MCQs
Choose the correct answer from the given four options (1 to 16):
Question 1
Point (-3, 5) lies in the
(a) first quadrant
(b) second quadrant
(c) third quadrant
(d) fourth quadrant
Sol :
Point (-3, 5) lies in second quadrant, (b)
Question 2
Point (0, -7) lies
(a) on the x-axis
(b) in the second quadrant
(c) on the y-axis
(d) the fourth quadrant
Sol :
Point (0, -7) lies on y-axis (as x = 0) (c)
Question 3
Abscissa of a point is positive in
I and II quadrants
I and IV quadrants
I quadrant only
II quadrant only
Sol :
Abscissa of a point is positive in first and fourth quadrants. (b)
Question 4
The point which lies on y-axis at a distance of 5 units in the negative direction of y- axis is
(a) (0, 5)
(b) (5, 0)
(c) (0, -5)
(d) (-5, 0)
Sol :
(0, -5) is the required point. (c)
Question 5
If the perpendicular distance of a point P from the x-axis is 5 units and the foot of perpendicular lies on the negative direction of x-axis, then the point P has
(a) x-coordinate = -5
(b) y-coordinate = 5 only
(c) y-coordinate = -5 only
(d) y-coordinate = 5 or -5
Sol :
Perpendicular distance for a point P on x- axis in negative direction.
It will has y = 5 and x = -5 (d)
Question 6
The points whose abscissa and ordinate have different signs will lie in
(a) I and II quadrants
(b) II and III quadrants
(c) I and III quadrants
(d) II and IV quadrants
Sol :
Point which has abscissa and ordinate having different signs will lie in second and fourth quadrants. (d)
Question 7
The points (-5, 2) and (2, -5) lie in
(a) same quadrant
(b) II and III quadrants respectively
(c) II and IV quadrants respectively
(d) IV and II quadrants respectively
Sol :
Points (-5, 2) and (2, -5) lie in second and fourth quadrants respectively. (b)
Question 8
If P (-1,1), Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4) are plotted on the graph paper, then point(s) in the fourth quadrant are
(a) P and T
(b) Q and R
(c) S only
(d) P and R
Sol :
Points P (-1, 1), Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4) are plotted on graph The points in 4th quadrant are Q and R (b)
Question 9
On plotting the points O (0, 0), A (3, 0), B (3, 4), C (0, 4) and joining OA, AB, BC and CO which of the following figure is obtained?
(a) Square
(b) Rectangle
(c) Trapezium
(d) Rhombus
Sol :
On plotting the points O (0, 0), A (3, 0), B (3, 4), C (0, 4)
OA, AB, BC and CO are joined
The figure so formed will a rectangle (b)
Question 10
Which of the following points lie on the graph of the equation :
3x-5y + 7 = 0?
(a) (1, -2)
(b) (2, 1)
(c) (-1, 2)
(d) (1, 2)
Sol :
3x-5y+7=0
Let (1,-2) subtracting the value of x=1, y=-2, then
$3 \times 1-5(-2)+7=3+10+7=17 \neq 0$
Similarly substituting the value of x=2, y=1 then
$3 \times 2-5 \times 1+7=6-5+7 \neq 0$
(-1,2)
$3 \times(-1)-(5 \times 2)+7$
$\Rightarrow-3-10+7 \neq 0$
and (1,2)
$3 \times 1-5 \times 2+7=0$
⇒3-10+7=10-10=0
∴(1,2) lies on 3x-5y+7=0
Ans (d)
Question 11
The pair of equation x – a and y = b graphically represents lines which are
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)
Sol :
x = a, y = 6
Which are intersecting at (a, b) (d)
Question 12
The distance of the point P (2, 3) from the x>axis is
(a) 2 units
(b) 3 units
(c) 1 unit
(d) 5 units
Sol :
The distance of the point P (2, 3) from x- axis is 3 units (as y = 3). (b)
Question 13
The distance of the point P (-4, 3) from the y-axis is
(a) 5 units
(b) -4 units
(c) 4 units
(d) 3 units
Sol :
The distance of the point P (-4, 3) from y- axis will be 4 units. (c)
Question 14
The distance of the point P (-6, 8) from the origin is
(a) 8 units
(b) 2√7 units
(c) 10 units
(d) 6 units
Sol :
The distance of point P(-6,8) from origin is $\sqrt{(6)^{2}+(8)^{2}}=\sqrt{36+64}$
$=\sqrt{100}$=10 units
Ans (c)
Question 15
The distance between the points A (0, 6) and B (0, -2) is
(a) 6 units
(b) 8 units
(c) 4 units
(d) 2 units
Sol :
$A B=\sqrt{(0-0)^{2}+(6+2)^{2}}=\sqrt{0^{2}+8^{2}}$
$=\sqrt{8^{2}}$
=8 units
Ans (b)
Question 16
The distance between the points (0, 5) and (-5, 0) is
(a) 5 units
(b) 5√2 units
(c) 2√7 units
(d) 10 units
Sol :
The distance between the points (0, 5) and (-5, 0) is
$=\sqrt{(-5-0)^{2}+(0-5)^{2}}=\sqrt{(-5)^{2}+(-5)^{2}}$
$=\sqrt{25+25}=\sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2}$
Ans (b)
AOBC is a rectangle whose three vertices are A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
(a) 5 units
(b) 3 units
(c) √34 units
(d) 4 units
Sol :
Length of its diagonals
$\mathrm{AB}=\sqrt{(5-0)^{2}+(0-3)^{2}}=\sqrt{(5)^{2}+(-3)^{2}}$
$=\sqrt{25+9}=\sqrt{34}$ units
Ans (c)
If the distance between the points (2, -2) and (-1, x) is S units, then one of the value of x is
(a) -2
(b) 2
(c) -1
(d) 1
Sol :
Distance between (2,-2) and (-1,x)=5 units
$\therefore \sqrt{(2+1)^{2}+(-2-x)^{2}}=5$
$\Rightarrow \sqrt{3^{2}+(-2-x)^{2}}=5$
Squaring
$\Rightarrow 3^{2}+4+x^{2}+4 x=25$
$\Rightarrow x^{2}+4 x+13-25=0$
$\Rightarrow x^{2}+4 x-12=0$
$\Rightarrow x^{2}+6 x-2 x-12=0$
$\Rightarrow x(x+6)-2(x+6)=0$
$\Rightarrow(x+6)(x-2)=0$
∴Either x+6=0, then x=-6
or x-2=0 , then x=2
One value of x=2
Ans (b)
The distance between the points (4, p) and (1, 0) is 5 units, then the value of p is
(a) 4 only
(b) -4 only
(c) ±4
(d) 0
Sol :
Distance between (4, p) and (1,0) is 5 units
$\therefore \sqrt{(4-1)^{2}+(p-0)^{2}}=5$
$\sqrt{3^{2}+p^{2}}=5 \Rightarrow 9+p^{2}=25$
$p^{2}=25-9=16$ (squaring)
$\therefore p=\pm 4$
Ans (c)
The points (-4, 0), (4, 0) and (0, 3) are the vertices of a
(a) right triangle
(b) isosceles triangle
(c) equilateral triangle
(d) scalene triangle
Sol :
Points A(-4,0), B(4,0) , C(0,3) are the vertices of a triangle
Now $\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(4+4)^{2}+(0)^{2}}=\sqrt{(8)^{2}}$
=8 units
$\mathrm{BC}=\sqrt{(0-4)^{2}+(3-0)^{2}}$
$=\sqrt{(-4)^{2}+(3)^{2}}=\sqrt{16+9}=\sqrt{25}$
=5 unints
$\mathrm{CA}=\sqrt{0+4)^{2}+(3-0)^{2}}$
$=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5$ units
∵Two sides are equal in length (∵BC=CA)
∴It is an isosceles triangle
Ans (b)
The area of a square whose vertices are A (0, -2), B (3, 1), C (0, 4) and D (-3, 1) is
(a) 18 sq. units
(b) 15 sq. units
(c) √18 sq. units
(d) √15 sq. units
Sol :
Vertices of a square are A(0,-2), B(3,1), C(0,4) and D(-3,1)
$\therefore \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}^{\prime}-y_{1}\right)^{2}}$
$=\sqrt{(3-0)^{2}+(1+2)^{2}}=\sqrt{3^{2}+3^{2}}$
$=\sqrt{9+9}=\sqrt{18}$
∴Area of square$=(\text { side })^{2}$
$=(\sqrt{18})^{2}=18$ sq unit
Ans (a)
In the given figure, the area of the triangle ABC is
(a) 15 sq. units
(b) 10 sq. units
(c) 7.5 sq. units
(d) 2.5 sq. units
Sol :
Vertices of a $\Delta \mathrm{ABC}$ are A(1,3), B(-1,0) ,C(4,0)
$\therefore \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(-1-1)^{2}+(0-3)^{2}}$
$=\sqrt{(-2)^{2}+(-3)^{2}}=\sqrt{4+9}=\sqrt{13}$
$\mathrm{BC}=\sqrt{(4+1)^{2}+(0+0)^{2}}=\sqrt{5^{2}+\overline{0}}$
$=\sqrt{5^{2}}$
=5 units
∵Coordinates of A are (1,3)
∴Distance from A is x-axis=3 units
$\therefore$ Area $=\frac{1}{2} \mathrm{BC} \times 3=\frac{1}{2} \times 5 \times 3$
$=\frac{15}{2}$=7.5 sq units
Ans (c)
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(a) 5 units
(b) 12 units
(c) 11 units
(d) 7+√5 units
Sol :
Vertices of a ΔABC are A(0,4), B(0,0) , C(3,0)
$\therefore \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(0-0)^{2}+(0-4)^{2}}$
$=\sqrt{0^{2}+(-4)^{2}}$
$=\sqrt{0+16}=\sqrt{16}=4$ units
$\mathrm{BC}=\sqrt{(3-0)^{2}+(0-0)^{2}}$
$=\sqrt{3^{2}+0^{2}}$
$=\sqrt{9+0}=\sqrt{9}=3$ units
and CA$=\sqrt{(3-0)^{2}+(0-4)^{2}}$
$=\sqrt{3^{2}+(-4)^{2}}=\sqrt{9+16}$
$=\sqrt{25}=5$ units
∴Perimeter of ΔABC=AB+BC+CA
=4+3+5=12 units
Ans (b)
If A is a point on the .y-axis whose ordinate is 5 and B is the point (-3, 1), then the length of AB is
(a) 8 units
(b) 5 units
(c) 3 units
(d) 25 units
Sol :
A is a point on y-axis whose ordinates is 4 and B is a point (-3,1) then length of coordinates of A will be (0,5)
$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(-3-0)^{2}+(1-5)^{2}}$
$=\sqrt{(-3)^{2}+(-4)^{2}}$
$=\sqrt{9+16}=\sqrt{25}=5$ units
Ans (b)
The point A (9, 0), B (9, 6), C (-9, 6) and D (-9, 0) are the vertices of a
(a) rectangle
(b) square
(c) rhombus
(d) trapezium
Sol :
A(9,0) , B(9,6), C(-9,6) and D(-9,0)
$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(9-9)^{2}+(6-0)^{2}}$
$=\sqrt{0^{2}+6^{2}}=\sqrt{0+36}=\sqrt{36}=6$ units
$\mathrm{BC}=\sqrt{(-9-9)^{2}+(6-6)^{2}}$
$=\sqrt{(-18)^{2}+0^{2}}=\sqrt{18^{2}+0^{2}}$
$=\sqrt{324}=18$ units
$\mathrm{CD}=\sqrt{[-9-(-9)]^{2}+(0-6)^{2}}$
$=\sqrt{(9-9)^{2}+(-6)^{2}}$
$=\sqrt{(0)^{2}+6^{2}}=\sqrt{36}=\sqrt{36}$
=6 units
$\mathrm{DA}=\sqrt{(-9-9)^{2}+(0-0)^{2}}$
$=\sqrt{(-18)^{2}+(0)^{2}}$
$=\sqrt{324+0}=\sqrt{324}=18$ units
$\because \mathrm{AB}=\mathrm{CD}$ and BC=DA and these are opposite sides
∴ABCD is a rectangle
Ans (a)
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