Mathematics Class XI

Question 1 of 24

1. Question
1 point(s)
Find the equation of the bisector of two lines 3x – 4y = 9 & 4x + 3y = 7
- 7x+y=16
- 7x-y=16
- 7x-2y=16
- 7x+2y=16
Correct

Incorrect

Hint

${L_1}:3x - 4y = 9$
${L_2}:4x + 3y = 7$
Equation of bisector
$\dfrac{{3x - 4y - 9}}{{\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} }} = \pm \dfrac{{4x + 3y - 7}}{{\sqrt {{4^2} + {3^2}} }}$
$\dfrac{{3x - 4y - 9}}{5} = \pm \dfrac{{4x + 3y - 7}}{5}$
(Taking negative signs)
$\Rightarrow 3x - 4y - 9 = -(4x+3y-7)$
$\Rightarrow 3x-4y-9+4x+3y-7=0$
$\Rightarrow 7x-y-16=0$
$\Rightarrow 7x-y=16$
Question 2 of 24

2. Question
1 point(s)
The line passes through (2,2) & is perpendicular to the line 3x+y=3, its y-intercept is
- $\dfrac{1}{3}$
- $\dfrac{2}{3}$
- 1
- $\dfrac{4}{3}$
Correct

Incorrect

Hint

Given, y=3-3x, slope= -3
slope of the required line= $\dfrac {1}{3}$
$\therefore$ equation of the required line is
$y - 2 = \dfrac{1}{3}(x - 2)$
$\Rightarrow 3y - 6 = x - 2$
$\Rightarrow x - 3y + 4 = 0$
for y-intercept put x=0
$\Rightarrow - 3y + 4 = 0$
$\Rightarrow y = \dfrac{4}{3}$
Question 3 of 24

3. Question
1 point(s)
The ratio in which the line 3x+4y+2=0 divides the distance between the lines 3x+4y+5=0 & 3x+4y-5=0 is
- $1:2$
- $2:7$
- $2:3$
- $7:3$
Correct

Incorrect

Hint

Perpendicular distance from A to 3x+4y+2=0
$\Rightarrow \dfrac{{\left| {3{x_1} + 4{y_1} + 2} \right|}}{{\sqrt {9 + 16} }} = \dfrac{{\left| { - 5 - 2} \right|}}{{\sqrt {25} }} = \dfrac{7}{5}$
Perpendicular distance from B to 3x+4y+2=0
$\Rightarrow \dfrac{{\left| {3{x_2} + 4{y_2} + 2} \right|}}{{\sqrt {9 + 16} }} = \dfrac{{\left| {5 - 2} \right|}}{{\sqrt {25} }} = \dfrac{3}{5}$
$\therefore$ Required ratio = $\dfrac{7}{5}:\dfrac{3}{5}$
=7:3
Question 4 of 24

4. Question
1 point(s)
The co-ordinates of the foot of perpendiculars from the point (2,3) on the line y=3x+4 is
- $\left( {\dfrac{{37}}{{10}},\dfrac{{ – 1}}{{10}}} \right)$
- $\left( {\dfrac{{ – 1}}{{10}},\dfrac{{37}}{{10}}} \right)$
- $\left( {\dfrac{{10}}{{37}}, – 10} \right)$
- none of these
Correct

Incorrect

Hint

Given, y=3x+4, slope $\left( {{m_1}} \right) = 3$
slope of OP $\left( {{m_2}} \right) = - \dfrac{1}{3}$
Equation of OP is
$y - 3 = - \dfrac{1}{3}\left( {x - 2} \right) \Rightarrow x + 3y - 11 = 0 - \left( i \right)$ using the value of y from given $e{q^n}\left( i \right)$
x+3(3x+4)-11=0
$\Rightarrow x + 9x + 12 - 11 = 0 \Rightarrow 10x + 1 = 0 \Rightarrow x = - \dfrac{1}{{10}}$
$\Rightarrow y = \dfrac{{37}}{{10}}$
$\therefore \left( {x,y} \right) = \left( { - \dfrac{1}{{10}},\dfrac{{37}}{{10}}} \right)$
Question 5 of 24

5. Question
1 point(s)
If a line is at a distance K and parallel to x-axis, then the equation of the line is
- y=k
- y=2k
- y=k+1
- y=k-1
Correct

Incorrect

Hint

Equation of the line is y=k
Question 6 of 24

6. Question
1 point(s)
If a line is parallel to y-axis and at a distance C, then the equation of line is
- x=2c
- x=-c
- x=c+1
- x=-c+1
Correct

Incorrect

Hint

L || Y-axis, AB=C
$\therefore$ Equation of L is
x=-c
Question 7 of 24

7. Question
1 point(s)
In $\triangle ABC$ vertices are A(3,6), B(7,9) & C(3,9) & a,b,c are the length of $\overline {BC}, \overline {AC} \& \overline {AB} $ respectively, find the co-ordinate of excentre to A.
- (-6,12)
- (6,-12)
- (-6,-12)
- (6,12)
Correct

Incorrect

Hint

$a = \left| {\overline {BC} } \right| = \sqrt {{4^2} + {0^2}} = 4$
$b = AC = \sqrt {{0^2} + {3^2}} = 3$
$c = AB = \sqrt {{4^2} + {3^2}} = 5$
$I_1$ is the excentre to A.
Then ${I_1} = \left( {\dfrac{{ - a{x_1} + b{x_2} + c{x_3}}}{{ - a + b + c}},\dfrac{{ - a{x_1} + b{x_2} + c{x_3}}}{{ - a + b + c}}} \right)$
$= \left( {\dfrac{{ - 4 \times 3 + 3 \times 7 + 5 \times 3}}{{ - 4 + 3 + 5}},\dfrac{{ - 4 \times 6 + 3 \times 9 + 5 \times 9}}{{ - 4 + 3 + 5}}} \right)$
$= \left( {\dfrac{{ - 12 + 21 + 15}}{4},\dfrac{{ - 24 + 27 + 45}}{4}} \right)$
=(6,12)
Question 8 of 24

8. Question
1 point(s)
In $\triangle ABC$ A(3,6), B(7,9) & C(3,9) are vertices & a,b,c represents the length of $\overline {BC,} \overline {AC} \& \overline {AB} $ respectively, find the co-ordinate of excentre to B.
- (1,7)
- (-1,7)
- (1,-7)
- (-1,-7)
Correct

Incorrect

Hint

Given A(3,6), B(7,9) & C(3,9)
are the vertices of $\triangle ABC$
Here a=4, b=3, c=5
$\threfore$ co-ordinate of excentre OB $(I_C)$
$= \left( {\dfrac{{a{x_1} - b{x_2} + c{x_3}}}{{a - b + c}},\dfrac{{a{x_1} - b{x_2} + c{x_3}}}{{a - b + c}}} \right)$
$= \left( {\dfrac{{4 \times 3 - 3 \times 7 + 5 \times 3}}{{4 - 3 + 5}},\dfrac{{4 \times 6 - 3 \times 9 + 5 \times 9}}{{4 - 3 + 5}}} \right)$
$= \left( {\dfrac{{12 - 21 + 15}}{6},\dfrac{{24 - 27 + 45}}{6}} \right)$
=(1,7)
Question 9 of 24

9. Question
1 point(s)
In $\triangle ABC$ A(3,6), B(7,9) & C(3,9) are vertices & a,b,c represents the length of $\overline {BC,} \overline {AC} \& \overline {AB} $ respectively, find the co-ordinate of excentre to C.
- (9,-3)
- (9,3)
- (-9,3)
- (-9,-3)
Correct

Incorrect

Hint

Here a=4, b=3, c=5
$\therefore$ co-ordinate of excentre to C (I_3)
$= \left( {\dfrac{{a{x_1} + b{x_2} - c{x_3}}}{{a + b - c}},\dfrac{{a{x_1} + b{x_2} - c{x_3}}}{{a + b - c}}} \right)$
$= \left( {\dfrac{{12 + 21 - 15}}{{4 + 3 - 5}},\dfrac{{24 + 27 - 45}}{{4 + 3 - 5}}} \right)$
$= \left( {\dfrac{{18}}{2},\dfrac{6}{2}} \right) = \left( {9,3} \right)$
Question 10 of 24

10. Question
1 point(s)
A(3,6), B(7,9) & C(3,9) are vertices in $\triangle ABC$ And a,b,c are length of $\overline {BC} ,\overline {AC} \& \overline {AB} $ respectively. Find the co-ordinates of incentre of $\triangle ABC$:
- (-4,8)
- (-4,-8)
- (4,8)
- (4,-8)
Correct

Incorrect

Hint

Let I be the incentre in $\triangle ABC$
Here a=4, b=3, c=5
$\therefore I = \left( {\dfrac{{a{x_1} + b{x_2} + c{x_3}}}{{a + b + c}},\dfrac{{a{y_1} + b{y_2} + c{y_3}}}{{a + b + c}}} \right)$
$= \left( {\dfrac{{4 \times 3 + 3 \times 7 + 5 \times 3}}{{4 + 3 + 5}},\dfrac{{4 \times 6 + 3 \times 9 + 5 \times 9}}{{4 + 3 + 5}}} \right)$
$= \left( {\dfrac{{12 + 21 + 15}}{{12}},\dfrac{{24 + 27 + 45}}{{12}}} \right)$
$= \left( {\dfrac{{48}}{{12}},\dfrac{{96}}{{12}}} \right) = \left( {4,8} \right)$
Question 11 of 24

11. Question
1 point(s)
The points (2,5) & (5,1) are two opposite vertices of a rectangle. If other two vertices are points on the Straight line y=2x+k, then the value of k is
- 1
- 4
- -4
- 3
Correct

Incorrect

Hint

C bisects AB
$= \left( {\dfrac{{5 + 2}}{2},\dfrac{{5 + 1}}{2}} \right) = \left( {\dfrac{7}{2},3} \right)$
since C line on the line y=2k+k
$\Rightarrow 3 = 2 \times \dfrac{7}{2} + k$
k= -4
Question 12 of 24

12. Question
1 point(s)
If a straight line perpendicular to the line 2x+y=3 is passing through (1,1). Its y-intercept is
- 1
- $\dfrac{1}{2}$
- $\dfrac{1}{3}$
- 2
Correct

Incorrect

Hint

Straight line perpendicular to 2x+y=3 is 2y-x+c=0
Since, it passes through (1,1)
2(1)-1+c=0
$\Rightarrow C= -1$
$\therefore$ $E{q^n}$ of the line is 2y-x-1=0
$\therefore$ so y-intrcept is $\dfrac{1}{2}$
Question 13 of 24

13. Question
1 point(s)
The ratio by which the line 2x+5y=7 divided the straight line joining the points (-4,7) & (6,-5) is
- $1:2$
- $3:1$
- $1:1$
- $2:3$
Correct

Incorrect

Hint

2x+5y-7=0-(i) Given points A(-4,7) & B(6,-5)
$\therefore y - {y_1} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)$ (two-point from)
$\Rightarrow y - 7 = \dfrac{{12}}{{ - 10}}\left( {x + 4} \right)$
$\Rightarrow$ 5y – 35 = -6x -24
$\Rightarrow$ 6x+3y=11 -(ii)
solving (i) & (ii) we get
x=1, & y=1 $\therefore$ P(x,y)=(1,1)
since P(1,1) Is midpoint of A(-4,7) & B(6,-5)
then the ratio is 1:1.
Question 14 of 24

14. Question
1 point(s)
The straight lines x+y=0, 5x+y=4 & x+5y=4 from
- an isosceles triangle
- an equilateral triangle
- a scalene triangle
- a right angle triangle
Correct

Incorrect

Hint

x+y=0 -(i)
5x+y=4 -(ii)
x+5y=4 -(iii)
solving above equation, vertices are $A\left( {1, - 1} \right),B\left( {\dfrac{2}{3},\dfrac{2}{3}} \right),C\left( { - 1,1} \right)$
$\therefore AB = \dfrac{{\sqrt {26} }}{3}$
$BC = \dfrac{{\sqrt {26} }}{3}$
$CA = 2\sqrt 2$
$\therefore AB = BC$
Question 15 of 24

15. Question
1 point(s)
Let A(2,-3) & B(-2,1) be vertices of a $\triangle ABC$. If the centroid of the triangle moves on the line 2x+3y=1, then the locus of the verter C is the line
- 2x+3y=9
- 2x-3y=9
- 3x-2y=9
- 3x+2y=5
Correct

Incorrect

Hint

Let $C\left( {{x_1},{y_1}} \right)$
$\therefore G = \left( {\dfrac{{2 - 2 + {x_1}}}{3},\dfrac{{ - 3 + 1 + {y_1}}}{3}} \right) = \left( {\dfrac{{{x_1}}}{3},\dfrac{{{y_1} - 2}}{3}} \right)$
G satisfies the equation 2x+3y=1
$\Rightarrow 2\left( {\dfrac{{{x_1}}}{3}} \right) + 3\left( {\dfrac{{{y_1} - 2}}{3}} \right) = 1$
$\Rightarrow 2{x_1} + 3{y_1} = 9$
$\therefore$ Hence the locus of C is
2x+3y=9
Question 16 of 24

16. Question
1 point(s)
Joint equation of pair of lines through (3,-2) & parallel to ${x^2} – 4xy + 3{y^2} = 0$ is
- ${x^2} + 3{y^2} – 4xy – 14x + 24y + 45 = 0$
- ${x^2} + 3{y^2} + 4xy + 14x + 24y + 45 = 0$
- ${x^2} + 3{y^2} – 4xy – 14x – 24y – 45 = 0$
- none of these
Correct

Incorrect

Hint

${m_1} + {m_2} = \dfrac{4}{3},{m_1}{m_2} = \dfrac{1}{3}$
Solving, ${m_1} = \dfrac{1}{3},{m_2} = 1$
$\therefore y = \dfrac{1}{3}x + {c_1}\& y = x + {c_2}$
The above lines pases through (3,-2)
$\therefore$ Required equation of pair of lines
(3x-x+9) (y-x+5)=0
${x^2} + 3{y^2} - 4xy - 14x + 24y + 45 = 0$
Question 17 of 24

17. Question
1 point(s)
If the lines x+3y-9=0, 4x+by-2=0 & 2-y-4=0 are concurrent, then b is equal to
- -5
- 1
- 5
- 0
Correct

Incorrect

Hint

Given, x+3y-9=0, 4x+by-2=0, 2x-y-4=0 are concurrent if
$\left| {\begin{array}{ccccccccccccccc}1&3&{ - 9}\\4&b&{ - 2}\\2&{ - 1}&{ - 4}\end{array}} \right| = 0$
$\Rightarrow$ (-4b-2)-3(-16+4)-9(-4-2b)=0
$\therefore$ b= -5
Question 18 of 24

18. Question
1 point(s)
If non-zero numbers, a,b,c are in HP, then the straight line $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{1}{c} = 0$ always passes through a fixed point, is
- $\left( {1,\dfrac{{ – 1}}{2}} \right)$
- (1,-2)
- (-1,-2)
- $\left( { – 1,\dfrac{1}{2}} \right)$
Correct

Incorrect

Hint

$\dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}$
$\Rightarrow \dfrac{1}{a} - \dfrac{2}{b} + \dfrac{1}{c} = 0$
so, the line $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{1}{c} = 0$ always passes through the fixed point (1,-2)
Question 19 of 24

19. Question
1 point(s)
The values of $\lambda $, for which the equation ${x^2} – {y^2} – x + \lambda y – 2 = 0$ represents a pair of straight lines is
- (-3,1)
- (-1,1)
- (3,-3)
- (3,1)
Correct

Incorrect

Hint

Given ${x^2} - {y^2} - x + \lambda y - 2 = 0$
Here a=1, b=-1, c=-2, h=0
$g = \dfrac{{ - 1}}{2},f = \dfrac{\lambda }{2}$
$abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$ represents pair of straight lines.
$2 + 0 - \dfrac{{{\lambda ^2}}}{4} + \dfrac{1}{4} = 0$
$\Rightarrow \lambda = - \sqrt 9 \Rightarrow \lambda = \pm 3$
$\therefore \lambda = ( - 3,3)$
Question 20 of 24

20. Question
1 point(s)
The equation of pair of lines joinning origin to the points of intersection of ${x^2} + {y^2} = 9\& x + y = 3$ is
- ${x^2} + {y^2} = 7$
- xy=0
- ${(3 + y)^2} + {y^2} = 9$
- ${({x^2} + ( – x))^2} = 9$
Correct

Incorrect

Hint

${x^2} + {y^2} = 9,x + y = 3$
From above two equation, we make a homogeneous equation.
${x^2} + {y^2} = {(x + y)^2}$
$\Rightarrow {x^2} + {y^2} = {x^2} + {y^2} + 2xy$
$\Rightarrow$ 2xy=0
$\Rightarrow xy=0$
Question 21 of 24

21. Question
1 point(s)
Find the area of the triangle with vertices A(5,0), B(5,5) & C(2,1)
- 12
- 14
- 15
- 18
Correct

Incorrect

Hint

Area $\left( \triangle \right) = \left| {\begin{array}{ccccccccccccccc}{{x_1}}&{{x_2}}&{{x_3}}\\{{y_1}}&{{y_2}}&{{y_3}}\\1&1&1\end{array}} \right|$
$= \left| {\begin{array}{ccccccccccccccc}5&5&2\\0&5&1\\1&1&1\end{array}} \right| = 5\left| {\begin{array}{ccccccccccccccc}5&1\\1&1\end{array}} \right| + 1\left| {\begin{array}{ccccccccccccccc}5&2\\5&1\end{array}} \right|$
$=5 \times 4+5-10=20-5=15 \text{ sq. units}$
Question 22 of 24

22. Question
1 point(s)
Find the co-ordinate of the centroid of the triangle ABC with vertices A(4,7), B(2,-1) & C(3,9)
- (3,4)
- (3,5)
- (5,3)
- (5,4)
Correct

Incorrect

Hint

${C_2} = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$
$= \left( {\dfrac{{4 + 2 + 3}}{3},\dfrac{{7 - 1 + 9}}{3}} \right)$
$= \left( {3,5} \right)$
Question 23 of 24

23. Question
1 point(s)
Find the distance from P(2,-5) to the equation of line 4x-3y+2=0
- 7
- 3
- 5
- 4
Correct

Incorrect

Hint

$L = 4x - 3y + 3 = 0,P\left( {x,1} \right) = \left( {2, - 5} \right)$
$d = \left| {\dfrac{{A{x_1} + B{y_1} + C}}{{\sqrt {{A^2} + {B^2}} }}} \right| = \left| {\dfrac{{4 \times 2 + \left( { - 3} \right)\left( { - 5} \right) + 2}}{{\sqrt {{4^2} + {{\left( { - 3} \right)}^2}} }}} \right|$
$= \left| {\dfrac{{8 + 15 + 2}}{5}} \right| = \left| {\dfrac{{25}}{5}} \right| = \left| 5 \right| = 5$
Question 24 of 24

24. Question
1 point(s)
Find the angle between the lines 2x+3y-5=0 & 3x-2y+11=0
- $\dfrac{\pi }{3}$
- $\dfrac{\pi }{4}$
- $\dfrac {\pi }{6}$
- $\dfrac{\pi}{2}$
Correct

Incorrect

Hint

$L_1$ : 2X+3Y-5=0
$L_2$ :3X-2Y+11=0
slope of ${L_1}\left( {{m_1}} \right) = - \dfrac{2}{3}$
slope of ${L_2}\left( {{m_2}} \right) = - \left( {\dfrac{{ - 3}}{2}} \right) = \dfrac{3}{2}$
${m_1}{m_2} = \left( {\dfrac{{ - 2}}{3}} \right) \times \left( {\dfrac{3}{2}} \right) = - 1$
$\therefore {L_1} \bot {L_2}$
$\therefore \theta = 90^\circ = \dfrac{\pi }{2}$