NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 10

Question:1

Find the equation of the straight line which passes through the point (1, – 2) and cuts off equal intercepts from axes.

Answer:

Equation of line in intercept form= $\frac{x}{a}+\frac{y}{b}=1$
Given that $a=b;\frac{x}{a}+\frac{y}{b}=1$
$\frac{x+y}{a}=1$
x+y=a……(1)
If equation (1) passes through (1,-2) we get 1+(-2) =a
1-2=a
a=-1 Putting the value of'a'in equation 1we get x+y=-1
Equation of straight line is x+y+1=0

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).
Slopeof 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 line passing through the point A is perpendicular to BC
m₁*m₂= -1
-4*m₂=-1
m₂=1/4
Equation of line passing through point A
Equation of line: y-y₁ = m(x-x₁₎
y-2=1/4 (x-5)
4y-8=x-5
x-5-4y+8=0
Equation of straight line passing through point A is x-4y+3=0

Question:3

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

Answer:

Given equations are $y=(2-\sqrt{3})(x+5)$
$y=(2-\sqrt{3})x+(2-\sqrt{3})5$ ................(1)
and $y=(2+\sqrt{3})(x-7)$
$y=(2+\sqrt{3})x-7(2+\sqrt{3})$............(2)
In equation 1 the slope is $(2-\sqrt{3})$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 $\tan \theta =\left|\frac{(m_1-m_2)}{1+m_1{m}_2}\right|$
Putting the values of m₁ and m₂ in above equation we get
$\tan \theta =\left|\frac{2-\sqrt{3}-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\right|$
$=|-\frac{2\sqrt{3}}{1+(4-3)}|$
$=|-\frac{2\sqrt{3}}{2}|$
$=\sqrt{3}$
$\theta ={\tan }^{-1}\sqrt{3}$
θ=60⁰

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 $\frac{x}{a}+\frac{y}{14-a}=1$
$\frac{x(14-a)+ay}{a(14-a)}=1$
$14x-ax+ay=14a-a^2$........................(1)
If equation 1 passes through (3,4) then $14\ast 3-a\ast 3+a\ast 4=14a-a^2$
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 inthe equation x+y=4
x₁+y₁=4…......(1)
Distance of point x₁,y₁from the equation 4x+3y=10
$d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$
$1=\frac{4x_1+3y_1-10}{\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}}=\left|\frac{4x_1+3y_1-10}{5}\right|$
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)
$\frac{x}{a}+\frac{y}{b}=1$
$\frac{y}{b}=1-\frac{x}{a}$
$y=b-\frac{b}{a}x$
$y=(-\frac{b}{a})x+b$
y=mx+b Slope of equation 1 is $m_1=-\frac{b}{a}$
Similarly for equation 2 , $-\frac{y}{b}=1-\frac{x}{a}$
$-y=b-\frac{b}{a}x$
$y=\left(\frac{b}{a}\right)x-b$
$y=\left(\frac{b}{a}\right)x+(-1)b$
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 $\tan \theta =\left|\frac{(m_1-m_2)}{1+m_1{m}_2}\right|$
Putting the values of m₁ and m₂ in above equation we get
$\tan \theta =\left|\frac{-\frac{b}{a}-\frac{b}{a}}{1+(-\frac{b}{a})\left(\frac{b}{a}\right)}\right|=\left|\frac{-2\left(\frac{b}{a}\right)}{1-\left(\frac{b^2}{a^2}\right)}\right|$
$=|-\frac{2ab}{a^2-b^2}|=\frac{2ab}{a^2-b^2}$

Question:7

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

Answer:

Given that line passes through (1,2) making an angle of 30⁰ with y axis.
Angle made by line with x axis is 60⁰ 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= $\sqrt{3}$
$y-2=\sqrt{3}(x-1)$
$y-2=\sqrt{3}x-\sqrt{3}$
$\sqrt{3}x-y-\sqrt{3}+2=0$

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 $(\frac{23}{5},-\frac{21}{5})$
Now the slope of the equation 3x+4y=7 m=$-\frac{3}{4}$
Then the equation of the line passing through the point $(\frac{23}{5},-\frac{21}{5})$ having slope $-\frac{3}{4}$ is

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

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 $\frac{a}{-8}x+\frac{b}{-8}y=1$ $\frac{x}{(-\frac{8}{a})}+\frac{y}{(-\frac{8}{b})}=1$
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) respectiively.
Using the section formula we find the value of a and b
$(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
$(-5,4)=(\frac{1\ast 0+2\ast a}{1+2},\frac{1\ast b+2\ast 0}{1+2})=(\frac{2a}{3},\frac{b}{3})$
$-5=\frac{2a}{3}$ and $4=\frac{b}{3}$
$-15=2a$ and $b=12$
$a=-\frac{15}{2}$ and $b=12$
Coordinates of A and B are $(-\frac{15}{2},0)$ and $(0,12)$

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

Hence, trequired 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 x-axis.

Answer:

Given that length of the perpendicular from the origin is 4 units and line makes an angle with positive direction of x-axis
∠BAX=120
∠BA0=180-120=60
∠MAO=60
Now in triangle AMO ∠MAO+∠AOM+∠OMA=180⁰
60⁰+θ+ 90⁰=180⁰
θ=30⁰ ∠AOM= 30⁰
x cosθ+ysinθ=p
xcos 30⁰+ysin30⁰=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 opposite vertex is (2,2)
Slope of the equation of hypotenuse is $-\frac{3}{4}$
Now let the slope of AC be m $\tan \theta =\left|\frac{(m_1-m_2)}{1+m_1{m}_2}\right|$
Putting the values of m₁and m₂in the equation $\tan {45}^0=\left|\frac{(m-(-\frac{3}{4}))}{1-\frac{3}{4}m}\right|$
$1=\left|\frac{m+\frac{3}{4}}{1-\frac{3}{4}m}\right|=\left|\frac{4m+3}{4-3m}\right|$
$1=\pm \left|\frac{4m+3}{4-3m}\right|$
$\frac{4m+3}{4-3m}=1$
4m+3=4-3m
4m+3m=4-3
7m=1 m=$\frac{1}{7}$
OR $-\frac{4m+3}{4-3m}=1$
$4m+3=-(4-3m)$
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 base, and altitude from A on BC meets at 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
$d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$
Now, length of perpendicular from vertex A(2,-1) to the line x+y=2
$AD=\frac{|1\ast 2+1(-1)-1|}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}}$
$\frac{\sqrt{3}}{2}AB=\left|\frac{2-1-2}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$
Squaring both the sides, we get $\frac{3}{4}A{B}^2=\frac{1}{2}$
$A{B}^2=\frac{4}{3}\ast \frac{1}{2}=\frac{2}{3}$
$AB=\sqrt{\frac{2}{3}}$

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
$d=\frac{|2\ast a+0\ast b-1|}{\sqrt{a^2+b^2}}=\frac{2a-1}{\sqrt{a^2+b^2}}$
Now perpendicular distance from B(0,2) = $\left|\frac{0\ast a+2\ast b-1}{\sqrt{a^2+b^2}}\right|$
Now, perpendicular distance from C(1,1)= $\left|\frac{1\ast a+1\ast b-1}{\sqrt{a^2+b^2}}\right|$
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 fixed point is (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
$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$\frac{\sqrt{6}}{3}=\sqrt{{(a-1)}^2+{(b-2)}^2}$
$\frac{6}{9}={(a-1)}^2+{(b-2)}^2$
$\frac{2}{3}=a^2+1-2a+b^2+4-4b$
$2=3a^2+3-6a+3b^2+12-12b$
$3a^2+3b^2+3-6a-12b+13=0..........(i))$
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 3a²+3(4-a)²-6a-12(4-a)+13=0
3a²+48+3a²-24a-6a-48+12a+13=0
6a²-18a+13=0
Using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$a=\frac{-(-18)\pm \sqrt{{(-18)}^2-4\ast 6\ast 13}}{2\ast 6}=\frac{18\pm \sqrt{324-312}}{12}$
$=\frac{18\pm \sqrt{12}}{12}=\frac{9\pm \sqrt{3}}{6}$
Putting the value of a in the equation we get
$b=4-\frac{(9\pm \sqrt{3})}{6}=\frac{15\pm \sqrt{3}}{6}$

Now putting the value of a and b in equation $\tan \theta =\frac{b-2}{a-1}$
$=\frac{\frac{(15\pm \sqrt{3})}{6}-2}{\frac{9\pm \sqrt{3}}{6}-1}=\frac{3\pm \sqrt{3}}{3\pm \sqrt{3}}$
$\tan \theta =\frac{\sqrt{3}+1}{\sqrt{3}-1}or\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\theta ={\tan }^{-1}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$
$\theta ={\tan }^{-1}\left(\sqrt{3}\right)-{\tan }^{-1}\left(1\right)$
θ= 60⁰-45⁰=15⁰
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= 60⁰+45⁰=105⁰

Question:16

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

Answer:

Equation of line in intercept form=xa+yb=1
where a and b are 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 line segment, we get and

Hence Using slope intercept form we get
$$\begin{gathered} \frac{x}{\frac{-32}{3}}+\frac{y}{\frac{24}{5}}=1 \\ 9x-20y+96=0 \end{gathered}$$

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, 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 3,2to the line is75then
$d=\frac{|m\ast \left(3\right)-2+3-2m|}{\sqrt{m^2+1^2}}$
$\frac{7}{5}=\frac{|3m+1-2m|}{\sqrt{m^2+1^1}}=\frac{m+1}{\sqrt{m^2+1^2}}$
Squaring both the sides, we get $\frac{49}{25}=\frac{{(m+1)}^2}{m^2+1}$
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 equation 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 distance 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 gives 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 $=(0,2+\frac{5\sqrt{3}}{2})$

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
$p^2=\frac{(a^2+b^2)}{2}$
$\frac{1}{p^2}=\frac{2}{a^2+b^2}..........(ii)$
From equation (i) and (ii) we get $\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{a^2+b^2}$
$\frac{b^2+a^2}{a^2{b}^2}=\frac{2}{a^2+b^2}$
$(a^2+b^2)(a^2+b^2)=2\left(a^2{b}^2\right)$
$a^4+b^4=0$
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 slope of a line, m=tanθ Slope of line, m=3/5
Since, the lines cut off intercepts -3 on y axis so the line is passing
through the point (0,-3)
So, the equation of 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 straight line passing through the point (3,2) and is perpendicular to
line y=x
Let the equation of line L is (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 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 lies 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 is 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 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 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 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 $\tan \theta =\left|\frac{(m_1-m_2)}{1+m_1{m}_2}\right|$
Putting the values of m1 and m2 in above equation, we get $\tan \theta =\left|\frac{(\frac{b}{a}-\frac{a}{b})}{1+\left(\frac{b}{a}\right)\left(\frac{a}{b}\right)}\right|$
$\tan \theta =\left|\frac{\frac{b^2-a^2}{ab}}{1+1}\right|=\left|\frac{(b^2-a^2)}{2ab}\right|=\left|\frac{b^2-a^2}{2ab}\right|$ 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 line when two points are given is y-y₁= $\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
Putting the values we get $y-(-3)=\frac{-5-(-3)}{4-2}(x-2)$
$y+3=\frac{-5+3}{2}(x-2)$
$y+3=-\frac{2}{2}(x-2)$
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:
and

Solving these lines, we get point of intersection as $(\frac{-20}{17},\frac{15}{17})$
therefore Distance of this point from the line

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 slope of the required line be m₂
Then $\tan \theta =\left|\frac{(m_1-m_2)}{1+m_1{m}_2}\right|$
Putting the values in the above equation we gettan ${60}^{\circ }=\left|\frac{-\sqrt{3}-m_2}{1+(-\sqrt{3})\ast m_2}\right|$
$\sqrt{3}=\left|\frac{-\sqrt{3}-m_2}{1+(-\sqrt{3})\ast m_2}\right|$
$\sqrt{3}=\pm \left(\frac{-\sqrt{3}-m_2}{1+(-\sqrt{3})\ast m_2}\right)$
Taking+sign we get $-\sqrt{3}-m_2=\sqrt{3}(1-\sqrt{3}{m}_2)$
$-\sqrt{3}-m_2=\sqrt{3}-3m_2$
$3m_2-m_2=\sqrt{3}+\sqrt{3}$
$2m_2=2\sqrt{3}$
$m_2=\sqrt{3}$
Taking -ve sign we get $\sqrt{3}+m_2=\sqrt{3}(1-\sqrt{3}{m}_2)$

$3m_2+m_2=0$
$4m_2=0$
m₂=0
Equation of line passing through (3,-2) with slope $\sqrt{3}$ is y-y₁=m(x-x₁ )
$y-(-2)=\sqrt{3}(x-3)$
$y+2=\sqrt{3}x-3\sqrt{3}$
$\sqrt{3}x-y-3\sqrt{3}-2=0$
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 $\sqrt{3}x-y-(3\sqrt{3}+2)=0$ 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
$d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$
$\frac{\sqrt{3}}{2}=\frac{|m\ast 0+(-1)\ast 0+(-m)|}{\sqrt{m^2+{(-1)}^2}}=|-\frac{m}{\sqrt{m^2+1}}|$
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. $\frac{|c_1-c_2|}{\sqrt{1+m^2}}$

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 the equations
Since, both of them have the same slope they are parallel lines.
We know that distance between two parallel lines Ax+By+C₁=0 and Ax+By+C₂=0
$d=\frac{|c_1-c_2|}{\sqrt{A^2+B^2}}=\frac{|c_1-c_2|}{\sqrt{m^2+{(-1)}^2}}=\frac{|c_1-c_2|}{\sqrt{m^2+1}}$
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 foot of the perpendicular
Solving equation (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 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 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 $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$
$y-0=\frac{1-0}{0-1}(x-1)$
y=-1(x-1)
y=-x+1
x+y=1
Equation of diagonal OB is $y-0=\frac{1-0}{1-0}(x-0)$
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 positive x axis and c is the intercept on 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 $(\frac{4+x}{2},\frac{1+y}{2})$
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 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
$1\ast \left(\frac{y-1}{x-4}\right)=-1$
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,qbe the point which is equidistant from the givenlines
Now, we find the distance of (p,q) from the given lines
Distance of point (x₁,y₁) from the equation Ax+By+C=0
$d=\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$
Distance of (p,q) from equation i is $d=\frac{|4p+4q+10|}{\sqrt{4^2+3^2}}=\frac{|4p+4q+10|}{5}$
Distance of (p,q) from equation ii is $d=\frac{|5p-12q+26|}{\sqrt{5^2+{(-12)}^2}}=\frac{|5p-12q+26|}{13}$
Distance of (p,q) from equation iii is $d=\frac{|7p-24q+50|}{\sqrt{7^2+{24}^2}}=\frac{|7p-24q+50|}{25}$
Given that (p,q) is equidistant from the given lines $\frac{|4p-3q+10|}{5}$=$\frac{|5p-12q+26|}{13}=\frac{|7p-24q+50|}{25}$
On putting the value of (p,q) as (0,0) we get $\left|\frac{10}{5}\right|=\left|\frac{26}{13}\right|=|-\frac{50}{25}|$
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 re written as y=-3x+3
Comparing it with the y=mx+b form of equation we get the slope as m= -3
So slope of the perpendicular line will be1/3
The line passes through (2,2) and has a slope1/3 is $y-2=\frac{1}{3}(x-2)$
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 coefficient of x and y are same equation i, ii and iii are parallel to each other
We know that in case of i and iii
distance between two parallel lines is $d=\frac{|c_1-c_2|}{\sqrt{A^2+B^2}}=\frac{|5-2|}{\sqrt{3^2+(4{)}^2}}=\frac{\left|3\right|}{\sqrt{9+16}}=\left|\frac{3}{5}\right|$
Similarly in case of ii and iii distance between two parallel lines is $d=\frac{|c_1-c_2|}{\sqrt{A^2+B^2}}=\frac{|-5-2|}{\sqrt{3^2+(4{)}^2}}=\frac{|-7|}{\sqrt{9+16}}=|-\frac{7}{5}|$
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
$(x,y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
$(0,0)=(\frac{1\ast a+2\ast p}{1+2},\frac{1\ast b+2\ast q}{1+2})$
$(0,0)=(\frac{a+2p}{3},\frac{b+2q}{3})$
$\frac{a+2p}{3}=0$ 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 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, slope of line BC is m_BC=-1
Since both lines rae perpendicular $\frac{b}{a}\ast (-1)=-1$
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