M.C.Q

MCQ 11 Mark

If A(2, 3) and B(-3, 4), then (abscissa of A) - (abscissa of B) is:

A
5
B
-1
C
-5
D
1

Answer

5
Solution:
Here we have, the abscissa of A = 2 and abscissa of B = -3.
So, according to question,
(abscissa of A) - (abscissa of B)
= 2 - (-3)
= 5

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MCQ 21 Mark

If a > 0 and b > 0 then the point (a, b) lies in quadrant.

A
IV
B
II
C
III
D
None of these.

Answer

II
Solution:
Since, x co-ordinate is negative and y co-ordinate is positive, the given point lies in Quadrant II.

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MCQ 31 Mark

The zero of the polynomial (x - 2)² - (x + 2)² is:

A
0
B
2
C
1
D
-2

Answer

0
Solution:
(x - 2)² - (x + 2)²= (x - 2 + x + 2) (x - 2 - x - 2) [Using identity]
a² - b² = (a + b) (a - b)
= (2x) (-4)
= -8x
Then the zero is,
-8x = 0
⇒ x = 0

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MCQ 41 Mark

Signs of the abscissa and ordinate of a point in the second quadrant are respectively.

A
(+, +)
B
(+, -)
C
(-, -)
D
(-, +)

Answer

(-, +)
Solution:
Abscissa means just the horizontal axis i.e. x axis. Ordinate means just the vertical axis i.e. y axis.
x-corrdinate and y-corrdinate make a point.
The signs of abscissa and ordinate of a point in quadrant II are (-, +).

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MCQ 51 Mark

Two point having same abscissae but different ordinates lie on:

A
x-axis
B
a-axis
C
A line parallel to y-axis
D
A line parallel to x-axis

Answer

A line parallel to y-axis
Solution:
Let two points be (a, b) and (a, c).
If abscissa is same = a
and ordinate is different then all such points will lie on a line parallel to Y
axis because value of X-intercept
i.e. abscissa is fixed.

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MCQ 61 Mark

The equation of y-axis is:

A
(1, 7)
B
(2, 10)
C
(4, 12)
D
(-1, 1)

Answer

(4, 12)
Solution:
y = 3x + 4
for x = 4, we have
= 3 × 4 + 4
= 12 + 4
= 16 ≠ 12
Hence, (4, 12) doesn't lie on the given line.

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MCQ 71 Mark

The distance of the point (3, 5) from x- axis is:

A
3 units
B
4 units
C
5 units
D
6 units

Answer

5 units
Solution:
Distance of any point from x-axis is y-coordinate of that given point:

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MCQ 81 Mark

The distance of the point (2, 3) from the y-axis:

A
5 units
B
13 units
C
3 units
D
2 units

Answer

2 units
Solution:
The distance from y-axis is equal to the x-coordinate, so distance = 2 units

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MCQ 91 Mark

If P(3, 9) and Q(-3, -4), then (abscissa of P) - (ordinate of Q) is:

A
-1
B
1
C
7
D
-7

Answer

7
Solution:
From the given data we have,
The abscissa of P = 3 and ordinate of Q = -4,
So, according to question,
(abscissa of P) - (ordinate of Q)
= 3 - (-4)
= 7

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MCQ 101 Mark

The area of the triangle formed by the points A(2, 0), B(6, 0) and C(4, 6) is:

A
24sq. unit
B
12sq. unit
C
10sq. unit
D
None of these

Answer

12sq. unit
Solution:

Let CD be perpendicular drawn from C to AB.
The length of the perpendicular will be equal to the ordinate of point C.
⇒ CD = 6 unit
AB = 4 unit
Now, area of $\triangle\text{ABC}=\frac{1}{2}\times\text{Base}\times\text{height}$
$\triangle\text{ABC}=\frac{1}{2}\times\text{5}\times\text{6}$
$12\text{sq. units}$

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MCQ 111 Mark

The ordinate of every point on the x-axis is:

A
1
B
-1
C
0
D
Any real numbers.

Answer

0
Solution:
The ordinate (y co-ordinate) of every point on the x-axis is 0.

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MCQ 121 Mark

Abscissa of all points on the x-axis is:

A
Any number.
B
0
C
-1
D
1

Answer

Any number.
Solution:
Abscissa of point in x axis is can be any number but ordinate will always be zero, because for any point to lie on x-axis its y-ordinate must be equal to zero.

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MCQ 131 Mark

The point O(0, 0) lies on:

A
x-axis
B
y-axis
C
both x-axis and y-axis
D
Any quadrant.

Answer

both x-axis and y-axis
Solution:
Point (0, 0) is the co-ordinate of origin and origin is the point of intersection of x and y-axis.
So, point O(0, 0) lies on both axis.

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MCQ 141 Mark

If x > 0 and y < 0, then the point (x, y) lies in:

A
IV Quadrant.
B
III Quadrant.
C
I Quadrant.
D
II Quadrant.

Answer

IV Quadrant.
Solution:
Since, x > 0 i.e, x is +ve,
y < 0 i.e, y is -ve,
Recall that (+, +) lies in I quadrant, (-, +) lies in II quadrant, (-, -) lies in III quadrant, (+, -) lies in IV quadrant.
So, (x, y) lies in IV quadrant.

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MCQ 151 Mark

The point (7, 0) lies:

A
On the positive direction of y-axis
B
On the positive direction of x-axis
C
In quadrant II
D
In quadrant IV

Answer

On the positive direction of x-axis
Solution:
Since value of y-ordinate is zero so, point lies on x-axis.
But value of x is +ve so it lies on +ve direction of x-axis.

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MCQ 161 Mark

Write the correct answer in the following:
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.

Answer

II and IV quadrants.
Solution:
The points whose abscissa and ordinate have different sings will be of the from (-x, y) or (x, -y) and these points will lie in II and IV quadrants.

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MCQ 171 Mark

A point whose abscissa is -3 and ordinate 2 lies in:

A
First quadrant.
B
Second quadrant.
C
Third quadrant.
D
Fourth quadrant.

Answer

Second quadrant
Solution:
If absciss = -3
Intercept on Y axis is = 2
Y > 0
So, Point is in Second Quadrant.

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MCQ 181 Mark

The area of $\triangle\text{AOB}$ having vertices A(0, 6), 0(0, 0) and B(6, 0) is:

A
36 sq units
B
18 sq units
C
24 sq units
D
12 sq units

Answer

18 sq units
Solution:
When we plot the given points in the graph paper then,
is the right angle triangle, where
OB = Base = 6 units
Height of triangle = OA = 6 units
$\therefore$ Area of $\triangle\text{AOB}=\frac{1}{2}\times\text{OA}\times\text{OB}$
$\Rightarrow$ Area of $\triangle\text{AOB}=\frac{1}{2}\times\text{6}\times\text{6}$
$\Rightarrow$ Area of $\triangle\text{AOB}=\frac{1}{2}\times\text{36}$
$\Rightarrow$ Area of $\triangle\text{AOB}=18\text{ square units}$

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MCQ 191 Mark

Abcissa of a point is positive in:

A
I and II quadrant.
B
I and IV quadrant.
C
I quadrant only.
D
II quadrant only.

Answer

I and IV quadrant.
Solution:
Absissa of a point is positive when the points are of the form (+, +) and (+, -).
So, the absissa of the point is positive in quadrant I and IV.

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MCQ 201 Mark

If x² + kx - 3 = ( x - 3) ( x + 1), then the value of 'k' is:

A
-3
B
2
C
-2
D
3

Answer

-2
Solution:
x² + kx - 3 = (x - 3) (x + 1)
⇒ x² + kx - 3 = x² + (-3 + 1) x + (-3) x 1
⇒ x² + kx - 3 = x² - 2x - 3
On comparing the term, we get k = -2

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MCQ 211 Mark

Write the correct answer in the following:
If the perpendicular distance of a point P from the x-axis is 5 units and the foot of the 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

Answer

y coordinate = 5 or -5
Solution:
We do know that perpendicular distance of a point from the X-axis Y-coordinate of that point. Here foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y coordinate = 5 or -5

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MCQ 221 Mark

If x < 0 and y > 0, then the point (x, y) lies in.

A
I Quadrant.
B
III Quadrant.
C
II Quadrant.
D
IV Quadrant.

Answer

II Quadrant.
Solution:
Here, x < 0 (i.e -ve) and y > 0, (i.e, +ve)
So in 2nd quadrant value of (x , y) is (-, +).
So the given point will lie in 2nd quadrant.

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MCQ 231 Mark

Ordinate of a point is negative in:

A
Quadrant III only
B
Quadrant III and IV
C
Quadrant I and II
D
Quadrant IV only

Answer

Quadrant III and IV
Solution:
Since, sign of point in 3rd quadrant is (-, -).
And in 4th quadrant, it is (+, -).
So, Ordinate of a point is -ve only in 3rd and 4th quadrant.

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MCQ 241 Mark

The perpendicular distance of the point P(-3, -6) from the x-axis is:

A
3 units
B
-3
C
6 units
D
-2

Answer

6 units
Solution:
Perpendicular distance of a point from x-axis is y-ordinate of the given point is (-3, - 6).
Distance = 6 unit

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MCQ 251 Mark

The point (-3, 0) lies:

A
In quadrant III
B
In quadrant IV
C
On the negative direction of x-axis
D
On the negative direction of y-axis

Answer

On the negative direction of x-axis
Solution:
Since value of y-ordinate is zero, so point lies on x-axis.
But value of x is -ve so, it lies on negative direction of x-axis.

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MCQ 261 Mark

The area of a triangle whose vertices are (0, 0), (4, 0) and (0, 6) is:

A
36 sq. units
B
6 sq. units
C
12 sq. units
D
24 sq. units

Answer

12 sq. units
Solution:
We have a point (0, 0) i.e; origin.
A point (4, 0) whose y-coordinate is zero.
So, this point is having 4 units in x-axis = base (let) A point (0, 6) i.e. 6 units in y-axis = height of a triangle so, these point forms a right angle triangle.
So, Area of a triangle $=\frac{1}{2}\times\text{Base}\times\text{Height}$
Area of a triangle $=\frac{1}{2}\times6\times4=12\text{ sq.}\text{units.}$

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MCQ 271 Mark

Write the correct answer in the following:
Abscissa of all the points on the x-axis is:

A
0
B
1
C
2
D
Any number.

Answer

Any number.
Solution:
Abscissa of all the points on the X-axis is any number because X-axis is a number line which contains many real numbers on it.

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MCQ 281 Mark

The equation of x-axis is:

A
x = 0
B
None of these.
C
y = x
D
y = 0

Answer

y = 0
Solution:
In co-ordinate, any point P is written as p(x, y).
When Y = 0 then the point P lies on x-axis.
So, Y = 0 is the equation of the x-axis.

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MCQ 291 Mark

The point whose abscissa is 4 and this point lies on the x-axis is:

A
(0, 4)
B
(4, 4)
C
(4, 0)
D
None of these.

Answer

(4, 0)
Solution:
Since the abscissa or x-coordinate of a point is 4 and this point lies on the x-axis. And the ordinate or y-coordinate of a point lying on the x-axis is 0.
Therefore the coordinate of the point is (4, 0).

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MCQ 301 Mark

Point (-10, 0) lies.

A
On the negative direction of the y-axis.
B
In the fourth quadrant.
C
In the third quadrant.
D
On the negative direction of the X-axis.

Answer

On the negative direction of the X-axis.
Solution:
In point (-10, 0) y-coordinate is zero, so it lies on X-axis and its x-coordinate is negative, so the point (-10, 0) lies on the X-axis in the negative direction.

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MCQ 311 Mark

A(-6, 3) be a point on the graph. Draw AL $\perp$ x-axis. The co-ordinates of L are:

A
(0, -6)
B
(-6, 0)
C
(0, 0)
D
(-6, 3)

Answer

(-6, 0)
Solution:
Since AL perpendicular to x-axis,
So, point L lies on x-axis, and we know that for any point on x-axis y-ordinate is zero.
So, we have L = (-6, 0)

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MCQ 321 Mark

Write the correct answer in the following:
The point whose ordinate is 4 and which lies on y-axis is:

A
(4, 0)
B
(0, 4)
C
(1, 4)
D
(4, 2)

Answer

(0, 4)
Solution:
Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

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MCQ 331 Mark

The point (-3, 0) lies.

A
In quadrant III
B
In quadrant IV
C
On the negative direction of y-axis.
D
On the negative direction of x-axis.

Answer

On the negative direction of x-axis.
Solution:
Since value of y-ordinate is zero, so point lies on x-axis,
But value of x is -ve so, it lies on negative direction of x-axis.

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MCQ 341 Mark

If O(0, 0), A(3, 0), B(3, 4) and C(0, 4) are the vertices of a quadrilateral OABC, then OABC is:

A
Trapezium.
B
Rectangle.
C
Square.
D
Rhombus.

Answer

Rectangle.
Solution:
As O(0, 0) and A(3, 0)
So, Length of side OA = 3 units
As A(3, 0) and B(3, 4)
So, Length of side AB = 4 units
As O(0, 0) and C(0, 4)
So, Length of side OC = 4 units
As B(3, 4) and C(0, 4)
So, Length of side BC = 3 units
As O(0, 0) and B(3, 4)
So, Length of diagonal OB = 5 units
As A(3, 0) and C(0, 4)
So, Length of diagonal AC = 5 units
In a quadrilateral opposite sides are equal and diagonals are also equal then it is a rectangle.
Therefore, OABC is a rectangle.

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MCQ 351 Mark

Point (-7, 0) lies.

A
On the negative direction of x-axis.
B
On the negative direction of y-axis.
C
In the III quadrant.
D
In the IV quadrant.

Answer

On the negative direction of x-axis.
Solution:
Point (-7, 0) lies on negative direction of the x-axis as its x co-ordinate is negative and y co-ordinate is zero.

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MCQ 361 Mark

The point which lies on x-axis at a distance of 3 units in the positive direction of x-axis is:

A
(0, -3)
B
(-3, 0)
C
(0, 3)
D
(3, 0)

Answer

(3, 0)
Solution:
Since it lies on x-axis so ordinate will be zero because the value of the y-coordinate in the x-axis is equal to zero.
Thus point will be (3, 0).

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MCQ 371 Mark

Ordinate of all points on the x-axis is:

A
Any number.
B
2
C
0
D
1

Answer

0
Solution:
In cartesian plane any point P is written as P(x, y) when the value of Y co-ordinate is equal to zero then the point P lies on x-axis.
So, for any point on x-axis, ordinate = 0

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MCQ 381 Mark

Which of the points P(0, 3), Q(1, 0), R(0, -1), S(-5, 0), T(1, 2) do not lie on the x-axis?

A
P, R and T
B
Q, S and T
C
Q and S only
D
P and R only

Answer

P, R and T
Solution:
We know that, if a point is of the form (x, 0) i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not.
Here, y-coordinates of points P(0, 3), R(0, -1) and T(1, 2) are not zero, so these points do not lie on the X-axis.

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MCQ 391 Mark

In the which quadrant does the point (-7, -4) lie?

A
IV
B
II
C
III
D
None of these.

Answer

III
Solution:
Since, both the coordinates of given point are negative, it lies in Quadrant III.

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MCQ 401 Mark

The points in which the abscissa and the ordinate have different signs will lie in.

A
Quadrant IV only
B
Quadrants I and III
C
Quadrants II and IV
D
Quadrant II only

Answer

Quadrants II and IV
Solution:
In 2nd quadrant and 4th quadrant sign of abscissa and ordinate both is opposite i.e, one is negative and other is positive.
In 2nd quadrant sign of co-ordinate are(-, +),
And in 4th quadrant sign of co-ordinate are (+, -)

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MCQ 411 Mark

A point both of whose co-ordinates are positive lies in:

A
Quadrant I and III
B
Quadrant III only
C
Quadrant I only
D
Quadrant II and IV

Answer

Quadrant I only
Solution:
In 1st quadrant only the sign of x and y coordinate is +ve. i.e, (+, +)

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MCQ 421 Mark

The radius of a circle whose radius is 5 units and whose centre lies on the origin. The co-ordinates of any point lies on the circle and on the y-axis are:

A
(5, 0) only
B
(0, 5) only
C
(5, 0) or (-5, 0)
D
(0, 5) or (0, -5)

Answer

(0, 5) or (0, -5)
Solution:
Since point lies on the y-axis so it can lie in two-point one in positive direction and another is negative direction of y-axis.
So point will be (0, 5) or (0, -5).

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MCQ 431 Mark

Write the correct answer in the following:
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.

Answer

Rectangle.
Solution:
Here, point O(0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B(3, 4) lies in 1^st quadrant and C(0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle.

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MCQ 441 Mark

Which of the following point does not lie on the line y = 2x + 3?

A
(-5, -7)
B
(3, 7)
C
(-1, 1)
D
(3, 9)

Answer

(3, 7)
Solution:
Let us put x = 3 in the give equation,
Then, y = 2(3) + 3
y = 6 + 3 = 9
So, the point will be (3, 9)
For x = 3, y = 9. But in the given option, y = 7
So, the given point (3, 7) will not lie on the line y = 2x + 3.

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MCQ 451 Mark

Write the correct answer in the following: In coordinates of P are:

A
(-4, 2)
B
(-2, 4)
C
(4, -2)
D
(2, -4)

Answer

(-2, 4)
Solution:
Here, given point P lies in II quadrant, so its abscissa will be negative and ordinate will be positive. Also its perpendicular distance from X-axis is 4 y-coordinate of P is 4 and its perpendicular distance from Y-axis is 2. So, x–coordinate is -2. Hence, coordinate of P are (-2, 4).

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MCQ 461 Mark

The distance of the point P(4, 3) from the origin is:

Answer

5
Solution:

Using pythagorous theorem: $\text{OP}^2=\text{OQ}^2=\text{QP}^2$
$\text{OP}^2-4^2+3^2$
$\text{OP}^2=\sqrt{16+9}=5$

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MCQ 471 Mark

Write the correct answer in the following:
If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (abscissa of P) – (abscissa of Q) is:

A
-5
B
1
C
-1
D
-2

Answer

1
Solution:
We have, points P(-2, 3) and Q(-3, 5)
Here, abscissa of P i.e., x-coordinate of P is -2 and abscissa of Q i.e., x-coordinate of Q is -3.
So, (Abscissa of P) - (Abscissa of Q)
= -2 - (-3) = -2 + 3 = 1

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MCQ 481 Mark

Ordinate of all points on the y-axis is.

A
Any number.
B
-1
C
0
D
1

Answer

Any number.
Solution:
In the cartesian plane any point P is written as p(x, y) when the value of x co-ordinate is equal to zero then the point P lies on y axis,
So, Ordinate of any point on y-axis can be any number but abscissa will be zero.

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MCQ 491 Mark

The name of the vertical line drawn to determine the position of any point in the Cartesian plane is:

A
x-axis
B
y-axis
C
Cartesian line.
D
None of these.

Answer

y-axis
Solution:
In the Cartesian plane, there are two axes. One is a horizontal line, that is called the x-axis and other is a vertical line, that is called y-axis.

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MCQ 501 Mark

If x > 0 and y > 0, then the point (x, y) lies in which quadrant?

A
III Quadrant.
B
I Quadrant.
C
II Quadrant.
D
IV Quadrant.

Answer

I Quadrant.
Solution:
Since, x > 0 and y > 0
So, both x and y co-ordinate are +ve,
And we know that in 1st quadrant value of both Abscissa and Ordinate is +ve so, the point will lie in 1st quadrant.

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