Maths Class X Question Bank

Question 1 of 29

1. Question
1 point(s)
A line intersects x and y–axes at P and Q respectively. If (2, –5) is the mid–point of PQ, then the coordinates of P and Q are respectively.
- (4, 0) and (0,–10)
- (2, 0) and (0, –5)
- (– 4, 0) and (0, 10)
- (0, -10) and (4, 0)
Correct

Incorrect

Hint

Mid point of $P(0, y)$ and $Q(x, 0)$ is
$\left(\dfrac{0+x}{2},\dfrac{y+0}{2}\right)$
A/Q $\dfrac{0+x}{2}=2\Rightarrow x=4$
$\dfrac{y+0}{2}=-5\Rightarrow y=-10$
Coordinate of P and Q are (0, -10) and (4, 0)
Question 2 of 29

2. Question
1 point(s)
The distance between the points $(\cos\theta,\sin\theta) $ and $(\sin\theta,-\cos\theta) $ is:
- $\sqrt3 $
- $\sqrt2 $
- $2 $
- $1 $
Correct

Incorrect

Hint

$\sqrt{(\sin\theta-\cos\theta)^2+(\cos\theta-\sin\theta)^2}\\ \\ =\sqrt{\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta+\cos^2\theta+\sin^2\theta+2\cos\theta\sin\theta}\\ \\ =\sqrt{1+1}=\sqrt2$
Question 3 of 29

3. Question
1 point(s)
If the points (1, 2), (–5, 6) and (a, – 2) are collinear, then a is equal to :
- -3
- 7
- 2
- -2
Correct

Incorrect

Hint

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0\\ \\ 1(6+2)+(-5)(-2-2)+a(2-6)=0\\ \\ 8+20-4a=0\\ \\ 28=4a\Rightarrow a=7$
Question 4 of 29

4. Question
1 point(s)
The points A(9, 0), B(9, 6), C(– 9, 6) and D(– 9, 0) are the vertices of a :
- Square
- Rectangle
- Rhombus
- Trapezium
Correct

Incorrect

Hint

$A(9,0),B(9,6),C(-9,6)$ and $D(-9,0)$
$AB=\sqrt{(9-9)^2+(6-0)^2}=6\\ \\ BC=\sqrt{(-9-9)^2+(6-6)^2}=18\\ \\ CD=\sqrt{(-9+9)^2+(0-6)^2}=6\\ \\ DA=\sqrt{(-9-9)^2+(0-0)^2}=18$
opposite sides are equal so it is a rectangle
Question 5 of 29

5. Question
1 point(s)
AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of its diagonal is :
- $5 $
- $3 $
- $\sqrt{34} $
- $4 $
Correct

Incorrect

Hint

$A(0,3),0(0,0),B(5,0)\\ \\ AB=\sqrt{(5-0)^2+(0-3)^2}\\ \\ =\sqrt{34}$
Question 6 of 29

6. Question
1 point(s)
The points (– 4, 0), (4, 0) and (0, 3) are vertices of a:
- Right triangle
- Isosceles triangle
- Equilateral triangle
- Scalene triangle
Correct

Incorrect

Hint

$A(-4,0),B(4,0),C(0,3)\\ \\ AB=\sqrt{8^2+0^2}=\sqrt{64}=8\\ \\ BC=\sqrt{(-4)^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\\ \\ AC=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$
It is an isosceles triangle
Question 7 of 29

7. Question
1 point(s)
The coordinates of the vertices of an equilateral triangle are $A(3, y), B (3,\sqrt3) $ and $C(0,0) $. The value of y is :
- 4
- 5
- -1
- None of these
Correct

Incorrect

Hint

$AB=BC\\ \\ \sqrt{(3-3)^2+(\sqrt3-y)^2}=\sqrt{(0-3)^2+(0-\sqrt3)^2}\\ \\ =3+y^2-2\sqrt3y=9+3\\ \\ \Rightarrow y^2-2\sqrt3y-9=0\\ \\ \Rightarrow y^2-3\sqrt3y+\sqrt3y-9=0\\ \\ \Rightarrow y(y-3\sqrt3)+\sqrt3(y-3\sqrt3)=0\\ \\ \Rightarrow y=-\sqrt3,3\sqrt3$
Question 8 of 29

8. Question
1 point(s)
The distance between the points $(a\cos\theta+b\sin\theta,0) $ and $(0,a\sin\theta-b\cos\theta) $ is:
- $a^2+b^2 $
- $a+b $
- $a^2-b^2 $
- $\sqrt{a^2+b^2} $
Correct

Incorrect

Hint

$\sqrt{(a\cos\theta+b\sin\theta-0)^2+(a\sin\theta-b\cos\theta-0)^2}\\ \\ =\sqrt{a^2\cos^2\theta+b^2\sin^2\theta+2ab\sin\theta\cos\theta+a^2\sin\theta+b^2\cos\theta-2ab\sin\theta\cos\theta}\\ \\ =\sqrt{a(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)}\\ \\ =\sqrt{a^2+b^2}$
Question 9 of 29

9. Question
1 point(s)
(–1, 2), (2, –1) and (3, 1) are three vertices of a parallelogram. The coordinates of the fourth vertex are :
- (0, 4)
- (– 2, 0)
- (– 2, 6)
- (6, 2)
Correct

Incorrect

Hint

$A(-1,2),B(2,-1),C(3,1),D(a,b)$
Let the fourth vertex be a,b
Mid point of $AC=BD\ \left(\dfrac{3-1}{2},\dfrac{2+1}{2}\right)=\left(\dfrac{a+2}{2},\dfrac{b-1}{2}\right)\\ \\ \Rightarrow \dfrac{a+2}{2},\dfrac{b-1}{2}=1,\dfrac{3}{2}\\ \\ \Rightarrow\dfrac{a+2}{2}=1\Rightarrow a=0\\ \\ \dfrac{b-1}{2}=\dfrac{3}{2}\Rightarrow b=4$
The fourth vertex = (0, 4)
Question 10 of 29

10. Question
1 point(s)
If A(5, 3), B(11, –5) and C(12, a) are the vertices of a right angled triangle, right angled at C, then the value of a is :
- – 2, 4
- –2, – 4
- 2, – 4
- 2, 4
Correct

Incorrect

Hint

$A(5,3),B(11,-5),C(12,a)\\ \\ AB^2=BC^2+CA^2\\ \\ AB^2=(11-5)^2+(-5-3)^2=100\\ \\ AC^2=(12-5)^2+(a-3)^2\\ \\ BC^2=(11-5)^2+(-5-3)^2$
So, $100=50+2a^2+34+4a\\ \\ \Rightarrow 2a^2+4a-16=0\\ \\ \Rightarrow a^2+2a-8=0\\ \\ \Rightarrow a=2,-4$
Question 11 of 29

11. Question
1 point(s)
The perimeter of the triangle with vertices (0, 4),(0, 0) and (3, 0) is :
- $5 $
- $12 $
- $11 $
- $7+\sqrt5 $
Correct

Incorrect

Hint

$A(0,4),B(0,0),C(3,0)\\ \\ AB=\sqrt{0^2+(-4)^2}=\sqrt{16}=4\\ \\ BC=\sqrt{3^2+0^2}=\sqrt9=3\\ \\ AC=\sqrt{3^2+(4)^2}=\sqrt{9+16}=5$
Perimeter $=3+4+5=12$
Question 12 of 29

12. Question
1 point(s)
The point which lies on the perpendicular bisector of the line segment joining the points A(– 2, –5) and B(2, 5) is :
- (0, 0)
- (0, 2)
- (2, 0)
- (– 2, 0)
Correct

Incorrect

Hint

$A(-2,-5),B(2,5)\\ \\ \dfrac{-2+2}{2},\dfrac{-5+5}{2}=(0,0)$
Question 13 of 29

13. Question
1 point(s)
The area of the triangle with vertices $(a, b + c),(b, c + a)$ and $(c, a + b)$ is :
- $(a+b+c)^2 $
- $0 $
- $(a+b+c) $
- $abc $
Correct

Incorrect

Hint

$\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\ \\ =\dfrac{1}{2}[a(c+a-a-b)+b(a+b-c)+c(b+c-c-a)]\\ \\ =\dfrac{1}{2}[ac-ab-bc+ab+bc-ac]\\ \\ =\dfrac{1}{2}\times0=0$
Question 14 of 29

14. Question
1 point(s)
If the centroid of the triangle formed by the points $(a, b), (b, c)$ and $(c, a)$ is at the origin, then $a^3 + b^3 + c^3$ is equal to :
- abc
- 0
- a + b + c
- 3abc
Correct

Incorrect

Hint

$\dfrac{a+b+c}{3},\dfrac{a+b+c}{3}=0,0\\ \\ \Rightarrow a+b+c=0\\ \\ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\\ \\ a^3+b^3+c^3-3abc=0\times(a^2+b^2+c^2-ab-bc-ca)\\ \\ \Rightarrow a^3+b^3+c^3=3abc$
Question 15 of 29

15. Question
1 point(s)
The line segment joining points (–3, – 4) and (1, – 2) is divided by y-axis in the ratio :
- 1 : 3
- 2 : 3
- 3 : 1
- 2 : 3
Correct

Incorrect

Hint

$A(-3,-4),\ B(1,-2)\\ \\ (x,y)=(0,y)\\ \\ x=\dfrac{m\times 1+n(-3)}{m+n}\\ \\ \Rightarrow 0=m-3n\\ \\ \Rightarrow m=3n\\ \\ \Rightarrow \dfrac{m}{n}=\dfrac{1}{3}$
Question 16 of 29

16. Question
1 point(s)
If points $(x,0),\ (0,y) $ are $(1,1) $ and collinear, then $\dfrac{1}{x}+\dfrac{1}{y} $ is equal to:
- 1
- 2
- 0
- -1
Correct

Incorrect

Hint

$\begin{vmatrix} x & 0 & 1\\ 0 & y & 1\\ 1 & 1 & 1 \end{vmatrix}=0\\ \\ \Rightarrow x(y-1)-0+1(0-y)=0\\ \\ \Rightarrow xy-x-y=0\\ \\ \Rightarrow xy=x+y\\ \\ \Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=1$
Question 17 of 29

17. Question
1 point(s)
If the ratio in which P divides the line segment joining $(x_1,y_1) $ and $(x_2,y_2) $ be k : 1, then the coordinates of the point P are :
- $\left(\dfrac{kx_1+x_2}{k+1},\dfrac{ky_2+y_1}{k+1}\right) $
- $\left(\dfrac{kx_2+x_1}{k+1},\dfrac{ky_2+y_1}{k+1}\right) $
- $\left(\dfrac{x_1+x_2}{k+1},\dfrac{y_1+y_2}{k+1}\right) $
- None of these
Correct

Incorrect

Hint

$x,y=\dfrac{mx_2+nx_2}{m+n},\dfrac{my_2+ny_1}{m+n}\\ \\ x=\dfrac{kx_2+x_1}{k+1},\ y=\dfrac{ky_2+y_1}{k+1}$
Question 18 of 29

18. Question
1 point(s)
Area of a triangle is taken :
- Always positive
- Always negative
- Sometimes positive sometimes negative
- None of these
Correct

Incorrect

Hint

Area of a triangle is taken always positive
Question 19 of 29

19. Question
1 point(s)
The area of the triangle formed by $(x, y + z),(y, z + x)$ and $(z, x + y)$ is :
- $x + y + z $
- $xyz $
- $(x + y + z)^2 $
- $0 $
Correct

Incorrect

Hint

$Area=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1+y_2)]\\ \\ =\dfrac{1}{2}[x(z+x+x-y)+y(x+y-y-z)+z(y+z-z-x)]\\ \\ =\dfrac{1}{2}[x(z-y)+y(x-z)+z(y-x)]\\ \\ \dfrac{1}{2}\times(xz-xy+yx-zy+zy-zx)\\ \\ =\dfrac{1}{2}\times0=0$
Question 20 of 29

20. Question
1 point(s)
The fourth vertex D of a parallelogram ABCD whose three vertices are A(–2, 3), B(6, 7) and C(8, 3) is :
- (0, 1)
- (0, –1)
- (–1, 0)
- (1, 0)
Correct

Incorrect

Hint

$A(-2,3),B(6,7),C(8,3)\\ \\ 3=\dfrac{6+x_4}{2},3=\dfrac{7+y_4}{2}\\ \\ \Rightarrow x_4=6-6,\ y_4=6-7=-1$
The fourth vertex = (0,-1)
Question 21 of 29

21. Question
1 point(s)
If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then :
- a = b
- a = 2b
- 2a = b
- a = –b
Correct

Incorrect

Hint

$A(1,2),B(0,0),C(a,b)\\ \\ m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{0-2}{0-1}=2\\ \\ m_2=\dfrac{b-0}{a-0}=\dfrac{b}{a}$
If $a,b,c$ are collinear
$m_1=m_2\Rightarrow 2=\dfrac{b}{a}\Rightarrow 2a=b$
Question 22 of 29

22. Question
1 point(s)
In the figure, OAB is a triangle. The coordinates of the point which is equidistant from the three vertices are :
- $(x, y) $
- $ (y, x) $
- $\left(\dfrac{x}{2},\dfrac{y}{2}\right) $
- $\left(\dfrac{y}{2},\dfrac{x}{2}\right) $
Correct

Incorrect

Hint

Let the points be $(a,b)$
$a^2+(b-2y)^2=a^2+b^2\\ \\ b^2+4y^2-4by=b^2\\ \\ \Rightarrow 4y^2=4by\\ \\ \Rightarrow y=b\\ \\ (a-2x)^2+b^2=a^2+b^2\\ \\ \Rightarrow 4x^2=4xa\\ \\ \Rightarrow x=a$
Points are $(x,y)$
Question 23 of 29

23. Question
1 point(s)
A circle drawn with origin as the centre passes through $\left(\dfrac{13}{2},0\right) $ The point which does not lie in the interior of the circle is :
- $\left(-\dfrac{3}{4},1\right) $
- $2,\dfrac{7}{3} $
- $\left(5,\dfrac{1}{2}\right) $
- $\left(-6,\dfrac{5}{2}\right) $
Correct

Incorrect

Hint

$\sqrt{(6.5)^2+0^2}=6.5$
(a) Distance between $(0,0)$ and $\left(\dfrac{-3}{4},1\right)$
$=\dfrac{5}{4}=1.25$
(b) Distance between $(0,0)$ and $\left(2,\dfrac{7}{3}\right)=3.1$
(c) Distance between $(0,0)$ and $\left(5,\dfrac{-1}{2}\right)$ is $=6.5$
So $\left(-6,\dfrac{5}{2}\right)$ does not be in the interior of the circle
Question 24 of 29

24. Question
1 point(s)
The coordinates of the centroid of the triangle with vertices (a, 0), (0, b) and (a, b) are :
- $\left(\dfrac{a}{2},\dfrac{b}{2}\right) $
- $\left(\dfrac{a}{3},\dfrac{b}{3}\right) $
- $\left(\dfrac{2a}{3},\dfrac{2b}{3}\right) $
- None of these
Correct

Incorrect

Hint

$x=\dfrac{a+0+a}{3}=\dfrac{2a}{3}\\ \\ y=\dfrac{0+b+b}{3}=\dfrac{2b}{3}$
Question 25 of 29

25. Question
1 point(s)
Area of the triangle with vertices (x, 0), (0, y) and (x, y) is :
- $x+y $
- $x-y $
- $\dfrac{x}{y} $
- $\dfrac{xy}{z} $
Correct

Incorrect

Hint

$Area=\dfrac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\ \\ =\dfrac{1}{2}[x(y-y)+0(y-0)+x(0-y)]\\ \\ =\dfrac{1}{2}[0+0-xy]=\dfrac{xy}{2}$
Question 26 of 29

26. Question
1 point(s)
If the area of a quadrilateral ABCD is zero, then the four points A, B, C and D are :
- Collinear
- Not collinear
- Nothing can be said
- None of these
Correct

Incorrect

Hint

The four points will be collinear it is because whenever all points are collinear the area of any polygon is 0.
Question 27 of 29

27. Question
1 point(s)
If the points (0, 0), (1, 2) and (x, y) are collinear then :
- x = y
- 2x = y
- x = 2y
- 2x = –y
Correct

Incorrect

Hint

$\dfrac{1}{2}[0(2-y)+1(y-0)+x(0-2)]=0\\ \\ \dfrac{1}{2}[0+y-2x]=0\\ \\ y=2x$
Question 28 of 29

28. Question
1 point(s)
The perpendicular distance of A (5, 12) from the y-axis is :
- 13 units
- 5 units
- 12 units
- 17 units
Correct

Incorrect

Hint

The point of the y-axis is (0, 12)
Distance between (5, 12) and (0, 12)
$=\sqrt{(0-5)^2+(12-12)^2}=\sqrt{25}=5$
Question 29 of 29

29. Question
1 point(s)
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is :
- 8
- 10
- 12
- 15
Correct

Incorrect

Hint

$A(0,4),B(0,0),C(3,0)\\ \\ AB=\sqrt{0^2+(-4)^2}=\sqrt6=4\\ \\ BC=\sqrt{3^2+0^2}=3\\ \\ CA=\sqrt{(3-0)^2+(0-4)^2}=\sqrt{9+16}=5$
Perimeter $=4+3+5=12$