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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.

Q 1.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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}