NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

Question 2

Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, – 1).

Answer:

Given points are A(5,2) B(2,3) and C(3,-1).
Slope of the line joining points B and C
= $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{3-2}=-\frac{4}{1}=-4$
It is given that the line passing through point A is perpendicular to BC
m₁×m₂= -1
-4×m₂=-1
m₂=$\frac 14$
Equation of a line passing through point A
Equation of line: y-y₁ = m(x-x₁₎
y-2=$\frac 14$ (x-5)
4y-8=x-5
x-5-4y+8=0
Equation of a straight line passing through point A is x-4y+3=0

Question 3

Find the angle between the lines $y=\left ( 2-\sqrt{3} \right )\left ( x+5 \right )$ and $y=\left ( 2+\sqrt{3} \right )\left ( x-7 \right )$

Answer:

Given equations are $y=\left ( 2-\sqrt{3} \right )\left ( x+5 \right )$
$y=\left ( 2-\sqrt{3} \right )x+\left ( 2-\sqrt{3} \right )5$ ................(1)
and $y=\left ( 2+\sqrt{3} \right )\left ( x-7 \right )$
$y=\left ( 2+\sqrt{3} \right ) x-7 \left ( 2+\sqrt{3} \right )$............(2)
In equation 1 the slope is $\left ( 2-\sqrt{3} \right )$as it is in the form of y=mx+b and in equation 2 it is $2+\sqrt{3}$
Let θ be the angle between the given m₁ and m₂ two lines $\tan \theta =\left |\frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |$
Putting the values of m₁ and m₂ in the above equation, we get
$\tan \theta =\left |\frac{2-\sqrt{3}-\left ( 2+\sqrt{3} \right )}{1+\left ( 2-\sqrt{3} \right )\left (2+\sqrt{3} \right )} \right |$
$=\left | -\frac{2\sqrt{3}}{1+\left ( 4-3 \right )} \right |$
$=\left | -\frac{2\sqrt{3}}{2} \right |$
$= \sqrt{3}$
$\theta =\tan ^{-1}\sqrt{3}$
θ=60^°

Question 4

Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

Answer:

Equation of line in intercept form = $\frac{x}{a}+\frac{y}{b}=1$
Given that a+b=14, b=14-a
So equation of line $\frac{x}{a}+\frac{y}{14-a}=1$
$\frac{x\left ( 14-a \right )+ay}{a\left ( 14-a \right )} =1$
$14x-ax+ay=14a-a^{2}$........................(1)
If equation 1 passes through (3,4), then $14 × 3 - a×3+a×4=14a-a^{2}$
42-3a+4a-14a+a²=0
a²-13a+42=0
a²-7a-6a+42=0
(a-6)(a-7)=0
a=6 or a=7
If a=6 then 6+b=14 b=8
If a=7 the 7+b=14 b=7
If a=6 and b=8 then equation of line is $\frac{x}{6}+\frac{y}{8}=1$
4x+3y=24
If a=7 and b=7 then equation of line $\frac{x}{7}+\frac{x}{7}=1$
x+y=7

Question 5

Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10.

Answer:

Let x₁, y₁ be any point lying in the equation x+y=4
x₁+y₁=4…......(1)
Distance of point x₁,y₁from the equation 4x+3y=10
$d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}$
$1=\frac{4x_{1}+3y_{1}-10}{\sqrt{\left ( 4 \right )^{2}+\left ( 3 \right )^{2}}}=\left | \frac{4x_{1}+3y_{1}-10}{5} \right |$
4x₁+3y₁-10=±5
4x₁+3y₁-10=5 or 4x₁+3y₁-10=-5
4x₁+3y₁=(15)...….(2) or 4x₁+3y₁=5….....(3)
From equation 1, we have y₁=4-x₁….. (4)
Putting the values of y₁in equation 2 we get 4x₁+3(4-x₁)=15
4x₁+12-3x₁=15
x₁=15-12 =3
Putting the value of x₁in equation 4, we get y₁=1
Now, 4x1+34-x1=5
4x₁+12-3x₁=5
x₁=5-12 x₁=-7
Putting the value in equation 4, we get y₁=4--7=4+7=11
Hence, the required points on the given line are (3,1) and (-7,11)

Question 6

Show that the tangent of an angle between the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{a}-\frac{y}{b}=1$ is $\frac{2ab}{a^{2}-b^{2}}$

Answer:

Equation of line in intercept form $\frac{x}{a}+\frac{y}{b}=1$............(i) and $\frac{x}{a}-\frac{y}{b}=1$..............(ii)
$\frac{x}{a}+\frac{y}{b}=1$
$\frac{y}{b}=1-\frac{x}{a}$
$y=b-\frac{b}{a}x$
$y=\left (-\frac{b}{a} \right )x+b$
y=mx+b Slope of equation 1 is $m_{1}=-\frac{b}{a}$
Similarly for equation 2 , $-\frac{y}{b}=1-\frac{x}{a}$
$-y=b-\frac{b}{a}x$
$y=\left (\frac{b}{a} \right )x-b$
$y=\left (\frac{b}{a} \right )x+\left (-1 \right )b$
Since the above equation is in y=mx+b form
Slope of the equation 2 is $m_{2}=\frac{b}{a}$
Let θ be the angle between the given two lines $\tan \theta =\left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |$
Putting the values of m₁ and m₂ in the above equation, we get
$\tan \theta =\left | \frac{-\frac{b}{a} -\frac{b}{a}}{1+\left ( -\frac{b}{a} \right )\left ( \frac{b}{a} \right )} \right |=\left | \frac{-2\left (\frac{b}{a} \right )}{1-\left (\frac{b^{2}}{a^{2}} \right )} \right |$
$=\left | -\frac{2ab}{a^{2}-b^{2}} \right |=\frac{2ab}{a^{2}-b^{2}}$

Question 7

Find the equation of lines passing through (1, 2) and making angle 30° with y-axis.

Answer:

Given that the line passes through (1,2), making an angle of 30^° with the y-axis.
Angle made by line with x axis is 60^° Slope of the line, m= $\sqrt{3}$
Equation of line passing through x₁,y₁ and having slope 'm' is y-y₁=m(x-x₁₎
Here, (x₁, y₁)=(1,2) and m= $\sqrt{3}$
$y-2= \sqrt{3}\left ( x-1 \right )$
$y-2= \sqrt{3}x- \sqrt{3}$
$\sqrt{3}x-y-\sqrt{3}+2=0$

Question 8

Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.

Answer:

. Given 2x+y=5…. (1) and x+3y=-8….(2)
Firstly, we find the point of intersection of both equations, we get $y=-\frac{21}{5}$ and $x=\frac{23}{5}$
Hence, the point of intersection is $\left (\frac{23}{5},-\frac{21}{5} \right )$
Now the slope of the equation 3x+4y=7 m=$-\frac{3}{4}$
Then the equation of the line passing through the point $\left (\frac{23}{5},-\frac{21}{5} \right )$ having slope $-\frac{3}{4}$ is

y-y₁=m(x-x₁)
$y-\left ( -\frac{21}{5} \right )=-\frac{3}{4}\left ( x- \frac{23}{5} \right )$
$y+\frac{21}{5} =-\frac{3}{4} x+\frac{69}{20}$
$\frac{3}{4}x+y =\frac{69}{20} -\frac{21}{5}$
$\frac{3x+4y}{4} = -\frac{15}{5}$
$3x+4y+3=0$

Question 9

For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x – 3y + 6 = 0 on the axes.

Answer:

Given equation is ax+by+8=0 or ax+by= -8
Now dividing by-8 to both sides $\frac{a}{-8}x+\frac{b}{-8}y = 1$ $\frac{x}{\left (-\frac{8}{a} \right )}+\frac{y}{\left (-\frac{8}{b} \right )} = 1$
So the intercepts are $-\frac{8}{a}$ and $-\frac{8}{b}$
Now, the second equation, which is given, is 2x-3y+6=0 or 2x-3y=-6
Dividing by-6 on both sides $-\frac{2}{-6}x-\frac{3}{-6}y=1$
$\frac{x}{-3}+\frac{y}{2}=1$
So, the intercepts are -3 and 2
Now, according to the question $-\frac{8}{a}=3$ and $-\frac{8}{b}=-2$
$a=-\frac{8}{3}$ and $b=4$

Question 10

If the intercept of a line between the coordinate axes is divided by the point (–5, 4) in the ratio 1:2, then find the equation of the line.

Answer:

. Let a and b be the intercepts on the given line
Coordinates of A and B are (a,0)and (0,b), respectively.
Using the section formula, we find the value of a and b
$\left ( x,y \right )=\left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right )$
$\left ( -5,4\right )=\left ( \frac{1×0+2×a}{1+2}, \frac{1×b+2×0}{1+2}\right )=\left (\frac{2a}{3},\frac{b}{3} \right )$
$-5=\frac{2a}{3}$ and $4=\frac{b}{3}$
$-15=2a$ and $b=12$
$a=-\frac{15}{2}$ and $b=12$
Coordinates of A and B are $\left ( -\frac{15}{2},0 \right )$ and $\left ( 0 , 12 \right )$

Equation of line AB $y-0 = \frac{12-0}{0-\left ( -\frac{15}{2} \right )}\left ( x- \left ( -\frac{15}{2} \right ) \right )$
$y = \frac{12}{\frac{15}{2}}\left ( x+\frac{15}{2} \right )$
$y =\frac{24}{15}\left ( x+\frac{15}{2} \right )$
$y =\frac{8}{5}x+15$
$5y=8x+60$

Hence, the required equation is 8x-5y+60=0

Question 11

Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of the x-axis.

Answer:

Given that the length of the perpendicular from the origin is 4 units, and the line makes an angle with the positive direction of the x-axis
∠BAX=120°
∠BAO=180°-120°=60°
∠MAO=60°
Now in triangle AMO ∠MAO+∠AOM+∠OMA=180°
60°+θ+ 90°=180°
θ=30° ∠AOM= 30°
x cosθ+ysinθ=p
xcos 30°+ysin30°=4
$x\left ( \frac{\sqrt{3}}{2} \right )+y\left ( \frac{\sqrt{1}}{2} \right )=4$
$\sqrt{3}x+y=8$

Question 12

Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).

Answer:

Equation of hypotenuse is 3x+4y=4, and the opposite vertex is (2,2)
Slope of the equation of hypotenuse is $-\frac{3}{4}$
Now let the slope of AC be m $\tan \theta =\left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |$
Putting the values of m₁and m₂in the equation $\tan 45^{0} =\left | \frac{\left ( m-\left ( -\frac{3}{4} \right ) \right )}{1-\frac{3}{4}m} \right |$
$1=\left | \frac{m+\frac{3}{4}}{1-\frac{3}{4}m} \right |=\left | \frac{4m+3}{4-3m} \right |$
$1=\pm \left | \frac{4m+3}{4-3m} \right |$
$\frac{4m+3}{4-3m} =1$
4m+3=4-3m
4m+3m=4-3
7m=1 m=$\frac{1}{7}$
OR $-\frac{4m+3}{4-3m}=1$
$4m+3=-\left ( 4-3m \right )$
4m+3=-4+3m
4m-3m=-4-3
m= -7
If m=1/7
equation of AC is y-y₁=m(x-x₁)
y-2=1/7(x-2)
7y-14=x-2
x-7y-2+14=0
x-7y+12=0
If m= -7 then equation of AC is y-2=-(7)(x-2)
y-2=-7x+14
7x+y=16
The required equations are x-7y+12=0 and 7x+y=16.

Question 13

If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, – 1), then find the length of the side of the triangle.

Answer:

Let ABC be an equilateral triangle, BC is the base, and the altitude from A on BC meets at the mid-point D.
Given: Equation of the base BC is x+y=2
$\sin 60^{0}=\frac{AD}{AB}$
$\frac{\sqrt{3}}{2}=\frac{AD}{AB}$
$AD=\frac{\sqrt{3}}{2}AB$
Distance of point (x₁,y₁) from the equation Ax+By+C=0
$d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}$
Now, length of perpendicular from vertex A(2,-1) to the line x+y=2
$AD=\frac{\left | 1×2+1\left ( -1 \right ) -1 \right |}{\sqrt{\left ( 1 \right )^{2} +\left ( 1 \right )^{2}}}$
$\frac{\sqrt{3}}{2}AB=\left | \frac{2-1-2}{\sqrt{2}} \right |=\frac{1}{\sqrt{2}}$
Squaring both the sides, we get $\frac{3}{4}AB^{2}=\frac{1}{2}$
$AB^{2}=\frac{4}{3}×\frac{1}{2}=\frac{2}{3}$
$AB=\sqrt{\frac{2}{3}}$

Question 14

A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2, 0), (0, 2) and (1, 1) on the line is zero. Find the coordinates of the point P.

Answer:

Let the variable line be ax+by=1
Length of perpendicular from (2,0) to the line ax+by-1=0
$d=\frac{\left | 2×a+0×b-1 \right |}{\sqrt{a^{2}+b^{2}}}=\frac{2a-1}{\sqrt{a^{2}+b^{2}}}$
Now perpendicular distance from B(0,2) = $\left | \frac{0×a+2×b-1}{\sqrt{a^{2}+b^{2}}} \right |$
Now, perpendicular distance from C(1,1)= $\left | \frac{1×a+1×b-1}{\sqrt{a^{2}+b^{2}}} \right |$
The algebraic sum of the perpendicular from the given points (2,0), (0,2)
and (1,1) to this line is zero.
d₁+d₂+d₃=0
$\frac{2a-1}{\sqrt{a^{2}+b^{2}}}+\frac{2b-1}{\sqrt{a^{2}+b^{2}}}+\frac{a+b-1}{\sqrt{a^{2}+b^{2}}}=0$
2a-1+2b-1+a+b-1=0
3a+3b-3=0
a+b-1=0
a+b=1
So, the equation ax+by=1 represents a family of straight lines passing
through a fixed point.
Comparing equation ax+by=1 and a+b=1, we get x=1 and y=1.
Coordinates of the fixed point are (1,1)

Question 15

In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance √6/3 from the given point.

Answer:

Let the given line x+y=4 and the required line 'l' intersect at B (a,b)
Slope of line'l'is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{b-2}{a-1}$
$m=\tan \theta =\frac{b-2}{a-1}$
Given that $AB=\frac{\sqrt{6}}{3}$
So, by distance formula for point A(1,2) and B(a,b) we get d
$=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$\frac{\sqrt{6}}{3}=\sqrt{\left ( a-1 \right )^{2}+\left (b-2 \right )^{2}}$
$\frac{6}{9}=\left ( a-1 \right )^{2}+\left (b-2 \right )^{2}$
$\frac{2}{3}=a^{2}+1-2a+b^{2}+4-4b$
$2=3a^{2}+3-6a+3b^{2}+12-12b$
$3a^{2}+3b^{2}+3-6a-12b+13=0..........(i))$

Point $B(a, b)$ also satisfies $x+y=4$

$a+b=4 ; b=4-a$
Putting the value of $b$ in equation (i), we get
$3 a^2+3(4-a)^2-6 a-12(4-a)+13=0$

$3 a^2+48+3 a^2-24 a-6 a-48+12 a+13=0$

$6 a^2-18 a+13=0$
Using the formula $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$a=\frac{-\left ( -18 \right )\pm \sqrt{\left ( -18 \right )^{2}-4×6×13}}{2×6}=\frac{18\pm \sqrt{324-312}}{12}$
$=\frac{18\pm \sqrt{12}}{12}=\frac{9\pm \sqrt{3}}{6}$
Putting the value of a in the equation, we get
$b=4-\frac{\left (9\pm \sqrt{3} \right )}{6} =\frac{15\pm \sqrt{3}}{6}$
Now putting the value of a and b in equation $\tan \theta =\frac{b-2}{a-1}$
$=\frac{\frac{\left (15 \pm \sqrt{3}\right )}{6}-2} {\frac{9\pm \sqrt{3} }{6}-1}=\frac{3 \pm \sqrt{3}}{3\pm \sqrt{3}}$
$\tan \theta =\frac{\sqrt{3}+1}{\sqrt{3}-1} or \frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\theta =\tan ^{-1}\left (\frac{\sqrt{3}-1}{\sqrt{3}+1} \right )$
$\theta =\tan ^{-1}\left (\sqrt{3} \right )-\tan ^{-1}\left (1 \right )$
θ= 60°-45°=15°
Similarly, taking $\theta =\tan ^{-1}\left (\frac{\sqrt{3}+1}{\sqrt{3}-1} \right )$
$\theta =\tan ^{-1}\left (\sqrt{3} \right )+\tan ^{-1}\left (1 \right )$
θ= 60°+45°=105°

Question 16

A straight line moves so that the sum of the reciprocals of its intercepts made on the axes is constant. Show that the line passes through a fixed point.

Answer:

Equation of a line in intercept form=xa+yb=1
where a and b are the intercepts on the axes
Given that $\frac{1}{a}+\frac{1}{b}=\frac{1}{k}$
Then, $\frac{k}{a}+\frac{k}{b}=1$
This shows that the line is passing through the fixed point that is (k,k)

Question 17

Find the equation of the line which passes through the point minus 4, 3 and the portion of the line intercepted between the axes is divided internally in the ratio 5 ratio 3 by this point?

Answer:

Let the line cut the $x$-axis at $(a, 0)$ and $Y$ axis at $(0, b)$.
Therefore
It is given that the point $(-4,3)$ divides the line internally in $5: 3$ ratio.
Hence, applying the formula for internal division of a line segment,
we get $\frac{3 \mathrm{a}}{8}=-4$ and $\frac{5 \mathrm{~b}}{8}=3$

Hence $\mathrm{a}=\frac{-32}{3}$ and $\mathrm{b}=\frac{24}{5}$
Using slope-intercept form, we get

$\frac{x}{\frac{-32}{3}}+\frac{y}{\frac{24}{5}}=9 x-20 y-96=0$

Question 18

Find the equations of the lines through the point of intersection of the lines x – y + 1 = 0 and 2x – 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5.

Answer:

Given two lines are: x-y+1=0 and 2x-3y+5=0
Solving these two equations gives us points of intersection, we get y=3 and x=2
(x,y)=(2,3)
Let m be the slope of the required line
Then, the equation of the line is y-3=mx-2
y-3=mx-2m
mx-y-2m+3=0….(1)
Since the perpendicular distance from the point (3, 2) to the line is 75, then
$d=\frac{\left | m×\left ( 3 \right ) -2 +3-2m \right |}{\sqrt{m^{2}+1^{2}}}$
$\frac{7}{5}=\frac{\left | 3m+1-2m \right |}{\sqrt{m^{2}+1^{1}}}=\frac{m+1}{\sqrt{m^{2}+1^{2}}}$
Squaring both the sides, we get $\frac{49}{25}=\frac{\left ( m+1 \right )^{2}}{m^{2}+1}$
49m²+1=25(m+1)²
49m²+49=25m²+25+50m
25m²+25+50m-49m²-49=0
-24m²+50m-24=0
-12m²+25m-12=0
Factorising, we get (3m-4)(4m-3)=0
3m-4=0 or 4m-3=0
3m=4 or 4m=3
m=4/3 or 3/4
Putting the value of m=4/3 in equation 1 we get 4x/3-y-2(4/3)+3=0
4x/3-y-8/3+3=0
4x/3-y=-1/3
4x-3y+1=0
Putting the value of m=3/4 in the equation we get 3x/4-y-2(3/4)+3=0
3x/4-y-3/2+3=0
3x/4-y+3/2=0
3x-4y+6=0
Hence, the required equations are 4x-3y+1=0 and 3x-4y+6=0

Question 19

If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point

Answer:

Let the coordinates of a moving point P be (a,b)
It is given that the sum of the distances from the axes to the point is always 1 |x|+|y|=1
±x±y=1 -x-y=1, x+y=1, -x+y=1 and x-y=1
Hence, these equations give us the locus of the point P, which is a square.

Question 20

P₁, P₂ are points on either of the two lines $y-\sqrt{3}\left | x \right |=2$ at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P₁, P₂ on the bisector of the angle between the given lines.

Answer:

Given lines are $y-\sqrt{3}\left | x \right |=2$
If x≥0, then $y-\sqrt{3} x =2$..............(i)
If x<0, then $y+\sqrt{3}x=2$.............(ii)
On adding both the equations we get $y-\sqrt{3}x +y+\sqrt{3}x=2+2$
2y=4
y=2
Putting the value of y=2 in equation (ii), we get $2+\sqrt{3}x=2$
$\sqrt{3}x=2-2=0$
Point of intersection of given lines is (0,2 )
Now we find the slopes of given lines Slope of equation (i) is $y+\sqrt{3}x=2$
Comparing the above equation with y=mx+b, we get $m=\sqrt{3}$
and we know that $m=\tan \theta =\sqrt{3}$
θ= 60⁰
Slope of equation (ii) is $y=-\sqrt{3}x+2$ , we get $m=-\sqrt{3}$
$\tan \theta =-\sqrt{3}$
θ=180⁰- 60⁰= 120⁰

In ACB,
$\cos 30^{\circ}=\frac{BA}{AC}$
$\frac{\sqrt{3}}{2}=\frac{BA}{5}$
$BA=\frac{5\sqrt{3}}{2}$
$OB=OA+AB=2+\frac{5\sqrt{3}}{2}$
Hence, the coordinates of the foot of perpendicular $=\left (0,2+\frac{5\sqrt{3}}{2} \right )$

Question 21

If p is the length of the perpendicular from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$ and a², p², b² are in A.P, then show that a⁴ + b⁴ = 0.

Answer:

Equation of line in intercept form=$\frac{x}{a}+\frac{y}{b}=1$
Since, p is the length of perpendicular drawn from the origin to the gievn line p= $\left | \frac{\frac{0}{a}+\frac{0}{b}-1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}} \right |$
Squaring both the sides, we have $p= \left | \frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}} \right |$
$\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}........(i)$
Since, a^2,b² and p² are in AP
2p²=a²+b²
$p^{2}=\frac{\left ( a^{2}+b^{2} \right )}{2}$
$\frac{1}{p^{2}}=\frac{2}{a^{2}+b^{2}}..........(ii)$
From equations (i) and (ii), we get $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{2}{a^{2}+b^{2}}$
$\frac{b^{2}+a^{2}}{a^{2}b^{2}}=\frac{2}{a^{2}+b^{2}}$
$\left ( a^{2}+b^{2} \right )\left ( a^{2}+b^{2} \right )=2\left ( a^{2}b^{2} \right )$
$a^{4}+b^{4}=0$
Hence, proved.

Question 22

A line cutting off intercept – 3 from the y-axis and the tangent at angle to the x-axis is 3/5, its equation is
A. 5y – 3x + 15 = 0
B. 3y – 5x + 15 = 0
C. 5y – 3x – 15 = 0
D. None of these

Answer:

. Given that tanθ=3/5
We know that the slope of a line, m=tanθ Slope of line, m=3/5
Since the lines cut off intercepts -3 on the y-axis so the line passes
through the point (0, -3)
So, the equation of a line is y-y₁=m(x-x₁)
y-(-3)=3/5(x-0)
y+3=3x/5
5y+15=3x
5y-3x+15=0
Hence, the correct option is (a)

Question 23

Slope of a line which cuts off intercepts of equal lengths on the axes is
A. – 1
B. – 0
C. 2
D. $\sqrt{3}$

Answer:

Equation of line in intercept form= $\frac{x}{a}+\frac{y}{b}=1$
Given that a=b ; $\frac{x}{a}+\frac{y}{a}=1$
$\frac{x+y}{a} =1$
x+y=a……(1)
y=-x+a
y=(-1)x+a
Since the above equation is in y=mx+b form
So, the slope of the line is (-1).

Question 24

The equation of the straight line passing through the point (3, 2) and perpendicular to the line y = x is
A. x – y = 5
B. x + y = 5
C. x + y = 1
D. x – y = 1

Answer:

Given that a straight line passing through the point (3,2) is perpendicular to
line y=x
Let the equation of line L be (y-y₁) =m(x-x₁)
Since L is passing through the point (3,2)
y-2=m(x-3)..…(i)
Now, given equation is y=x
Comparing it with y=mx+b
We get the slope of the equation as 1.
It is also given that line L and y=x are perpendicular to each other.
We know that when two lines are perpendicular, then m₁×m₂= -1 m×1=-1
m=-1
Putting this value of m in equation (i), we get y-2=-(1)(x-3)
y-2=-x+3
x+y=5
Hence, the correct option is b

Question 25

The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 is
A. y – x + 1 = 0
B. y – x – 1 = 0
C. y – x + 2 = 0
D. y – x – 2 = 0

Answer:

Given that line passing through the point (1,2) and perpendicular to the line
x+y+1=0
Let the equation of line L be x-y+k=0…..(i)
Since L is passing through the point (1,2)
1-2+k=0
k=1
Putting this value of k in the equation, (i) we get x-y+1=0 or y-x-1=0
Hence, the correct option is b

Question 26

The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a, respectively, is
A. $\frac{a^{2}-b^{2}}{ab}$
B. $\frac{b^{2}-a^{2}}{2}$
C. $\frac{b^{2}-a^{2}}{2ab}$
D. None of these

Answer:

First equation of a line in intercept form =
$\frac{x}{a}+\frac{y}{-b}=1$
$\frac{x}{a}-\frac{y}{b}=1$
bx-ay=ab……. (i)
Let the second equation of the line having intercepts on the axes b, -a is $\frac{x}{b}+\frac{y}{-a}=1$
$\frac{x}{b}-\frac{y}{a}=1$
ax-by=ab…..(ii)
Now we find the slope in the first equation bx-ay=ab
ay=bx-ab
$y=\frac{b}{a}x-b$
Slope of the equation $m_{1}=\frac{b}{a}$
Now, we find the slope of equation (ii) ax-by=ab
by=ax-ab
$y=\frac{a}{b}x-a$
Slope of the equation $m_{2}=\frac{a}{b}$
Let θ be the angle between the given two lines $\tan \theta =\left |\frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2} } \right |$
Putting the values of m1 and m2 in above equation, we get $\tan \theta =\left |\frac{\left ( \frac{b}{a}-\frac{a}{b} \right )}{1+\left ( \frac{b}{a} \right )\left (\frac{a}{b} \right )} \right |$
$\tan \theta =\left | \frac{\frac{b^{2}-a^{2}}{ab}}{1+1} \right |=\left | \frac{\left (b^{2}-a^{2} \right )}{2ab} \right |=\left | \frac{b^{2}-a^{2} }{2ab} \right |$ is the required angle.
Hence, the correct option is (c).

Question 27

If the line $\frac{x}{a}+\frac{y}{b}=1$ passes through the points (2, –3) and (4, –5), then (a, b) is
A. (1, 1)
B. (– 1, 1)
C. (1, – 1)
D. (– 1, –1)

Answer:

Given points are (2, -3) and (4, -5)
Firstly, the equation of line is found
We know that the equation of a line when two points are given is y-y₁= $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left ( x-x_{1} \right )$
Putting the values we get $y-\left ( -3 \right )=\frac{-5-\left ( -3 \right )}{4-2}\left ( x-2 \right )$
$y+3=\frac{-5+3 }{2}\left ( x-2 \right )$
$y+3=-\frac{2 }{2}\left ( x-2 \right )$
y+3= -1( x-2)
y+3= -x+2
x+y=2-3
x+y=-1
$\frac{x}{-1}+\frac{y}{-1}=1$ in intercept form
Comparing the equaton with intercept form of equation that is $\frac{x}{a}+\frac{y}{b}=1$
The value of a=-1 and b=-1
Hence, the correct option is (d)

Question 28

The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0 is
A. $\frac{130}{17\sqrt{29}}$
B. $\frac{13}{7\sqrt{29}}$
C. $\frac{130}{7}$
D. None of these

Answer:

(a) Given lines are:

$2 x-3 y+5=0 \text { and } 3 x+4 y=0$

Solving these lines, we get point of intersection as $\left(\frac{-20}{17}, \frac{15}{17}\right)$
Therefore Distance of this point from the line $5 x-2 y=0$

$=\frac{\left|5 \times\left(-\frac{20}{17}\right)-2\left(\frac{15}{17}\right)\right|}{\sqrt{25+4}}=\frac{\left|\frac{-100}{17}-\frac{30}{17}\right|}{\sqrt{29}}=\frac{130}{17 \sqrt{29}}
$

Question 29

The equations of the lines which pass through the point (3, –2) and are inclined at 60° to the line $\sqrt{3}x+y=1$ is
A. y + 2 = 0, $\sqrt{3}x-y-2-3\sqrt{3}=0$
B. x – 2 = 0, $\sqrt{3}x-y+2+3\sqrt{3}=0$
C. $\sqrt{3}x-y-2-3\sqrt{3}=0$
D. None of these

Answer:

Given equation is $\sqrt{3}x+y=1$ and θ= 60⁰
Slope of the equation $\sqrt{3}x+y=1$
$y=1-\sqrt{3}x$
Slope of the equation $m_{1}=-\sqrt{3}$
Let the slope of the required line be m₂
Then $\tan \theta = \left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |$
Putting the values in the above equation we gettan $60^{\circ}=\left | \frac{-\sqrt{3}-m_{2}}{1+\left ( -\sqrt{3} \right )×m_{2}} \right |$
$\sqrt{3}=\left | \frac{-\sqrt{3}-m_{2}}{1+\left ( -\sqrt{3} \right )×m_{2}} \right |$
$\sqrt{3}= \pm\left ( \frac{-\sqrt{3}-m_{2}}{1+\left ( -\sqrt{3} \right )×m_{2}} \right )$
Taking+sign we get $-\sqrt{3}-m_{2}=\sqrt{3}\left ( 1-\sqrt{3}m_{2} \right )$
$-\sqrt{3}-m_{2}=\sqrt{3}-3m_{2}$
$3m_{2}-m_{2}=\sqrt{3}+\sqrt{3}$
$2m_{2}=2\sqrt{3}$
$m_{2}=\sqrt{3}$
Taking -ve sign we get $\sqrt{3}+m_{2}=\sqrt{3}\left ( 1-\sqrt{3}m_{2} \right )$
$\sqrt{3}+m_2=\sqrt{3}-3 m_2$

$3m_{2}+m_{2} =0$
$4m_{2} =0$
m₂=0
Equation of line passing through (3,-2) with slope $\sqrt{3}$ is y-y₁=m(x-x₁ )
$y-\left ( -2 \right )=\sqrt{3}\left ( x-3 \right )$
$y+2=\sqrt{3}x-3\sqrt{3}$
$\sqrt{3}x-y-3\sqrt{3}-2=0$
and equation of line passing through (3, -2) with slope 0 is y-y₁=m(x-x₁)
y-(-2)=0(x-3)
y+2=0
Hence, the required equations are $\sqrt{3}x-y-\left ( 3\sqrt{3}+2 \right )=0$ and y+2=0
Hence, the correct option is (a).

Question 30

The equations of the lines passing through the point (1, 0) and at a distance $\frac{\sqrt{3}}{2}$ from the origin, are
A. $\sqrt{3}x+y-\sqrt{3}=0,\sqrt{3}x-y-\sqrt{3}=0$
B. $\sqrt{3}x+y+\sqrt{3}=0,\sqrt{3}x-y+\sqrt{3}=0$
C. $x+\sqrt{3}y-\sqrt{3}=0,x-\sqrt{3}y-\sqrt{3}=0$
D. None of these.

Answer:

Let the equation of any line passing through the point (1,0) is y-y₁=m(x-x₁)
y-0=m(x-1)
y=mx-m
mx-m-y=0….........(i)
Distance from the origin of the line is $\frac{\sqrt{3}}{2}$ Distance of point (x₁,y₁) from the equation Ax+By+C=0
$d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}$
$\frac{\sqrt{3}}{2}=\frac{\left | m×0+\left ( -1 \right )×0+\left ( -m \right ) \right |}{\sqrt{m^{2}+\left ( -1 \right )^{2}}}=\left |- \frac{m}{\sqrt{m^{2}+1}} \right |$
Squaring both the sides we get $\frac{3}{4}=\frac{m^{2}}{m^{2}+1}$
3(m²+1)=4m²
3m²+3=4m²
4m²-3m²=3
m²=3
$m = \pm \sqrt{3}$
Putting the value of m= $\sqrt{3}$ in equation (i) we get $\sqrt{3}$x-y-$\sqrt{3}$=0
Now, putting the value of m= -$\sqrt{3}$ in the same equation we get-$\sqrt{3}$x-y+$\sqrt{3}$=0
Hence, the correct option is (a)

Question 31

The distance between the lines y = mx + c₁ and y = mx + c₂ is
A. $\frac{c_{1}-c_{2}}{\sqrt{m^{2}+1}}$

B. $\frac{\left |c_{1}-c_{2} \right |}{\sqrt{1+m^{2}}}$

C. $\frac{c_{2}-c_{1} }{\sqrt{1+m^{2}}}$

D. 0

Answer:

Given equations are y=mx+c₁….(i) and y=mx+c₂…….(ii)
Firstly, we find the slope of both equations
Since both of them have the same slope, they are parallel lines.
We know thatthe distance between two parallel lines Ax+By+C₁=0 and Ax+By+C₂=0
$d=\frac{\left | c_{1}-c_{2} \right |}{\sqrt{A^{2}+B^{2}}}=\frac{\left | c_{1}-c_{2} \right |}{\sqrt{m^{2}+\left ( -1 \right )^{2}}}=\frac{\left | c_{1}-c_{2} \right |}{\sqrt{m^{2}+1}}$
Hence, the correct option is b.

Question 32

The coordinates of the foot of perpendiculars from the point (2, 3) on the line y = 3x + 4 is given by
A. $\frac{37}{10},\frac{-1}{10}$
B. $\frac{-1}{10},\frac{37}{10}$
C. $\frac{10}{37},-10$
D. $\frac{2}{3},-\frac{1}{3}$

Answer:

Given equations are y=3x+4….(i)
Comparing this equation with y=mx+b form, the slope of the equation is 3.
Let the equation of any line passing through the point (2,3) is y-y₁=m(x-x₁)
y-3=m(x-2)……(ii)
Given that equation (i) is perpendicular to equation (ii)
And we know that, if two lines are perpendicular then m₁m₂= -1
3×m₂=-1
m₂=-1/3, which is the slope of the required line
Putting the value of slope in equation ii we get y-3=-1/3(x-2)
3y-9=-x+2
x+3y-9-2=0
x+3y-11=0……(iii)
Now we have to find the coordinates of the foot of the perpendicular
Solving equations (i) and (iii), we get x+3(3x+4)-11=0
x+9x+12-11=0
10x+1=0
x=-1/10
Putting the value of x in equation i , we get y=3(-1/10)+4
y= -3/10+4
y=37/10
So the required coordinates are (-1/10,37/10)
Hence, the correct option is (b)

Question 33

If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be
A. 2x + 3y = 12
B. 3x + 2y = 12
C. 4x – 3y = 6
D. 5x – 2y = 10
Answer:

Let the given line meets the axes at A(a,0) and B(0,b)
Given that (3,2) is the midpoint $3=\frac{0+a}{2}$
a=6 and$2=\frac{0+b}{2}$
b=4
Intercept form of the line AB is $\frac{x}{a}+\frac{y}{b}=1$
Putting the value of a and b in the above equation, we get $\frac{x}{6}+\frac{y}{4}=1$
$\frac{2x+3y}{12}=1$
2x+3y=12
Hence, the correct option is (a)

Question 34

Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is
A. y + 2 = x + 1
B. y + 2 = 3 (x + 1)
C. y – 2 = 3 (x – 1)
D. y – 2 = x – 1

Answer:

Given equation of the line is y=3x-1
Now we find the slope of the above equation by comparing it with y=mx+b form
So m=3
Now, we find the equation of the line passing through the point (1,2) and parallel to the given line
with slope=3
y-y₁=m(x-x₁)
y-2=3(x-1)
Hence, the correct option is (c)

Question 35

Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are
A. y = x, y + x = 1
B. y = x, x + y = 2
C. 2y = x, y + x = 1/3
D. y = 2x, y + 2x = 1

Answer:

It is given that the lines x=0, y=0 , x=1 and y=1 form a square of side 1 unit
Let us form a square OABC having corners O(0,0) from the given lines
with A(1,0), B(1,1) and C(0,1)

Now we have to find the equation of the diagonal AC
Equation of a line is found out by $y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left ( x-x_{1} \right )$
$y-0=\frac{1-0}{0-1}\left ( x-1 \right )$
y=-1(x-1)
y=-x+1
x+y=1
Equation of diagonal OB is $y-0=\frac{1-0}{1-0}\left ( x-0 \right )$
y=x
y=x
Hence, the correct option is (a).

Question 36

For specifying a straight line, how many geometrical parameters should be known?
A. 1
B. 2
C. 4
D. 3

Answer:

Equation of straight line in intercept form=$\frac{x}{a}+\frac{y}{b}=1$
where a and b are the intercepts on the axis.
In intercept form, we need 2 parameters, a and b to specify a straight line
Slope-Intercept Form y=mx+c where m=tanθ
and θ is the angle made with the positive x-axis, and c is the intercept on the y-axis
So, we need two parameters, 'm' and 'c' to specify a straight line
Hence, the correct option is b.

Question 37

The point (4, 1) undergoes the following two successive transformations:
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive x-axis. Then the final coordinates of the point are
A. (4, 3)
B. (3, 4)
C. (1, 4)
D. $\frac{7}{2}, \frac{7}{2}$

Answer:

Let Q(x,y) be the reflection of P(4,1) about the line y=x, then
midpoint of PQ $\left ( \frac{4+x}{2},\frac{1+y}{2} \right )$
which lies on y=x $\frac{4+x}{2}=\frac{1+y}{2}$
4+x=1+y
x-y+3=0……(i)
Now, we find the slope of the given equation y=x
Since this equation is in y=mx+b form
So the slope=m=1
Slope of $PQ=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y-1}{x-4}$
Since PQ is perpendicular to y=x
And when two lines rae perpendicular, then m₁m₂=-1
$1×\left ( \frac{y-1}{x-4} \right )=-1$
y-1=-(x-4)
x+y-5=0…..(ii)
On adding equation i and equation ii we get x-y+3+x+y-5=0
2x-2=0
x-1=0
x=1
Putting the value of x=1 in equation i we get 1-y+3=0 -y+4=0 y=4
It is given that translation through a distance of 2 units along the positive x-axis takes place
The point after translaton is (1+2,4)=(3 ,4)
Hence, the correct option is (b).

Question 38

A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is
A. (1, –1)
B. (1, 1)
C. (0, 0)
D. (0, 1)

Answer:

Given equations are 4x+3+10=0 …..(i)
5x-12y+26=0…......(ii)
and 7x+24y-50=0
Let p, q be the points which are equidistant from the given lines
Now, we find the distance of (p,q) from the given lines
Distance of point (x₁,y₁) from the equation Ax+By+C=0
$d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}$
Distance of (p,q) from equation i is $d=\frac{\left | 4p+4q+10 \right |}{\sqrt{4^{2}+3^{2}}}=\frac{\left | 4p+4q+10 \right |}{5}$
Distance of (p,q) from equation ii is $d=\frac{\left | 5p-12q+26 \right |}{\sqrt{5^{2}+\left (-12 \right )^{2}}}=\frac{\left | 5p-12q+26 \right |}{13}$
Distance of (p,q) from equation iii is $d=\frac{\left | 7p-24q+50 \right |}{\sqrt{7^{2}+24^{2}}}=\frac{\left | 7p-24q+50\right |}{25}$
Given that (p,q) is equidistant from the given lines $\frac{\left | 4p-3q+10 \right |}{5}$=$\frac{\left | 5p-12q+26 \right |}{13}=\frac{\left | 7p-24q+50\right |}{25}$
On putting the value of (p,q) as (0,0), we get $\left | \frac{10}{5} \right |=\left | \frac{26}{13} \right |=\left | -\frac{50}{25} \right |$
Hence, the correct option is C.

Question 39

A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is
A. 1/3
B. 2/3
C. 1
D. 4/3

Answer:

Given line is 3x+y=3. It can be rewritten as y=-3x+3
Comparing it with the y=mx+b form of equation, we get the slope as m= -3
So the slope of the perpendicular line will be 1/3
The line passes through (2,2) and has a slope1/3 is $y-2=\frac{1}{3}\left ( x-2 \right )$
3y-6=x-2
3y=x+4
$y=\frac{1}{3}x + \frac{4}{3}$
So, the y intercept is $\frac{4}{3}$
Hence, the correct option is (d).

Question 40

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the lines 3x + 4y + 5 = 0 and 3x + 4y – 5 = 0 is
A. 1 : 2
B. 3 : 7
C. 2 : 3
D. 2 : 5

Answer:

Given lines are 3x+4y+5=0…....(i)
3x+4y-5=0…....(ii)
and 3x+4y+2=0….......(iii)
Since the coefficients of x and y are the same, equations i, ii and iii are parallel to each other
We know that in the case of i and iii
distance between two parallel lines is $d=\frac{\left | c_{1}-c_{2} \right |}{\sqrt{A^{2}+B^{2}}}=\frac{\left | 5-2 \right |}{\sqrt{3^{2}+(4)^{2}}}=\frac{\left | 3 \right |}{\sqrt{9+16}}=\left | \frac{3}{5} \right |$
Similarly in case of ii and iii distance between two parallel lines is $d=\frac{\left | c_{1}-c_{2} \right |}{\sqrt{A^{2}+B^{2}}}=\frac{\left | -5-2 \right |}{\sqrt{3^{2}+(4)^{2}}}=\frac{\left | -7 \right |}{\sqrt{9+16}}=\left |- \frac{7}{5} \right |$
Ratio between the distance is $\frac{3}{5}:\frac{7}{5}=3:7$
Hence, the correct option is b

Question 41

One vertex of the equilateral triangle with centroid at the origin and one side as x + y – 2 = 0 is
A. (–1, –1)
B. (2, 2)
C. (–2, –2)
D. (2, –2)

Answer:

Let ABC be an equilateral triangle with vertex A (a,b)
Let AD be perpendicular to BC, and let (p,q) be the coordinates of D
Given that the centroid P lies at the origin (0,0)
We know that the centroid of a triangle divides the median in the ratio 1:2
Now, using the section formula, we get
$\left ( x,y \right )=\left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right )$
$\left ( 0,0 \right )=\left ( \frac{1×a+2×p}{1+2} , \frac{1×b+2×q}{1+2}\right )$
$\left ( 0,0 \right )=\left ( \frac{a+2p}{3} , \frac{b+2q}{3}\right )$
$\frac{a+2p}{3}=0$ and $\frac{b+2q}{3}=0$
a+2p=0 and b+2q=0…A…
a+2p=b+2q
2p-2q=b-a …i
It is given that BC=x+y-2=0
Since the above equation passes through p,q
p+q-2=0
Now we find the slope of the line AP
$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{b-0}{a-0}=\frac{b}{a}$
Equation of line BC is x+y-2=0
y= -x+2
y=-(1)x+2
Since the above equation is in y=mx+b form
So, the slope of line BC is m_BC=-1
Since both lines rae perpendicular $\frac{b}{a}×\left ( -1 \right )=-1$
b=a
Putting this value in equation (i )we get 2p-q=b-b=0
p=q
Now putting this value in equation (i) we get p+q-2=0
2p=2
p=1
q=1
Putting the value of p and q in equation A, we get a+2×1=0 and b+2×1=0
a=-2 and b=-2
So, the coordinates of vertex A (a,b) is (-2,-2)
Hence, the correct option is c

Question 42

Fill in the blanks
If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through ____.
Answer:

Given that a,b,c are in AP $b=\frac{a+c}{2}$
2b=a+c
a-2b+c=0….i
Now comparing equation i with the given equation ax+by+c=0 we get x=1, y=-2
So, the line will pass through (1,-2 ).

Question 43

Fill in the blanks
The line which cuts off equal intercept from the axes and pass through the point (1, –2) is ____.

Answer:

Equation of straight line in intercept form=$\frac{x}{a}+\frac{y}{b}=1$ where a and b are intercepts
Given that a=b $\frac{x}{a}+\frac{y}{a}=1$
$\frac{x+y}{a}=1$
x+y=a….(i)
If equation ipasses through the point (1, -2) we get 1+(-2)=a
1-2=a
a=-1
Putting the value of a in equation i, we get x+y=-1
x+y+1=0

Question 44

Fill in the blanks
Equations of the lines through the point (3, 2) and making an angle of 45° with the line x – 2y = 3 are ____.

Answer:

Given equation is x-2y=3
x-3=2y
$y=\frac{1}{2}x+\left ( -\frac{3}{2} \right ).................(i)$
The slope of the equation can be found by comparing with y=mx+b form
So, $m_{1}=\frac{1}{2}$
We have to find an equation which is passing through the point (3,2)
A line passing through the point x₁,y₁ has an equation y-y₁=m(x-x₁)
So, here x₁=3 and y₁=2
y-2=m(x-3)....…(ii)
Now, it is given that the angle between the given two lines is 45⁰
$\tan \theta =\left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |$
Putting the values of m₁and m₂ in above equation we get tan45^{0 = $\left | \frac{ m-\frac{1}{2} }{1+m×\frac{1}{2}} \right |$}
$1=\left | \frac{ 2m-1 }{2+m} \right |$
$1=\pm \frac{ 2m-1 }{2+m}$
2m-1=2+m or-(2m-1)=2+m
2m-m=2+1 or -2m+1-m=2
m=3 or-3m=1 or m=-1/3
Putting the value of m=3 in equation (ii) we get y-2=3(x-3)
y-2=3x-9
3x-y-9+2=0
3x-y-7=0
Putting the value of m=-1/3 in equation (ii) we get y-2= -1/3(x-3)
3(y-2)=3-x
3y-6=3-x
x+3y-6-3=0
x+3y-9=0

Question 45

Fill in the blanks
The points (3, 4) and (2, – 6) are situated on the ____ of the line 3x – 4y – 8 = 0.

Answer:

Given line 3x-4y-8=0 and given points are (3,4) and (2,-6)
For point (3,4)
3(3)-4(4)-8=9-16-8
=9-24= -15<0
Forpoint (2, -6) 3(2)-4(-6)-8 =6+24-8
=30-8=22>0
So the points (3,4)and (2,-6) are situated on the opposite sides of 3x-4y-8=0.

Question 46

Fill in the blanks
A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is ____.

Answer:

Given point is (3, -2) and equation of line is 5x-12y=3
Let (p,q) be any moving point
Distance between them (p,q) and (3,-2) $d_{1}=\sqrt{\left ( p-3 \right )^{3}+\left ( q-\left ( -2 \right ) \right )^{2}}$
(d₁)²=(p-3)²+(q+2)²
Now, the distance of the point (p,q) from the given line 5x-12y-3=0
$d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}$
$d_{2}=\frac{\left | 5p-12q-3 \right |}{\sqrt{(5)^{2}+(12)^{2}}}= \frac{\left | 5p-12q-3 \right |}{\sqrt{25+144}}=\frac{\left | 5p-12q-3 \right |}{\sqrt{169}}$
= $=\frac{\left | 5p-12q-3 \right |}{13}$
Taking numerical values only, we have (p-3)²+(q-2)²=$=\frac{5p-12q-3 }{13}$
13[(p-3)²+(q+2)²]=5p-12q-3
13[p²+9-6p+q²+4+4q]=5p-12q-3
On solving, we get 13p²+13q²-83p+64q+172=0
A point moves so that the square of its distance from the point (3,-2) is numerically equal to its distance from the line 5x-12y=3.
The equation of its locus is 13p²+13q²-83p+64q+172=0

Question 47

Fill in the blanks
Locus of the mid-points of the portion of the line x sin θ + y cos θ = p intercepted between the axes is ____.

Answer:

Given equation of the line is x sinθ+y cosθ=p…..(i)
Let Ph,kbe the midpoint of the given line where it meets the two axes at a,0and 0,b.
Since (a,0) lies on equation (i), then asinθ+0=p a= $\frac{p}{\sin \theta }$…..(ii)
(0,b) also lies on the equation i then 0+bcosθ=p
$b=\frac{p}{\cos \theta }$……(iii)
Since P (h,k) is the midpoint of the given line $h=\frac{a+0}{2}=\frac{a}{2}$
2h=a and $k=\frac{0+b}{2}=\frac{b}{2}$
2k=b
Putting the value of a=2h in equation (ii) we get $2h=\frac{p}{\sin \theta }$
$\sin \theta =\frac{p}{2k }$…..(iv)
Putting the value of b=2k in equation (ii) we get $2k =\frac{p}{\cos \theta }$
$\cos \theta =\frac{p}{2k }$….(v)
Squaring and adding equations (iv) and (v), we get $\sin^{2}\theta + \cos^{2}\theta=\left ( \frac{p}{2h} \right )^{2}+\left ( \frac{p}{2k} \right )^{2}$
$1= \frac{p^{2}}{4h^{2}}+ \frac{p^{2}}{4k^{2}}$ or $1= \frac{p^{2}}{4x^{2}}+ \frac{p^{2}}{4y^{2}}$
or 4x²y²=p²y²+p²x²
or 4x²y²=p²(x²+y²) is the locus of the mid-points of the portion of the line intercepted between the axes.

Question 48

State whether the statements are true or false.
If the vertices of a triangle have integral coordinates, then the triangle can’t be equilateral.

Answer:

Let ABC be a triangle with vertices A(x₁,y₁), B (x₂,y₂) and C (x₃, y₃) where x_i, y_i, i=1,2,3 are integers
The area of $ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2}-y_{3} \right ) +x_{2}\left ( y_{3}-y_{1} \right ) +x_{3}\left ( y_{1}-y_{2} \right ) \right ]$
Since xi and yi are all integers, but $\frac{1}{2}$ is a rational number.
So, the result comes out to be a rational number. i.e . Area of ABC=a rational number
Suppose, ABC be an equilateral triangle, then the area of ABC is=$\frac{\sqrt{3}}{4}\left ( AB \right )^{2}$
It is given that the vertices are integral coordinates, it means the value of the coordinates is in whole
number. Therefore, the value of (AB)² is also an integer.
$\frac{\sqrt{3}}{4}$ (positive integer)
But, $\sqrt{3}$ is an irrational number
Area of triangle ABC=an ir-rational number
This contradicts the fact that the area is a rational number
Hence, the given statement is true.

Question 49

State whether the statements are true or false.
The points A (– 2, 1), B (0, 5), C (– 1, 2) are collinear.

Answer:

Given points are A(-2,1), B (0,5) and C(-1,2)
There are two ways to find whether the given points are collinear or not.
The first is that if 3 points are collinear, then the slope of any two pairs of points will be equal.
Second way is that if the value of the area of the triangle formed by the 3 points is zero, then the points
are collinear.
Slope of AB $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-1}{0-\left ( -2 \right )}=2$
Slope of BC $m==\frac{2-5}{-1-0}=\frac{3}{-1}=3$
Slope of CA is$m==\frac{1-2}{-2-\left ( -1 \right )}=\left (-\frac{1}{-2+1} \right )=1$
Since the slopes are different.
So, the given points are not collinear
Hence, the given statement is False.

Question 50

State whether the statements are true or false.
Equation of the line passing through the point (a cos3θ, a sin3θ) and perpendicular to the line x sec θ + y cosec θ = a is x cos θ – y sin θ = a sin 2θ.

Answer:

Let the equation of the line y=mx+c….(i)
So, the slope of the above equation is m
Given equation of line is x secθ+y cosecθ=a sin 2θ.
$\frac{\sec \theta x }{cosec \theta }+y=-\frac{a \sin 2 \theta}{cosec \theta }$
Since the slope of the equation is m'=$y=-\frac{\sec \theta }{cosec \theta }$
Given that equation (i) is perpendicular to x secθ+y cosec θ=a sin 2θ
m×m'= -1
$m×\left (-\frac{\sec \theta }{cosec \theta } \right )=-1$
$m=\frac{cosec \theta }{\sec \theta }$
Putting the value of m in equation (i), we get $y=\frac{cosec \theta }{\sec \theta }x+c$
$y=\frac{cosec \theta +c\left ( sec \theta \right ) }{\sec \theta }$
ysecθ=x cosec θ+c secθ
x cosec θ-y secθ=k….(ii)
If equation (ii) passes through the point (a cos³θ, a sin³θ)
(a cos³θ )cosecθ-(a sin³θ) secθ=k
$\frac{a\cos ^{3} \theta}{\sin \theta}- \frac{a\sin ^{3} \theta}{\cos \theta}=x \, cosec\, \, \theta -y\sec \theta$
$\frac{a\cos ^{4} \theta -a\sin ^{4}\theta}{\sin \theta \cos \theta}=\frac{x}{\sin \theta }-\frac{x}{\cos \theta }$
a[(cos²θ-sin²θ)(cos²θ+sin²θ)]=x cosθ-ysinθ
a[cos²θ-sin²θ]=xcosθ-ysinθ
a[cos2θ]=xcos θ-y sinθ
The given statement is FALSE.

Question 51

State whether the statements are true or false.
The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y – 10 = 0 and 2x + y + 5 = 0.

Answer:

Given equations are x+2y-10=0….(i) and 2x+y+5=0…..(ii)
The point of intersection is obtained by solving them together, i.e. $\left ( -\frac{20}{3} ,\frac{25}{3}\right )$
If the given line 5x+4y=0 passes through the point $\left ( -\frac{20}{3} ,\frac{25}{3}\right )$then $5\left ( -\frac{20}{3} \right )+4\left ( \frac{25}{3} \right )=0$
$-\frac{100}{3}+ \frac{100}{3} =0$
0=0
So, the given line passes through the point of intersection of the given lines.
Hence, the given statement is True.

Question 52

State whether the statements are true or false.
The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are $y-3=\left ( 2\pm \sqrt{3} \right )\left ( x-2 \right )$.

Answer:

Let ABC be an equilateral triangle with vertex (2,3), and the equation of the opposite side is x+y=2
In the case of an equilateral triangle, θ=60
Let the slope of line AB is m, and the slope of the given equation x+y=2 is m₂=-1
We know that $\tan \theta =\left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |$
Putting the values of m₁ and m₂ in the above equation, we get $\tan 60^{\circ}=\left | \frac{\left ( m-\left ( -1 \right ) \right )}{1+m\left ( -1 \right )} \right |$
$\sqrt{3}=\left | \frac{m+1}{1-m}\right |$
$\sqrt{3}=\pm \left (\frac{m+1}{1-m} \right )$
$\sqrt{3}= \left (\frac{m+1}{1-m} \right ) 0r -\left (\frac{m+1}{1-m} \right )$
$\left ( 1-m \right )\sqrt{3}=1+m \, \, or \, \, \left ( 1-m \right )\left (-\sqrt{3} \right )=-1-m$
$\sqrt{3}-\sqrt{3}m=1+m \, \, or\, \, -\sqrt{3}+\sqrt{3}m=-1-m$
$\sqrt{3}-1=m \left ( 1+\sqrt{3} \right )\, \, or\, \, -\left (\sqrt{3}-1 \right )=-\left ( \sqrt{3} +1\right )$
$m=\frac{\sqrt{3}+1}{\sqrt{3}-1}...........(i)\, \, \, or\, \, m=\frac{\sqrt{3}-1}{\sqrt{3}+1}...........(ii)$
Rationalizing both the equations we get$m=2+ \sqrt{3} \, \, \, or\, \, m=2- \sqrt{3}$
So, the slope of line AB is $2\pm + \sqrt{3}$
Thus, the equations of the other two lines joining the point (2, 3) are $y-3=2\pm \sqrt{3}\left ( x-2 \right )$
Hence, the given statement is True.

Question 53

State whether the statements are true or false.
The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

Answer:

Given two lines are 4x+y-1=0…(i ) and 7x-3y-35=0…..(ii)
Now, the point of intersection of these lines can be found as x=2 and y=-7
To find the equation of the line joining the points (3,5) and (2,-7)
$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left ( x-x_{1} \right )$
$y-5=\frac{-7-5}{2-3}\left ( x-3 \right )$
$y-=-\frac{12}{-1}\left ( x-3 \right )$
y-5=12x-3
y-5=12x-36
12x-y-31=0……iv
Now, the distance of equation (iv) from the point (0,0) is
d =$\frac{\left |Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}$
$d=\frac{\left |12(8)-34-31 \right |}{\sqrt{12^{2}+(-1)}}$

$d=\frac{\left | 31\right |}{\sqrt {145}}$
Now the distance of
equation iv from the point (8,34) is $\frac{\left |Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}}$
$d=\frac{\left |12(8)-34-31 \right |}{\sqrt{12^{2}+(-1)}}=\frac{\left | 31\right |}{\sqrt {145}}$
Hence, the equation of line 12x-y-31=0 is equidistant from (0,0)and (8,34)
Hence, the given statement is True.

Question 54

State whether the statements are true or false.

The line $\frac{x}{a}+\frac{y}{b}=1$ moves in such a way that $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}$, where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^{2}+y^{2}=c^{2}$

Answer:

Equation of line $\frac{x}{a}+\frac{y}{b}=1$……i
Equation of a line passing through the origin and perpendicular to the given line $\frac{x}{a}-\frac{y}{b}=0$….ii
Now the foot of a perpendicular from the origin on the line (i) is the point of intersection of lines (i) and ii
So, to find its locus, we have to eliminate the variables a and b
Squaring and adding both the equations we get$\left (\frac{x}{a}+\frac{y}{b} \right )^{2}+\left (\frac{x}{a}-\frac{y}{b} \right )^{2}=1+0$
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{2xy}{ab}+\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{2xy}{ab}=1$
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$x^{2}\left (\frac{1}{a^{2}}+\frac{1}{b^{2}} \right )+y^{2}\left (\frac{1}{a^{2}}+\frac{1}{b^{2}} \right )=1$
$\left (x^{2}+y^{2} \right )+\left (\frac{1}{a^{2}}+\frac{1}{b^{2}} \right )=1$
$\left (x^{2}+y^{2} \right )+\left (\frac{1}{c^{2}} \right )=1$
x²+y²=c²
Hence, the given statement is true.

Question 55

State whether the statements are true or false.
The lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent if a, b, c are in G.P.

Answer:

Given that ax+2y+1=0 bx+3y+1=0, and cx+4y+1=0 are concurrent.
For the lines to be concurrent $\begin{vmatrix} a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1 \end{vmatrix}=1$
Now expanding along the first column, we get
a[3-4]-b[2-4]+c[2-3]=0
-a+2b-c=0
2b=a+c and we know that if a, b ,c are in AP then$b=\frac{a+c}{2}$
This means given lines are in AP not in GP
Hence, the given statement is False.

Question 56

State whether the statements are true or false.
Line joining the points (3, – 4) and (– 2, 6) is perpendicular to the line joining the points (–3, 6) and (9, –18).

Answer:

Given points are (3,-4), (-2,6), (-3,6) and (9,-18)
Now we find the slope since the lines are perpendicular, the product of the slopes is -1 i.e. m₁m₂= -1
Slope of the line joining the points (3,-4) and (-2,6)
$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Here, x₁=3, x₂=-2 , y₁=-4 and y₂=6
$m_{1}=\frac{6-\left ( -4 \right )}{-2-3}=\frac{6+4}{-5}=\frac{10}{-5}=-2$
Now, slope of the line joining the points -(3,6) and (9,-18)
$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Here, x₁=-3, x₂=9, y₁=6 and y₂=-18
$m_{2}=\frac{-18-6}{9-(-3)}=\frac{24}{9+3}=-\frac{24}{12}=-2$
m₁=m₂=-2
and m₁m₂= -2×(-2)=-4 ≠-1
So, the lines are parallel and not perpendicular
Hence, the given statement is False

Question 57

Match the questions given under Column C1 with their appropriate answers given under the Column C2.

Answer:

Let P (x₁,y₁) be any point on the given line x+5y=13
x₁+5y₁=13
5y₁=13-x₁….(i)
Distance of the point P(x₁,y₁)from the equation 12x-5y+26=0
$d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B ^{2}}}$
$2=\frac{\left | 12x_{1}+5y_{1}+26 \right |}{\sqrt{\left ( 12 \right )^{2}+\left ( -5 \right )^{2}}}$
$2=\frac{\left | 12x_{1}-\left ( 13-x_{1} \right )+26 \right |}{\sqrt{144+25}} = \frac{\left | 12x_{1}-13+x_{1} +26 \right |}{13}$
$2=\frac{\left | 13x_{1}+13 \right |}{13}$
2=|x₁+1|
2= ±(x₁+1)
So x₁=1….(ii) or x₁=-3…. (iii)
Putting the value in equation (i) we get 5y₁=13-1=12
$y_{1}=\frac{12}{5}$
Putting the value of x₁=-3 in the same equation we get 5y₁=13-(-3)=16
$y_{1}=\frac{16}{5}$
Hence, the required points on the given line are$\left ( 1,\frac{12}{5} \right ) \, \, and\, \, \left ( -3,\frac{16}{5} \right )$and -3,165
Hence, (a)-(iii)
Let P x₁, y₁ be any point lying in the equation x+y=4
x₁+y₁=4…. i
Now, the distance of the point from the equation is $d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+ B ^{2}}}$
$1=\frac{\left | 4x_{1}+3y_{1}-10 \right |}{\sqrt{(4)^{2}+(3)2}}=\frac{\left | 4x_{1}+3y_{1}-10 \right |}{\sqrt{16-9}}$
$1=\left |\frac{ 4x_{1}+3y_{1}-10 }{5}\right |$
4x₁+3y₁-10= ±5
either 4x₁+3y₁-10=5 or 4x₁+3y₁-10=-5
4x₁+3y₁=15 ….(ii) or 4x₁+3y₁=5…..(iii)
From equation i, we have y₁=4-x₁….(iv)
Putting the value of y₁ in equation II we get 4x₁+34-x₁=15
4x₁+12-3x₁=15
x₁=3 Putting the value of x₁ in equation (iv) we get y₁=4-3=1
Putting the value of y₁ in equation iii, we get 4x₁+34-x₁=5
4x₁+12-3x₁=5
x₁=5-12=-7
Putting the value of x₁ in equation (iv) , we get y₁=4-(-7)
y₁=4+7=11
Hence, the required points on the given line are (3,1) and (-7,11)
Hence, b-i
c Given that AP=PQ=QB and given points are A(-2,5) and B(3,1)
Firstly, we find the slope of the line joining the points (-2,5) and (3,1)
Slope of line joining two points=$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m_{AB}=\frac{1-5}{3-\left ( -2 \right )}=-\frac{4}{3+2}=-\frac{4}{5}$
Now equation of line passing through the point (-2, 5) y-5=-4/5[x-(-2)]
5y-25=-4(x+2)
4x+5y-17=0 Let P (x_1,y₁) and Q (x₂,y₂) be any two points on the AB
P(x₁,y₁) divides the line AB in the ratio 1:2
$x_{1}=\frac{1×3+2×(-2)}{1+2}=\frac{3-4}{3}=\frac{1}{3}$
$y_{1}=\frac{1×1+2×5}{1+2}=\frac{1+10}{3}=\frac{11}{3}$
Now, Q (x₂,y₂) is the midpoint of PB $x_{2}=\frac{3+\left (-\frac{1}{3} \right )}{2}=\frac{8}{6}=\frac{4}{3}$
$y_{2}=\frac{1+{11}{3}}{2}=\frac{3+11}{6}=\frac{14}{6}=\frac{7}{3}$
Hence, the coordinates of Q (x₂,y₂) is (4/3,7/3)
Hence, c-ii.

Question 58

The value of the λ, if the lines (2x + 3y + 4) + λ (6x – y + 12) = 0 are

Answer:

a ) Given equation is (2x+3y+4)+ λ(6x-y+12)=0
2x+3y+4+6λx-λy+12λ=0
(2+6λ)x+(3-λ)y+4+12λ=0…. (i)
If equation i is parallel to the y-axis, then 3-λ=0
λ=3
Hence, iv.
b) Given equation is 2x+3y+4+λ6x-y+12=0
2x+3y+4+6λx-λy+12λ=0
(3-λ)y= -4-12λ-(2+6λ)x
$y=-\left (\frac{2+6\lambda }{3-\lambda } \right )x+\left ( -1 \right )\left (\frac{4+12\lambda }{3-\lambda } \right )$
Since the above equation is in y=mx+b form
So the slope of equation (i) is $m_{1}=-\left (\frac{2+6\lambda }{3-\lambda } \right )$
Now the second equation is 7x+y-4=0…..(ii)
y=-7x+4
So, the slope of equation (ii) is m₂=-7
Now equation (i) is perpendicular to equation (ii)
m₁m₂=-1
$-\left (\frac{2+6\lambda }{3-\lambda } \right )×\left ( -7 \right )=-1$
(2+6λ)×7=-(3-λ)
On solving we get λ=-17/41
Hence, (b)-(iii)
c) Given equation is (2x+3y+4)+λ(6x-y+12)=0
If the above equation passes through the point (1,2), then [2×1+3×2+4]+λ[6×1-2+12]=0
2+6+4+λ(6+10)=0
12+16λ=0
12=-16λ
λ=-12/16=-3/4
Hence, (c)-(i)
d) Given equation is (2x+3y+4)+λ(6x-y+12)=0
2x+3y+4+6λx-λy+12λ=0
(2+6λ)x+(3-λ)y+4+12λ=0…..(i)
If equation (i) is parallel to the x-axis, then 2+6λ=0 6λ=-2
λ=-1/3
(d)-(ii)

Question 59

The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and

Answer:

.a) Given equations are 2x-3y=0….(i) and 4x-5y=2….(ii )
Equation of line passing through equations i and (ii), we get 2x-3y+λ4x-5y-2=0….(iii)
If the above equation passes through the point (2,1), we get (2×2-3×1)+λ(4×2-5×1-2)=0
(4-3)+λ(8-5-2)=0
1+λ=0
λ=-1
Putting the value of λ in equation iii, we get (2x-3y)+(-1)(4x-5y-2)=0
2x-3y-4x+5y+2=0
-2x+2y+2=0
x-y-1=0
Hence, (a)-(iii)
b) Given equations are 2x-3y=0….(i) and 4x-5y=2….(ii)
Equation of line passing through equations i and ii, we get 2x-3y+λ4x-5y-2=0…(iii)
2x-3y+4λx-5λy-2λ=0
x(2+4λ)-y(3+5λ)-2λ=0
-y(3+5λ)= -(2+4λ)+2λ
$y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )+\left ( -\frac{2}{3+5\lambda } \right )$
Slope of equation iii is m₁=$\left ( \frac{2+4\lambda }{3+5\lambda } \right )$
Now, we find the slope of the given line x+2y+1=0…..(iv)
2y=-x-1 y=-1/2x+(-1/2)
So the slope of equation (iv) is m₂=(-1/2)
We know that, if two lines are perpendicular to each other, then the product of their slopes is equal to -1
So, m₁×m₂=-1
$\left ( \frac{2+4\lambda }{3+5\lambda } \right )×\left (-\frac{1}{2} \right )=-1$
2+4λ=2×(3+5λ)
2+4λ=6+10λ
-6λ=4
λ=-4/6= -2/3
Putting the value of λ in equation (iii), we get (2x-3y)+(-2/3)(4x-5y-2)=0
6x-9y-8x+10y+4=0
-2x+y+4=0
2x-y=4
b-i
c) Given equations are 2x-3y=0….(i) and 4x-5y=2…..(ii)
Equation of line passing through equations i and (ii), we get (2x-3y)+ λ(4x-5y-2)=0….iii
2x-3y+4λx-5λy-2λ=0
x(2+4λ)-y(3+5λ)-2λ=0
$y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )x+\left ( -\frac{2}{3+5\lambda } \right )$
So slope of equation iii is m₁= $\left ( \frac{2+4\lambda }{3+5\lambda } \right )$
Now, we find the slope of the given line 3x-4y+5=0
3x+5=4y y=3/4x+5/4
So slope of the equation is 3/4
If 2 lines are parallel, then their slopes are also equal.
So $\left ( \frac{2+4\lambda }{3+5\lambda } \right )=\frac{3}{4}$
4(2+4λ)=3(3+5λ)
8+16λ=9+15λ
λ=1
Putting the values of λ in equation iii, we get (2x-3y)+(4x-5y-2)=0
2x-3y+4x-5y-2=0
6x-8y-2=0
3x-4y-1=0
Hence, c-iv
d) Given equations are 2x-3y=0…(i) and 4x-5y=2….(ii)
Equation of line passing through equations (i) and (ii), we get (2x-3y)+λ(4x-5y-2)=0
x(2+4λ)-y(3+5λ)-2λ=0
-y(3+5λ)= -(2+4λ)+2λ
$y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )x+\left (\frac{-2 }{3+5\lambda } \right )$
Slope of equation iii is $\left ( \frac{2+4\lambda }{3+5\lambda } \right )$
Since the equation is equally inclined with the axes, it means that the line makes equal angles with both the
coordinate axes.
It will make an angle of 45⁰ or 135⁰
m₂=tan45⁰ andtan135⁰
= 1 and-1
$y=\left ( \frac{2+4\lambda }{3+5\lambda } \right )=-1 \, \, or\, \, \left (\frac{2+4\lambda }{3+5\lambda } \right )=1$
2+4λ=-3+5λ or 2+4λ=3+5λ
So λ=-5/9 or-1
Putting the value in equation iii we get (2x-3y)+(-5/9)(4x-5y-)2=0
18x-27y-20x+25y+10=0
-2x-2y+10=0
x+y-5=0
If λ=-1, then the required equation is (2x-3y)+(-1)( 4x-5y-2)=0
2x-3y-4x+5y+2=0
-2x+2y+2=0
x-y-1=0
d-ii, iii