}{9}} \right)}^2}{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{{23}}{9}} \right)}^2}} unit$
$AB = \frac{{23}}{9}\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + 1} unit$
$AB = \frac{{23}}{9}\sqrt {\frac{5}{4}} unit$
$AB = \frac{{23}}{9} \times \frac{{\sqrt 5 }}{2}unit$ $AB = \frac{{23\sqrt 5 }}{{18}}units$ Thus, the required distance is $\frac{{23\sqrt 5 }}{{18}}$

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

Let y = mx + c be the line through point (–1, 2).
Accordingly, 2 = m (–1) + c.
⇒ 2 = –m + c
⇒ c = m + 2
∴ y = mx + m + 2 … (1)
The given line is
x + y = 4 … (2)
On solving equations (1) and (2), we obtain
$x = \frac{{2 - m}}{{m + 1}} \; and \;y = \frac{{5m + 2}}{{m + 1}}$
$\therefore \; \left( {\frac{{2 - m}}{{m + 1}},\frac{{5m + 2}}{{m + 1}}} \right)$ is the point of intersection of line (1) and (2)
Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,
$\sqrt {{{\left( {\frac{{2 - m}}{{m + 1}} + 1} \right)}^2} + {{\left( {\frac{{5m + 2}}{{m + 1}} - 2} \right)}^2}} = 3$
${\left( {\frac{{2 - m}}{{m + 1}} + 1} \right)^2} + {\left( {\frac{{5m + 2}}{{m + 1}} - 2} \right)^2} = {3^2}$
$\Rightarrow {\left( {\frac{{2 - m + m + 1}}{{m + 1}}} \right)^2} + {\left( {\frac{{5m + 2 - 2m - 2}}{{m + 1}}} \right)^2} = {3^2}$
$\Rightarrow \frac{9}{{{{\left( {m + 1} \right)}^2}}} + \frac{{9{m^2}}}{{{{\left( {m + 1} \right)}^2}}} = 9$
$\Rightarrow \frac{{1 + {m^2}}}{{{{\left( {m + 1} \right)}^2}}} = 1$
1+ m2 = m2+1+2m
⇒ 2m = 0
⇒ m = 0
Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.

Question (17)

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

Solution

Let A(1,3) and B(−4,1) be the coordinates of the end points of the hypotenuse.
Now, plotting the line segment joining the points A(1,3) and B(−4,1) on the coordinate plane, we will get two right triangles with AB as the hypotenuse. Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle having AB as the hypotenuse can be either P or Q.
CASE 1: When ∠APB is taken. The perpendicular sides in ∠ APB are AP and PB. Now, side PB is parallel to x-axis and at a distance of 1 units above x-axis. So, equation of PB is, y=1 or y−1=0. The side AP is parallel to y-axis and at a distance of 1 units on the right of y-axis. So, equation of AP is x=1 or x−1=0.
CASE 2: When ∠AQB is taken. The perpendicular sides in ∠AQB are AQ and QB. Now, side AQ is parallel to x-axis and at a distance of 3 units above x-axis. So, equation of AQ is, y=3 or y−3=0. The side QB is parallel to y-axis and at a distance of 4 units on the left of y-axis. So, equation of QB is x=−4 or x+4=0. Hence, the equation of the legs are : x=1, y=1 or x=−4, y=3

Question (18)

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

Solution

The equation of the given line is x + 3y = 7 .... (1)
Let point B (a, b) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.
Slope of $AB = \frac{{b - 8}}{{a - 3}}$ while the slope of line (1) $= - \frac{1}{3}$
Since line (1) is perpendicular to AB,
$\left( {\frac{{b - 8}}{{a - 3}}} \right) \times \left( { - \frac{1}{3}} \right) = - 1$
$\Rightarrow \frac{{b - 8}}{{3a - 9}} = 1$
→ b - 8 = 3a - 9
→ 3a - b = 1 ...(2)
Mid-point of AB $= \left( {\frac{{a + 3}}{2},\frac{{b + 8}}{2}} \right)$
The mid-point of line segment AB will also satisfy line (1).
Hence, from equation (1), we have
$\frac{{a + 3}}{2} + 3\left( {\frac{{b + 8}}{2}} \right) = 7$
→ a+3+3b+24 = 14
→ a + 3b = -13 ....(3)
On solving equations (2) and (3), we obtain a = –1 and b = –4.
Thus, the image of the given point with respect to the given line is (–1, –4).

Question (19)

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

Solution

The equations of the given lines are
y = 3x + 1 … (1)
2y = x + 3 … (2)
y = mx + 4 … (3)
Slope of line (1), m1 = 3
Slope of line (2), ${m_2} = \frac{1}{2}$
Slope of line (3), m3 = m
It is given that lines (1) and (2) are equally inclined to line (3). This means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).
$\left| {\frac{{{m_1} - {m_3}}}{{1 + {m_1}{m_3}}}} \right| = \left| {\frac{{{m_2} - {m_3}}}{{1 + {m_2}{m_3}}}} \right|$
$\Rightarrow \left| {\frac{{3 - m}}{{1 + 3m}}} \right| = \left| {\frac{{\frac{1}{2} - m}}{{1 + \frac{1}{2}m}}} \right|$
$\Rightarrow \left| {\frac{{3 - m}}{{1 + 3m}}} \right| = \left| {\frac{{1 - 2m}}{{m + 2}}} \right|$
$\Rightarrow \frac{{3 - m}}{{1 + 3m}} = \pm \left( {\frac{{1 - 2m}}{{m + 2}}} \right)$
$\Rightarrow \frac{{3 - m}}{{1 + 3m}} = \frac{{1 - 2m}}{{m + 2}} \;or \; \frac{{3 - m}}{{1 + 3m}} = - \left( {\frac{{1 - 2m}}{{m + 2}}} \right)$
If $\frac{{3 - m}}{{1 + 3m}} = \frac{{1 - 2m}}{{m + 2}}$, then
(3-m)(m+2) = (1-2m)(1+3m)
⇒ -m2 + m + 6 = 1 + m -6m2
⇒ 5m2 + 5 = 0
⇒ (m2 + 1) = 0
$\Rightarrow m = \sqrt { - 1}$, which is not real
Hence, this case is not possible
If $\frac{{3 - m}}{{1 + 3m}} = - \left( {\frac{{1 - 2m}}{{m + 2}}} \right)$, then
⇒ (3 - m) (m +2 ) = - (1-2m) (1+3m)
⇒ -m2 + m + 6 = -(1+m-6m2)
⇒ 7m2 - 2m - 7 = 0
$\Rightarrow m = \frac{{2 \pm \sqrt {4 - 4\left( 7 \right)\left( { - 7} \right)} }}{{2\left( 7 \right)}}$
$\Rightarrow m = \frac{{2 \pm 2\sqrt {1 + 49} }}{{14}}$
$\Rightarrow m = \frac{{1 \pm 5\sqrt 2 }}{7}$
Thus, the required value of m is $\frac{{1 \pm 5\sqrt 2 }}{7}$

Question (20)

If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Solution

The equations of the given lines are
x + y – 5 = 0 ... (1)
3x – 2y + 7 = 0 ... (2)
The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by
${d_1} = \frac{{\left| {x + y - 5} \right|}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }} \; and \; {d_2} = \frac{{\left| {3x - 2y + 7} \right|}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 2} \right)}^2}} }}$
$\text \; {i.e} {d_1} = \frac{{\left| {x + y - 5} \right|}}{{\sqrt 2 }} \;and \; {d_2} = \frac{{\left| {3x - 2y + 7} \right|}}{{\sqrt {13} }}$
It is given that d1 + d2 = 10
$\therefore \frac{{\left| {x + y - 5} \right|}}{{\sqrt 2 }} + \frac{{\left| {3x - 2y + 7} \right|}}{{\sqrt {13} }} = 10$
$\Rightarrow \sqrt {13} \left| {x + y - 5} \right| + \sqrt 2 \left| {3x - 2y + 7} \right| - 10\sqrt {26} = 0$
$\Rightarrow \sqrt {13} \left( {x + y - 5} \right) + \sqrt 2 \left( {3x - 2y + 7} \right) - 10\sqrt {26} = 0$ [Assuming (x+y-5) and (3x -2y +7) are positive]
$\Rightarrow \sqrt {13} x + \sqrt {13} y - 5\sqrt {13} + 3\sqrt 2 x - 2\sqrt 2 y + 7\sqrt 2 - 10\sqrt {26} = 0$
$\Rightarrow x\left( {\sqrt {13} + 3\sqrt 2 } \right) + y\left( {\sqrt {13} - 3\sqrt 2 } \right) + \left( {7\sqrt 2 - 5\sqrt {13} - 10\sqrt {26} } \right) = 0$
which is the equation of a line.
Similarly, we can obtain the equation of line for any signs of (x+y-5) and (3x-2y+7)
Thus, point P must move on a line.

Question (21)

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

Solution

l1 : 9x + 6y – 7 = 0
l2: (3x + 2y + 6 = 0)× 3 ⇒ 9x+6y+18 = 0
l1||l2
Let the equation of line parallel to l1 and l2 respectively
p1 = p2
$\frac{{\left| {k - \left( { - 7} \right)} \right|}}{{\sqrt {{9^2} + {6^2}} }} = \frac{{\left| {k - 18} \right|}}{{\sqrt {{9^2} + {6^2}} }}$
$\frac{{\left| {k + 7} \right|}}{{\sqrt {81 + 36} }} = \frac{{\left| {k - 18} \right|}}{{\sqrt {81 + 36} }}$
|k+7| = |k-18|
k+7 = ± ( k - 18 )
k +7 = k -18
∴ 7 = -18
Which is not possible
∴ k+7 = -( k - 18)
k+7 = -k +18
2k = 11
$k = \frac{{11}}{2}$
∴ The equation of line l will be $9x + 8y + \frac{{11}}{2} = 0$
$18x + 16y + 11 = 0$

Question (22)

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

Solution

Let ray passing through B. incident on x-axis at A and reflected throuhj C(5, 3)
Since A is on x-axis the co-ordinate of A(a, 0)
Draw normal at A it will be parallel to y-axis
Let line $\overleftrightarrow{AB}$ make an angle of θ with +ve direction of x-axis
i.e. ∠ CAD = θ
θ ∠ PAC = 90 -θ
Since angle of incident = angle of reflection
∠BAP = ∠ PAC
∠BAP = 90-θ
∴ ∠BAD = θ + 90 -θ + 90 - θ
∠BAD = 180- θ
Since $\overleftrightarrow{AB}$ make an angle θ
slope of AC = tanθ
∴ slope of AC $\frac{{3 - 0}}{{5 - a}}$
$\frac{{30}}{{5 - a}} = \tan \theta - - - (1)$
A(a, 0), B(1, 2)
Slope of $AB = \frac{{2 - 0}}{{1 - a}} = \frac{2}{{1 - a}}$
Line AB makes an angle of 180-θ with x-axis
Slop of AB = tan(180 - θ)
$\therefore \frac{2}{{1 - a}} = - \tan \theta$
$\tan \theta = - \frac{2}{{1 - a}} - - - (2)$
From (1) and (2)
$\frac{3}{{5 - a}} = \frac{{ - 2}}{{1 - a}}$
3)1 - a) = -2(5-a)
3 - 3a = -10 + 2a
13 = 5a
$a = \frac{{13}}{5}$
So coordinate of A(a, 0) $= \left( {\frac{{13}}{5},0} \right)$

Question (23)

Prove that the product of the lengths of the perpendiculars drawn from the points
$\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ and $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to be line $\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$ is b2

Solution

$l:\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$
l: bcosθ x + a sinθ y - ab = 0
Let p1 ne ⊥ distance of l from $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$
${p_1} = \frac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
${p_1} = \frac{{\left| {b\cos \theta \sqrt {{a^2} - {b^2}} + a\sin \theta \left( 0 \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}$
${p_1} = \frac{{\left| {b\cos \theta \sqrt {{a^2} - {b^2}} - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}$
p2 be ⊥ distance from $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$
${p_2} = \frac{{\left| {b\cos \theta \left( { - \sqrt {{a^2} - {b^2}} } \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}$

p1p2
$= \frac{{\left| {b\cos \theta \left( {\sqrt {{a^2} - {b^2}} } \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }} \times \frac{{\left| {b\cos \theta \left( { - \sqrt {{a^2} - {b^2}} } \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}$
$= {\frac{{{{\left( { - ab} \right)}^2} - \left( {b\cos \theta \sqrt {{a^2} - {b^2}} } \right)}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}^2}$
$= \frac{{{a^2}{b^2} - {b^2}\left( {{a^2} - {b^2}} \right){{\cos }^2}\theta }}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$
$= \frac{{{b^2}\left[ {{a^2} - {a^2}{{\cos }^2}\theta + {b^2}{{\cos }^2}\theta } \right]}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$
$= \frac{{{b^2}\left[ {{a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}{{\cos }^2}\theta } \right]}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$
$= \frac{{{b^2}\left[ {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right]}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }} = {b^2}$

Question (24)

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

Solution

A person is standing at the crossing of lines l1: 2x-3y+4=0 and l2: 3x+4y-5 = 0, l3: 6x-7y+8 = 0 is equation of 3rd road.
We has to each to l3 for that we will walk across l4 which is passess through point point of intersection of l1 and l2 which will be ⊥ to l3
Equation of the which passes through the point of intersection of l1, l2 is given by
l1 + λl2 = 0
(2x - 3y +4) + λ(3x+4y-5) = 0 ----(1)
(3+3λ)x + (-3+4λ)y + 4 - 5λ = 0
∴ slope of line by $= \frac{{ - a}}{b} = - \frac{{\left( {2 + 3\lambda } \right)}}{{\left( { - 3 + 4\lambda } \right)}}$
l3: 6x - 7y + 8 = 0
slope of l3 $= \frac{{ - a}}{b} = \frac{{ - 6}}{{ - 7}} = \frac{6}{7}$
$\overleftrightarrow{l_3}$ ⊥ $\overleftrightarrow{l_4}$
∴ slope of l3 × slope of l4 = -1
$\frac{{\left( {2 + 3\lambda } \right)}}{{\left( { - 3 + 4\lambda } \right)}} \times \frac{6}{7} = 1$
∴ 6(2+3λ) = 7(-3+4λ)
12+18λ = -21 + 28λ
33 = 10λ
$\lambda = \frac{{33}}{{10}}$
Replace value of λ in (1)
$\left( {2x - 3xy + 4} \right) + \frac{{33}}{{10}}\left( {3x + 4y - 5} \right) = 0$
119x + 102y -125 = 0