Mathematics Class X

Question 1 of 29

1. Question
1 point(s)
The center of a circle is (2a,a-7) the values of a if the circle passes through the point (11,-9) and has diameter $10\sqrt 2 $units are?
- 5,3
- -5,-3
- 0,1
- 4,5
Correct

Incorrect

Hint

Radius (OP)
$\begin{array}{l} = \sqrt {{{(11 - 2a)}^2} + {{( - 9 - a + 7)}^2}} \\ = \sqrt {{{(11 - 2a)}^2} + {{( - 2 - a)}^2}} \end{array}$
Radius $= \dfrac{{10\sqrt 2 }}{2} = 5\sqrt 2$
$\begin{array}{l}{(11 - 2a)^2} + {( - 2 - a)^2} = {(5\sqrt 2 )^2}\\121 + 4{a^2} - 44a + 4 + {a^2} + 4a = 50\\ \Rightarrow {a^2} - 8a + 15 = 0\\ \Rightarrow a = 3,5\end{array}$
Question 2 of 29

2. Question
1 point(s)
A point on the line 3x+5y=15 and equivalent from the coordinate axis lies in?
- None of the quadrants
- Quadrant 1 only
- Quadrant 2
- Quadrant 1, 2 and 3
Correct

Incorrect

Hint

$\begin{array}{l}3x + 5y = 15\\ \Rightarrow \dfrac{x}{5} + \dfrac{y}{3} = 1\end{array}$
If the point P (x=y) lie on a straight line and equidistant from the coordinate axis then its coordinates will be $\left( {\dfrac{{15}}{8},\dfrac{{15}}{8}} \right)$
So, the point lies in quadrant 1 only.
Question 3 of 29

3. Question
1 point(s)
If the coordinates of opposite vertices of a square area (1,3) and (6,0), the length of the side of a square is?
- $\sqrt {34} $
- $\sqrt {17} $
- 12
- 3
Correct

Incorrect

Hint

Let the length of one side of the square is a and diagonal is $a\sqrt 2$
$\begin{array}{l}\sqrt {{{(1 - 6)}^2} + {{(3 - 0)}^2}} = a\sqrt 2 \\ = \sqrt {25 + 9} = a\sqrt 2 \\\sqrt {34} = a\sqrt 2 \\ = \sqrt 2 \times \sqrt {17} = a\sqrt 2 \\ \Rightarrow a = \sqrt {17} \end{array}$
Question 4 of 29

4. Question
1 point(s)
If the points (0,4), (4,0) and (5,p) are collinear, the value of p is?
- 1
- -1
- 2
- 3
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{1}{2}\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right] = 0\\\dfrac{1}{2}\left[ {0(0 - p) + 4(p - 4) + 5(4 - 0)} \right] = 0\\ \Rightarrow 4p - 16 + 20 = 0\\ \Rightarrow p = - 1\end{array}$
Question 5 of 29

5. Question
1 point(s)
The line joining (-1,0) and (-2,- $\sqrt 3 $) makes with the x-axis an angle equal to?
- $30^\circ $
- $45^\circ $
- $15^\circ $
- $60^\circ $
Correct

Incorrect

Hint

Tan Q= $\begin{array}{l}\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{ - \sqrt 3 - 0}}{{ - 2 - ( - 1)}}\\ = \dfrac{{\sqrt 3 }}{{ - 1}} = \sqrt 3 \end{array}$
Q= ${\tan ^{ - 1}}(\sqrt {3)}$
$= 60^\circ$
Question 6 of 29

6. Question
1 point(s)
What is the distance between the points x (a, b) and y(-a, -b)?
- $2\sqrt {{a^2} + {b^2}} $
- $\sqrt {{a^2} + {b^2}} $
- $2{a^2} + {b^2}$
- $2{a^2} + 2{b^2}$
Correct

Incorrect

Hint

Distance
$\begin{array}{l} = \sqrt {{{\left( { - a - a} \right)}^2} + {{\left( { - b - b} \right)}^2}} \\ = \sqrt {{{\left( { - 2a} \right)}^2} + {{\left( { - 2b} \right)}^2}} \\ = 2\sqrt {{a^2} + {b^2}} \end{array}$
Question 7 of 29

7. Question
1 point(s)
For what value of m are the points (5, 5), (m, 7) and (8, 8) collinear?
- 6
- 7
- 8
- 5
Correct

Incorrect

Hint

Area
$\begin{array}{l} = \dfrac{1}{2}\left| {5\left( {7 - 8} \right) + m\left( {8 - 5} \right) + 8\left( {5 - 7} \right)} \right| = 0\\ \Rightarrow 3m - 21 = 0\\ \Rightarrow m = 7\end{array}$
Question 8 of 29

8. Question
1 point(s)
Find the centroid of the triangle whose vertices are (-3, 2), (1, 5) and (11, -19).
- (3, 4)
- (3, -4)
- (-3, -4)
- (-3, 4)
Correct

Incorrect

Hint

Centroid
$\begin{array}{l} = \left( {\dfrac{{ - 3 + 1 + 11}}{3},\dfrac{{2 + 5 - 19}}{3}} \right)\\ = \left( {\dfrac{9}{3},\dfrac{{ - 12}}{3}} \right) = \left( {3, - 4} \right)\end{array}$
Question 9 of 29

9. Question
1 point(s)
Find the number of point an x-axis which are at a distance ‘c’ units from (2, 3)(c<3)
- 1
- 2
- 0
- 3
Correct

Incorrect

Hint

Let the points on x-axis be (x,0)
$\begin{array}{l}C = \sqrt {{{\left( {2 - x} \right)}^2} + {{\left( {3 - 0} \right)}^2}} \\ = \sqrt {{{\left( {2 - x} \right)}^2} + 9} \ge 3\end{array}$
But as it is given that C<3, no such points costs.
Question 10 of 29

10. Question
1 point(s)
Find a point on the x-axis which is equidistant from the points (5, 4) and (-3, 3)
- (2,0)
- (0,3)
- (-2,2)
- (3,0)
Correct

Incorrect

Hint

AP=BP
$\begin{array}{l} \Rightarrow A{P^2} = B{P^2}\\ \Rightarrow {\left( {x - 5} \right)^2} + {\left( {0 - 4} \right)^2} = {\left( {x + 2} \right)^2} + {\left( {0 - 3} \right)^2}\\ \Rightarrow x = 2\end{array}$
Question 11 of 29

11. Question
1 point(s)
Let$\Delta ABC$be a triangle whose vertices are (0,0), (a,5) and (-5,5) respectively. If the triangle is right angled at A. find the value of a
- 3
- 5
- 2
- 6
Correct

Incorrect

Hint

$B{C^2} = A{C^2} + A{B^2}$
Let A (0, 0), B (a, 5), C (-5, 5) be the vertices of the triangle.
$\begin{array}{l} \Rightarrow a - {\left( 5 \right)^2} + {\left( {5 - 5} \right)^2} = {\left( { - 5 - 0} \right)^2} + {\left( {5 - 0} \right)^2} + \left\{ {{{\left( {a - 0} \right)}^2} + {{\left( {5 - 0} \right)}^2}} \right\}\\ \Rightarrow a = 5\end{array}$
Question 12 of 29

12. Question
1 point(s)
The extremities of the diagonal of a parallelogram are (3,-4) and (-6,5). Third vertex is the point (-2,1). Find its fourth vertex.
- (1,1)
- (1,1)
- (0,1)
- (-1,0)
Correct

Incorrect

Hint

$\begin{array}{l}A{B^2} = C{D^2}\\ \Rightarrow {\left( {3 + 2} \right)^2}{\left( { - 4 - 1} \right)^2} = {\left( {x + 6} \right)^2} + {\left( {y - 5} \right)^2}\\ \Rightarrow {\left( {x + 6} \right)^2} + {\left( {y + 4} \right)^2} = 50...(1)\end{array}$
Similarly,
$\begin{array}{l}B{C^2} = A{D^2}\\ \Rightarrow {\left( {x - 3} \right)^2} + {\left( {y + 4} \right)^2} = 32...(2)\end{array}$
D(-1, 0) satisfies both the equations.
Question 13 of 29

13. Question
1 point(s)
Find the third vertex of an equilateral triangle whose two vertices are (2,4) and (2,6)
- $\left( {\sqrt 3 ,5} \right)$
- $\left( {2\sqrt 3 ,5} \right)$
- $\left( {2 + \sqrt 3 ,5} \right)$
- (2,5)
Correct

Incorrect

Hint

$\begin{array}{l} = \left\{ {\dfrac{{\left( {2 + 2} \right) \pm \sqrt 3 \left( {6 - 4} \right)}}{2},\dfrac{{6 + 4 \pm \sqrt 3 \left( {2 - 2} \right)}}{2}} \right\}\\ = \left( {2 \pm \sqrt 3 ,5} \right)\end{array}$
Question 14 of 29

14. Question
1 point(s)
In a parallelogram PQRS, P=(-1,1) Q=(8,0), R=(7,5). What are the coordinates of S?
- (-1,4)
- (-2,4)
- (-2,7/2)
- (-3/2,4)
Correct

Incorrect

Hint

$\begin{array}{l}{x_4} = \left( {{x_1} + {x_3} - {x_2}} \right)\\ = - 1 + 7 - 8 = - 2\\{y_4} = \left( {{y_1} + {y_3} - {y_2}} \right)\\ = - 1 + 5 - 0 = \left( { - 2,4} \right)\end{array}$
Question 15 of 29

15. Question
1 point(s)
If the distance between the points (3,k) and (4,1) is$\sqrt {10} $. Find the value of k.
- 4 or 2
- -4 or 2
- 3 or -4
- 4 or -2
Correct

Incorrect

Hint

Let A and B denote the points (3, k) and (4, 1) respectively.
$\begin{array}{l}AB = \sqrt {10} \Rightarrow A{B^2} = 10\\{\left( {4 - 3} \right)^2} + {\left( {1 - k} \right)^2} = 10\\ \Rightarrow 1 - k = \pm 3\\1 - k = 3,1 - k = - 3\\k = - 2,k = 4\end{array}$
Question 16 of 29

16. Question
1 point(s)
The point on the y-axis which is equidistant from A(-5, -2) and B(3, 2) is
- (0,-2)
- (0,2)
- (-2,0)
- (0,-4)
Correct

Incorrect

Hint

AP=BP
$\begin{array}{l} \Rightarrow A{P^2} = B{P^2}\\ \Rightarrow {\left( {0 + 5} \right)^2} + {\left( {y + 2} \right)^2} = {\left( {0 - 3} \right)^2} + {\left( {y - 2} \right)^2}\\ \Rightarrow 8y + 25 - 9 = 0\\ \Rightarrow 8y = - 16\\ \Rightarrow y = - 2\end{array}$
The required value point is (0, -2).
Question 17 of 29

17. Question
1 point(s)
The vertices of a triangle are (-2,0), (2,3), (1,-3) then the type of triangle is
- Right triangle
- Equilateral triangle
- Isosceles triangle
- Scalene triangle
Correct

Incorrect

Hint

$\begin{array}{l}AB = \sqrt {{{\left( {2 + 2} \right)}^2} + {{\left( { - 3 - 0} \right)}^2}} = 5\\BC = \sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} = \sqrt {37} \\AC = \sqrt {{{\left( { - 2 - 1} \right)}^2} + {{\left( {0 + 3} \right)}^2}} = 3\sqrt 2 \\AB \ne BC \ne AC\end{array}$
So, it is a scalene triangle.
Question 18 of 29

18. Question
1 point(s)
The points which are not collinear are
- (0,1), (8,3), (6,7)
- (7,3), (5,1), (1,9)
- (-3,2), (1,-2), (9,-10)
- None of these
Correct

Incorrect

Hint

$\begin{array}{l}PQ = \sqrt {68} = 2\sqrt {17} \\QR = \sqrt {4 + 16} = 2\sqrt 5 \\PR = \sqrt {36 + 36} = 6\sqrt 2 \end{array}$
So, the points are not collinear.
Question 19 of 29

19. Question
1 point(s)
The coordinates of the point which divides the line segment joining (-7, 4) and (-6, -5) in the ratio 7:2 is
- (56/9, 3)
- (56/9, -3)
- (-61/9, -2)
- (61/9, -2)
Correct

Incorrect

Hint

$\begin{array}{l}x = \dfrac{{7\left( { - 6} \right) + 2\left( { - 7} \right)}}{{7 + 2}} = \dfrac{{ - 42 - 14}}{9} = \dfrac{{ - 56}}{9}\\y = \dfrac{{7\left( { - 5} \right) + 2\left( 4 \right)}}{{7 + 2}} = \dfrac{{ - 35 + 8}}{9} = - 3\end{array}$
$\left( {\dfrac{{ - 56}}{9}, - 3} \right)$ is the required point.
Question 20 of 29

20. Question
1 point(s)
If the coordinate of the centroid of a triangle are (1, 3) and two of its vertices are (-7, 6) and (6,5) then the third vertex is
- $\left( {\dfrac{2}{3},\dfrac{{14}}{3}} \right)$
- $\left( {\dfrac{{ – 2}}{3},\dfrac{{ – 14}}{3}} \right)$
- (2, -2)
- (-2, 2)
Correct

Incorrect

Hint

$\begin{array}{l}\dfrac{{ - 7 + 8 + x}}{3} = 1\\ \Rightarrow x = 2\\\dfrac{{6 + 5 + y}}{3} = 3 \Rightarrow y = - 2\end{array}$
Question 21 of 29

21. Question
1 point(s)
The area of the triangle whose vertices are (a,a), (a+1, a+1), (a+2, a) is
- ${a^3}$
- 1
- 2a
- ${2^{\frac{1}{2}}}$
Correct

Incorrect

Hint

$\begin{array}{l}D = \left( {\dfrac{{a + a + 2}}{2},\dfrac{{a + a}}{2}} \right)\\ = a + 1,a\\AC = 2\\BD = \sqrt {\left( {a + 1} \right) - {{\left( {a + 1} \right)}^2} + {{\left( {a + 1 - a} \right)}^2}} = 1\end{array}$
Area of the triangle
$\begin{array}{l} = \dfrac{1}{2} \times AC \times BD\\ = \dfrac{1}{2} \times 2 \times 1 = 1\end{array}$
Question 22 of 29

22. Question
1 point(s)
The ratio in which the line segment joining (3, 4) and (-2, -1) is divided by the x-axis is
- $3:2$
- $4:1$
- $4:3$
- None of these
Correct

Incorrect

Hint

Let the ratio be k:1
$\begin{array}{l}x = \dfrac{{ - 2 + 3}}{{k + 1}},y = \dfrac{{ - k + 4}}{{k + 1}}\\y = 0\\ \Rightarrow \dfrac{{ - k + 4}}{{k + 1}} = 0 \Rightarrow k = 4\end{array}$
Question 23 of 29

23. Question
1 point(s)
Find the centroid of the triangle whose vertices are (3, -5), (-3, 4) and (9, 2)
- (0,0)
- (3,1)
- (2,3)
- (3,-1)
Correct

Incorrect

Hint

$\begin{array}{l}x = \dfrac{{{x_1} + {x_2} + {x_3}}}{3} = \dfrac{{3 + \left( { - 3} \right) + 9}}{3} = 3\\y = \dfrac{{{y_1} + {y_2} + {y_3}}}{3} = \dfrac{{ - 5 + 4 - 2}}{3} = - 1\end{array}$
Question 24 of 29

24. Question
1 point(s)
Find the value of p for which these points are collinear (7, -2), (5,1), (3,p)
- 2
- 4
- 3
- -4
Correct

Incorrect

Hint

Area=0
$\begin{array}{l} \Rightarrow \dfrac{1}{2}\left| {7\left( {1 - p} \right) + 5\left( {p + 2} \right) + 3\left( { - 2 - 1} \right)} \right| = 0\\ \Rightarrow 7 - 7p + 5p + 10 - 9 = 0\\ \Rightarrow p = 2\end{array}$
Question 25 of 29

25. Question
1 point(s)
Determine the ratio in which the line 2x+y-4=0 divides the line segment joining the points A(2,-2) and B(3,7)
- 2:9
- 1:9
- 1:2
- 2:3
Correct

Incorrect

Hint

$\begin{array}{l}x = \dfrac{{3m + 2n}}{{m + n}},y = \dfrac{{7m - 2}}{{m + n}}\\2\left( {\dfrac{{3m + 2n}}{{m + n}}} \right) + \dfrac{{7m - 2n}}{{m + n}} - 4 = 0\\ \Rightarrow m:n = 2:9\end{array}$
Question 26 of 29

26. Question
1 point(s)
If the midpoint of the line segment joining the points A(3,4) and B(a,4) is P(x, y) and x+y-20=0 then find the value of a.
- 0
- 1
- 40
- 45
Correct

Incorrect

Hint

$\dfrac{{3 + a}}{2},\dfrac{{4 + 4}}{2}$
Now,
$\begin{array}{l} \Rightarrow \dfrac{{3 + a}}{2} - 4 - 20 = 0\\ \Rightarrow 3 + a = 48\\ \Rightarrow a = 45\end{array}$
Question 27 of 29

27. Question
1 point(s)
Find a point on the y-axis which is equidistant from A(6,5) and B(-4,3)
- (0,9)
- (9,0)
- (8,0)
- (0,1)
Correct

Incorrect

Hint

Let the point on y-axis be (0, y)
$\begin{array}{l}\sqrt {{{\left( {0 - 6} \right)}^2} + {{\left( {y + 5} \right)}^2}} = \sqrt {{{\left( {0 - 4} \right)}^2} + {{\left( {y - 3} \right)}^2}} \\ \Rightarrow 36 + {y^2} + 25 - 10y = 16 + {y^2} + 9 - 6y\\ \Rightarrow 4y = 36\\ \Rightarrow y = 9\end{array}$
Question 28 of 29

28. Question
1 point(s)
Determine the ratio in which the line 2x+y-4=0 divides the line segment joining A(2,-2) and B(3,7)
- 2:9
- 9:2
- 1:1
- 1:9
Correct

Incorrect

Hint

Coordinates = $\dfrac{{3k + 2}}{{k + 1}},\dfrac{{7k - 2}}{{k + 1}}$
$\begin{array}{l}\sqrt {{{\left( {0 - 6} \right)}^2} + {{\left( {y - 5} \right)}^2}} = \sqrt {{{\left( {0 - 4} \right)}^2} + {{\left( {y - 3} \right)}^2}} \\ \Rightarrow 36 + {y^2} + 25 - 10y = 16 + {y^2} + 9 - 6y\\ \Rightarrow 4y = 36\\ \Rightarrow y = 9\end{array}$
Question 29 of 29

29. Question
1 point(s)
Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
- X+3y-7=0
- x-3y+7=0
- X+3y+7=0
- None of these
Correct

Incorrect

Hint

Area
$\begin{array}{l} = \dfrac{1}{2}\left[ {x\left( {2 - 0} \right) + 1\left( {0 - y} \right) + 7\left( {y - 2} \right)} \right]\\0 = \dfrac{1}{2}\left[ {2x + 6y - 14} \right]\\x + 3y - 7 = 0\end{array}$