# SSC > Straight Lines

Explore popular questions from Straight Lines for SSC. This collection covers Straight Lines previous year SSC questions hand picked by experienced teachers.

General Intelligence and Reasoning
General Awareness
Quantitative Aptitude
English Comprehension
Q 1.

Correct2

Incorrect-0.5

The equation of the straight line joining the origin to the point of intersection of {tex} y-x+7=0\ and\ y+2x-2=0 {/tex}is:

A

{tex} 3x+4y=0 {/tex}

B

{tex} 3x-4y=0 {/tex}

C

{tex} 4x-3y=0 {/tex}

{tex} 4x+3y=0 {/tex}

##### Explanation

The intersection point of {tex} y-x+7=0 {/tex} and {tex} y+2x-2=0 {/tex} is {tex} \left(3,\ -4\right) {/tex}

Therefore, Equation of straight line joining from origin to the point {tex} \left(3,\ -4\right) {/tex} is

{tex} y-0 = \frac{-4}{3} \left(x-0\right) {/tex}

=> {tex} 3y = -4x => 4x+3y=0 {/tex}

Q 2.

Correct2

Incorrect-0.5

The angle between the straight lines {tex} x-y\sqrt{3}=5\ and\ \sqrt{3}x+y=7 {/tex} is:

A

{tex} 90 ^{\circ} {/tex}

{tex} 60 ^{\circ} {/tex}

C

{tex} 75 ^{\circ} {/tex}

D

{tex} 30 ^{\circ} {/tex}

##### Explanation

Given equation is compared with {tex} a_{1}x+b_{1}y=0\ and\ a_{2}x+b_{2}y=0 {/tex}

Now, {tex} a_{1}a_{2}+b_{1}b_{2}= \left(1\right) \left(\sqrt{3}\right)+ \left(-\sqrt{3}\right) \left(1\right)=0 {/tex}

Line are perpendicular

{tex} \theta-90 ^{\circ} {/tex}

Q 3.

Correct2

Incorrect-0.5

The three straight lines {tex} ax+by=c,\ bx+cy=a\ and\ cx+ay=b {/tex} are collinear if:

{tex} a+b+c = 0 {/tex}

B

{tex} c+a=b {/tex}

C

{tex} b+c=a {/tex}

D

{tex} a+b=c {/tex}

##### Explanation

We have {tex} ax+by=c {/tex} ...(i)

{tex} bx+cy=a {/tex} ...(ii)

and {tex} cx+ay=b {/tex} ...(iii)

On adding equation (i), (ii) and (iii), we get

{tex} ax+by+bx+cy+cx+ay = a+b+c {/tex}

=> {tex} \left(a+b+c\right)x+ \left(a+b+c\right)y = \left(a+b+c\right) {/tex}

On comparing with {tex} ox+oy=0 {/tex} (for collinearity)

We get{tex} a+b+c = 0 {/tex}

Q 4.

Correct2

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If a tangent to the curve {tex} y=6x-x^{2} {/tex} is parallel to the line {tex} 4x-2y-1=0 {/tex} then the point of tangency on the curve is: ' _

A

(6, 1)

B

(8, 2)

(2, 8)

D

(4, 2)

##### Explanation

Since, tangent is parallel to {tex} y = 2x-\frac{1}{2} {/tex}

Therefore, Equation of tangent is {tex} y = 2x+h {/tex}

The point of tangency will be the point of intersection of tangent and curve but in the given equation of options only option (a) satisfied the equation of curve then {tex} \left(2,\ 8\right) {/tex} will be the point oftangency.

Q 5.

Correct2

Incorrect-0.5

Distance between the lines {tex} 5x+3y-7=0\ and\ 15x+9y+14=0 {/tex} is:

A

{tex} \frac{35}{\sqrt{34}} {/tex}

B

{tex} \frac{1}{\sqrt{34}} {/tex}

C

{tex} \frac{35}{2\sqrt{34}} {/tex}

{tex} \frac{35}{3\sqrt{34}} {/tex}

##### Explanation

Given equation of lines are

{tex} 5x+3y-7=0 {/tex} ...(i)

and {tex} 15x+9y+14=0\ or\ 5x+3y+\frac{14}{3}=0 {/tex} ...(ii)

Therefore, Lines (i) and (ii) are parallel and {tex} C_{1} {/tex} and {tex} C_{2} {/tex} and of opposite signs, therefore these lines are on opposite sides of the origin, so the distance between them is

{tex} |\frac{C_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}| + |\frac{C_{2}}{\sqrt{a_{1}^{2}|+b_{2}^{2}}}| = |\frac{7}{\sqrt{5^{2}+3^{2}}}| + |\frac{14}{3\sqrt{5^{2}+3^{2}}}|{/tex} {tex} = |-\frac{7}{\sqrt{34}}| + |\frac{14}{3\sqrt{34}}| = \frac{35}{3\sqrt{34}} {/tex}

Q 6.

Correct2

Incorrect-0.5

The equation of the sides of a triangle are {tex} x-3y=0.\ 4x+3y=5\ and\ 3x+y=0 {/tex}. The lines {tex} 3x-4y=0 {/tex} passage through

A

the incentre

B

the centroid

C

the orthocentre

the circumcentre

##### Explanation

Two sides {tex} x-3y=0\ and\ 3x+y=0 {/tex} are perpendicular to each other. Therefore its orthocentre is the point of intersection of {tex} x-3y=0 {/tex} i.e. {tex} \left(0,\ 0\right) {/tex} so, the line {tex} 3x-4y=0 {/tex} passes through the orthocentre of triangle.

Q 7.

Correct2

Incorrect-0.5

If (-4, -5) is one vertex and {tex} 7x-y+8=0 {/tex} is one diagonal of a square then the equation of second diagonal is:

A

{tex} x+3y=21 {/tex}

{tex} 2x-3y=7 {/tex}

C

{tex} x+7y=31 {/tex}

D

{tex} 2x+3y=21 {/tex}

##### Explanation

Equation of perpendicular line to {tex} 7x-y+8=0 {/tex} is {tex} x+7y=h {/tex} which is passes through {tex} \left(-4,\ 5\right) {/tex}

h = 31

So, equation of another diagonal is {tex} x+7y=31 {/tex}

Q 8.

Correct2

Incorrect-0.5

The number of the straight which is equally inclined to both the axes is

4

B

3

C

5

D

1

##### Explanation

There are four possible straight line which are equally inclined in both the axes.

i.e., Ist, IInd, IIIrd and IVth quadrant.

Q 9.

Correct2

Incorrect-0.5

The line passing. through {tex} \left(1-\frac{ \pi }{2}\right) {/tex} and perpendicular to {tex} \sqrt{3} \sin \theta + 2\cos \theta =\frac{4}{r} {/tex}, is:

{tex} 2=\sqrt{3}r\cos \theta-2r\sin \theta {/tex}

B

{tex} 5=2\sqrt{3}r\sin \theta +4r\cos \theta {/tex}

C

{tex} 2=\sqrt{3}r\cos \theta + 2r\sin \theta {/tex}

D

{tex} 5=2\sqrt{3}r\sin \theta + 4r\cos \theta {/tex}

##### Explanation

Any line which is perpendicular to

{tex} \sqrt{3}\sin \theta + 2\cos \theta = \frac{4}{r} {/tex} is

{tex} \sqrt{3}\sin \left(\frac{ \pi }{2}+\theta \right) + 2\cos \left(\frac{ \pi }{2}+\theta \right)

= \frac{k}{r} {/tex} ...(i)

Since, it is passing through {tex} \left(-1,\ \frac{ \pi }{2}\right) {/tex}

{tex} \sqrt{3}\sin \pi + 2\cos \pi = \frac{k}{-1}=k=2 {/tex}

Putting k = 2 in eq. (i) we get

{tex} \sqrt{3}\cos \theta - 2\sin \theta = \frac{2}{r} {/tex}

=> {tex} 2 = \sqrt{3}r \cos \theta - 2r \sin \theta {/tex}

Q 10.

Correct2

Incorrect-0.5

If the straight line {tex} ax+by+c=0 {/tex} always passes through {tex} \left(1-2\right) {/tex} then abc are

in AP

B

in HP

C

in GP

D

none of these

##### Explanation

Since {tex} ax+by+c=0 {/tex} is always passes through {tex} \left(1-2\right) {/tex}

{tex} a-2b+c=0 {/tex}

=> {tex} 2b = a+c {/tex}

Therefore a, b and c are in AP.

Q 11.

Correct2

Incorrect-0.5

The equation of pair of lines joining origin to the points of intersection of {tex} x^{2}+y^{2}=9 {/tex} and {tex} x+y=3 {/tex} is:

A

{tex} x^{2}+ \left(3-x^{2}\right)=9 {/tex}

B

{tex} \left(3+y\right)^{2}+y^{2}=9 {/tex}

{tex} xy=0 {/tex}

D

{tex} \left(x-y\right)^{2}=9 {/tex}

##### Explanation

We have {tex} x^{2}+y^{2}=9 {/tex} ...(i)

and {tex} x+y=3 {/tex} ...(ii)

From equation (i) we get

{tex} x^{2}+y^{2}=3^{2} {/tex}

=> {tex} x^{2}+y^{2}= \left(x+y\right)^{2} {/tex} [using equation. (ii)]

=> {tex} x^{2}+y^{2}=x^{2}+y^{2}+2xy {/tex}

=> {tex} 2xy = 0 {/tex}

=> {tex} xy = 0 {/tex}

Q 12.

Correct2

Incorrect-0.5

If the {tex} \angle \theta {/tex} is acute, then the acute angle between {tex} x^{2} \left(\cos \theta -\sin \theta \right)+2xy \cos \theta +y^{2} \left(\cos \theta +\sin \theta \right)=0 {/tex} is

A

{tex} 2 \theta {/tex}

B

{tex} \frac{ \theta }{3} {/tex}

{tex} \theta {/tex}

D

{tex} \frac{ \theta }{2} {/tex}

##### Explanation

Comparing the given equation, we get

{tex} a=\cos \theta - \sin \theta ,\ b=\cos \theta + \sin \theta ,\ h=\cos \theta {/tex}

{tex} tan\phi =\frac{2\sqrt{h^{2}-ab}}{a+b} {/tex}

=>{tex} \tan \phi = \frac{2\sqrt{\cos^{2} \theta - \left(\cos^{2} \theta - \sin^{2} \theta \right) }}{\cos \theta - \sin \theta +\cos \theta +\sin \theta } = \frac{2\sin \theta }{2\cos \theta } {/tex}

=> {tex} \tan \phi = \tan \theta => \phi = 0 {/tex}

Q 13.

Correct2

Incorrect-0.5

Set of lines {tex} \left(x-2y+1\right)+h \left(x+y\right)=0 {/tex} (where h is a parameter) passing through a fixed point:

A

{tex} \left(\frac{1}{3},\ -\frac{1}{3}\right) {/tex}

{tex} \left(-\frac{1}{3},\ \frac{1}{3}\right) {/tex}

C

(1, 1)

D

none of these

##### Explanation

Set of lines passes through intersection point of

{tex} x-2y+1=0\ and\ x+y=0 {/tex} which is {tex} \left(-\frac{1}{3},\ \frac{1}{3}\right) {/tex}

Q 14.

Correct2

Incorrect-0.5

The equation of the line equidistant from the lines {tex} 2x+3y-5=0\ and\ 4x+6y=11 {/tex} is

A

{tex} 2x+3y-1=0 {/tex}

B

{tex} 4x+6y-1=0 {/tex}

{tex} 8x+12y-1=0 {/tex}

D

none of these

##### Explanation

Since, the lines {tex} 2x+3y+5=0\ and\ 2x+3y-\frac{11}{2}=0 {/tex} are parallel

Let required lines is {tex} 2x+3y+n=0,\ n=\frac{C_{1}+C_{2}}{2} {/tex}

{tex} n=\frac{5-\frac{11}{2}}{2} = \frac{-1}{4} {/tex}

So {tex} 8x+12y-1=0 {/tex} is required line.

Q 15.

Correct2

Incorrect-0.5

The equation of line parallel to lines {tex} L_{1}=x+2y-5=0\ and\ x+2y+9=0 {/tex} and dividing the distance between {tex} L_{1} {/tex} and {tex} L_{2} {/tex} in the ratio 1 : 6 (internally) is:

{tex} x+2y-3=0 {/tex}

B

{tex} x+2y+2=0 {/tex}

C

{tex} x+2y+7=0 {/tex}

D

none of these

##### Explanation

Let line be {tex} x+2y+n=0 {/tex}

{tex} n=\frac{-5 \times 6+1 \times 9}{7}=-3 {/tex} {tex} \ Here \ n = \frac{mc_{2}+nc_{1}}{m+n} {/tex}

So, required line is {tex} x+2y-3=0 {/tex}.

Q 16.

Correct2

Incorrect-0.5

The family of lines making an angle {tex} 30 ^{\circ} {/tex} with the line {tex} \sqrt{3}y=x+1 {/tex} is:

A

{tex} x=h {/tex} (h is parameter)

B

{tex} y=-\sqrt{3}x+h {/tex} (h is parameter)

{tex} y=\sqrt{3}+h {/tex}

D

none of these

##### Explanation

Slope of given line is {tex} \frac{1}{\sqrt{3}} {/tex} it's angle from positive x-axis is {tex} 30 ^{\circ} {/tex}. Now lines making an angle {tex} 30 ^{\circ} {/tex} from it either x-axis (i.e. y = 0) or makes and angle {tex} 60 ^{\circ} {/tex} with positive x-axis (i.e. {tex} y=\sqrt{3}x+n {/tex}

Q 17.

Correct2

Incorrect-0.5

A square of area 25 sq unit is formed by talking two sides as {tex} 3x+4y=k_{1}\ and\ 3x+4y=k_{2} {/tex} then {tex} |k_{1}-k_{2}| {/tex} is:

A

5

B

1

25

D

none of these

##### Explanation

Each side of square is 5 unit, distance between given lines is 5 unit.

i.e., {tex} |\frac{k_{1}-k_{2}}{5}|=5 => |k_{1}-k_{2}|=25 {/tex}

Q 18.

Correct2

Incorrect-0.5

The lines {tex} ax+by+c=0,\ bx+cy+a=0\ and\ cx+ay+b=0 {/tex} {tex} a \ne b \ne c {/tex} are concurrent if:

A

{tex} a^{3}+b^{3}+c^{3}+3abc=0 {/tex}

B

{tex} a^{2}+b^{2}+c^{2}-3abc=0 {/tex}

{tex} a+b+c=0 {/tex}

D

none of these

##### Explanation

Since, the given lines are concurrent
\begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix} = 0 =>{tex} a^{3}+b^{3}+c^{3}-3abc=0 {/tex}

{tex} => \left(a + b + c\right) \left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right) = 0 {/tex}

{tex} => \frac{ \left(a + b + c\right) }{2}\{ \left(a-b\right)^{2} + \left(b - c\right)^{2} + \left(c - a\right)^{2} = 0 \} {/tex}

{tex} => a + b + c = 0 {/tex}

Q 19.

Correct2

Incorrect-0.5

A straight line through the point (2, 2) intersects the lines {tex} \sqrt{3}x+y=0\ and\ \sqrt{3}x-y=0 {/tex} at the points A and B. The equation to the line AB so that the triangle OAB is equilateral is:

A

{tex} x-2=0 {/tex}

{tex} y-2=0 {/tex}

C

{tex} x+y-4=0 {/tex}

D

none of these

##### Explanation

From the given equations, we get

{tex} m^{2}+am+2=0 {/tex}

Since, m is real {tex} a^{2}\ge 8,\ |a| \ge 2\sqrt{2} {/tex}

So least value of {tex} |a|\ is\ 2\sqrt{2} {/tex}

{tex} \sqrt{3}x+y=0 {/tex} makes an angle of {tex} 120 ^{\circ} {/tex} with OX and {tex} \sqrt{3}x-y=0 {/tex} makes an angle {tex} 60 ^{\circ} {/tex} with OX.

So the required line is {tex} y-2=0 {/tex}

Q 20.

Correct2

Incorrect-0.5

Equation to the straight line cutting of an intercept from the negative direction of the axis of y and inclined at {tex} 30 ^{\circ} {/tex} to the positive direction of axis of x is:

A

{tex} y+x-\sqrt{3}=0 {/tex}

B

{tex} y-x+2=0 {/tex}

C

{tex} y-\sqrt{3}x-2=0 {/tex}

{tex} \sqrt{3}y-x+2\sqrt{3}=0 {/tex}

##### Explanation

Let the equation of line is y = mx + C {tex} m = tan 30 ^{\circ} = \frac{1}{\sqrt{3}} {/tex} and c = -2 [Because, It is intercepted in negative axes of y with an angle of {tex} 30 ^{\circ} {/tex}] The required line {tex} y = \frac{x}{\sqrt{3}} - 2 => \sqrt{3}y - x + 2\sqrt{3} = 0 {/tex}

Q 21.

Correct2

Incorrect-0.5

Two consecutive side of parallelogram are {tex} 4x+5y=0\ and\ 7x+2y=0 {/tex}. One diagonal of the parallelogram is {tex} 11x+7y=9 {/tex} the other diagonal is{tex} ax+by+c=0 {/tex}, then

A

{tex} a=-1,\ b=-1,\ c=2 {/tex}

{tex} a=1,\ b=-1,\ c=0 {/tex}

C

{tex} a=-1,\ b=-1,\ c=0 {/tex}

D

{tex} a=1,\ b=1,\ c=1 {/tex}

##### Explanation

Since, the co-ordinates of three vertices A, B and C are {tex} \left(\frac{5}{3},\ -\frac{4}{3}\right),\ \left(0,\ 0\right)\ and\ \left(-\frac{2}{3},\ \frac{7}{3}\right) {/tex} respectively. Also the mid point of AC is {tex} \left(\frac{1}{2},\ \frac{1}{2}\right) {/tex}.

Therefore, the equation of line passing through {tex} \left(\frac{1}{2},\ \frac{1}{2}\right) {/tex} and {tex} \left(0,\ 0\right) {/tex} is given by {tex} x-y=0 {/tex}, which is the required equation of another diagonal. {tex} a=1,\ b=-1\ and\ c=0 {/tex}

Q 22.

Correct2

Incorrect-0.5

The line parallel to the x-axis and passing through the intersection of the lines {tex} ax+2by+3b=0\ and\ bx-2ay-3a=0 {/tex}, where {tex} (a, b) \ne (0, 0) {/tex}, is

A

above the x-axis at a distance of {tex} \frac{2}{3} {/tex} from it.

B

above the x-axis at a distance of {tex} \frac{3}{2} {/tex} from it.

C

below the x-axis at a distance of {tex} \frac{2}{3} {/tex} from it.

below the x-axis at a distance of {tex} \frac{3}{2} {/tex} from it.

##### Explanation

Equation of a line passing through the intersection of lines

{tex} ax+2by+3b=0\ and\ bx-2ay-3a=0 {/tex} is

{tex} \left(ax+2by+3b\right)+n \left(bx-2ay-3a\right)=0 {/tex} ...(i)

Now, this line is parallel to x-axis so cofficient of x = 0

=> {tex} a+nb=0 => n = -\frac{a}{b} {/tex}

On putting this value in equation (i) we get

{tex} b \left(ax+2by+3b\right)-a \left(bx-2ay-3a\right)=0 {/tex}

=> {tex} 2b^{2}y+3b^{2}+2a^{2}y+3a^{2}=0 {/tex}

=> {tex} 2 \left(b^{2}+a^{2}\right)y+3 \left(a^{2}+b^{2}\right) = 0 {/tex}

=> {tex} y = -\frac{3}{2} {/tex}

Q 23.

Correct2

Incorrect-0.5

The number of integral values of m, for which the x-coordinate of the point of intersection of the line {tex} 3x+4y=9\ and\ y=mx+1 {/tex} is also an integer is:

2

B

0

C

4

D

1

##### Explanation

Solving {tex} 3x+4y=9\ and\ y=mx+1 {/tex}, we get

{tex} x=\frac{5}{3+4m} {/tex}

x is an integer

{tex} 3+4m=1,\ -1,\ 5, -5 {/tex}

=> {tex} m=\frac{-2}{4},\ \frac{-4}{4},\ \frac{2}{4},\ \frac{-8}{4} {/tex} so, m has two intergral values.

Q 24.

Correct2

Incorrect-0.5

The arc (in sq unit) of the quadrilateral formed by two pairs of lines {tex} l^{2}x^{2}-m^{2}y^{2}-n \left(lx+my\right)=0\ and\ l^{2}m^{2}-m^{2}y^{2}+n \left(lx-my\right)=0 {/tex} is

{tex} \frac{n^{2}}{2|lm|} {/tex}

B

{tex} \frac{n^{2}}{|lm|} {/tex}

C

{tex} \frac{n}{2|lm|} {/tex}

D

{tex} \frac{n^{2}}{4|lm|} {/tex}

##### Explanation

Given equation can be reduced in product 0 linear equation {tex} \left(lx+my\right) \left(lx-my-n\right)=0\ and\ \left(lx-my\right) \left(lx+my+n\right)=0 {/tex}

and {tex} \left(lx-my\right) \left(lx+my+n\right)=0 {/tex}

=> {tex} lx+my=0,\ lx-my-n=0 {/tex}

and {tex} lx-my=0,\ lx+my+n=0 {/tex}

Area= {tex} |\frac{ \left(c_{1}-d_{1}\right) \left(c_{2}-d_{2}\right) }{ \left(a_{1}b_{2}-a_{2}b_{1}\right) }|=|\frac{ \left(0+n\right) \left(0-n\right) }{ \left(-lm-lm\right) }|=|\frac{n^{2}}{2|lm|}{/tex}

Q 25.

Correct2

Incorrect-0.5

If the pair of lines {tex} ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 {/tex} intersect on the y-axis then:

{tex} 2fgh=bg^{2}+ch^{2} {/tex}

B

{tex} bg^{2} \ne ch^{2}{/tex}

C

{tex} abc = 2fgh {/tex}

D

none of these

##### Explanation

Let {tex} f \left(x,\ y\right)=ax^{2}+by^{2}+2hxy+2gx+2fy+c=0 {/tex} ...(i)

Point of intersection of lines {tex} \frac{af \left(x,\ y\right) }{ax} = 0 {/tex}

=> {tex} 2ax+2hy+2g=0 {/tex}

Since, it intersects on y-axis i.e. x = 0

{tex} y = -\frac{g}{h} {/tex}

Thus, putting this value in equation (i), we get

{tex} \frac{bg^{2}}{h^{2}}+2f \left(-\frac{g}{h}\right)+c=0 {/tex}

=> {tex} bg^{2}+ch^{2}=2fgh {/tex}