Coordinate Geometry — Multiple Choice Questions

Point (-3, 5) lies in the

1. first quadrant

2. second quadrant

3. third quadrant

4. fourth quadrant

Answer

In second quadrant,

x-coordinate is negative and y-coordinate is positive.

∴ P(-3, 5) lies in second quadrant.

Hence, Option 2 is the correct option.

Point (0, -7) lies

1. on the x-axis

2. in the second quadrant

3. on the y-axis

4. in the fourth quadrant

Answer

x-coordinate of any point on y-axis = 0.

∴ P(0, -7) lies on y-axis.

Hence, Option 3 is the correct option.

Abscissa of a point is positive in

1. I and II quadrants

2. I and IV quadrants

3. I quadrant only

4. II quadrants only

Answer

Abscissa (or x-coordinate) is positive in I and IV quadrants.

Hence, Option 2 is the correct option.

The point which lies on y-axis at a distance of 5 units in the negative direction of y-axis is

1. (0, 5)

2. (5, 0)

3. (0, -5)

4. (-5, 0)

Answer

Let point be P(x, y).

Since,

Point lies on y-axis.

∴ x = 0.

Also, point is at a distance of 5 units in the negative direction of y-axis.

∴ y = -5.

P = (x, y) = (0, -5).

Hence, Option 3 is the correct option.

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

1. x-coordinate = -5

2. y-coordinate = 5 only

3. y-coordinate = -5 only

4. y-coordinate = 5 or -5

Answer

In graph, let 1 block = 1 unit.

Since,

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.

Let foot of perpendicular be B.

So, position of point P can be A or C.

From graph,

The y-coordinate of A and C = 5 and -5.

Hence, Option 4 is the correct option.

The points whose abscissa and ordinate have different signs will lie in

1. I and II quadrants

2. II and III quadrants

3. I and III quadrants

4. II and IV quadrants

Answer

In second quadrant,

x-coordinate is negative and y-coordinate is positive.

In fourth quadrant,

x-coordinate is positive and y-coordinate is negative.

Hence, points having opposite signs lie in II and IV quadrants.

Hence, Option 4 is the correct option.

The points (-5, 2) and (2, -5) lie in

1. same quadrant

2. II and III quadrants respectively

3. II and IV quadrants respectively

4. IV and II quadrants respectively

Answer

In second quadrant,

x-coordinate is negative and y-coordinate is positive.

∴ (-5, 2) lies in II quadrant.

In fourth quadrant,

x-coordinate is positive and y-coordinate is negative.

∴ (2, -5) lies in IV quadrant.

Hence, Option 3 is the correct option.

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

1. P and T

2. Q and R

3. S only

4. P and R

Answer

From graph,

Q and R lies in fourth quadrant.

Hence, Option 2 is the correct option.

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?

1. Square

2. Rectangle

3. Trapezium

4. Rhombus

Answer

From graph,

OA = BC = 3 units

OC = AB = 4 units

∴ Figure obtained is a rectangle.

Hence, Option 2 is the correct option.

Which of the following points lie on the graph of the equation :

3x - 5y + 7 = 0?

1. (1, -2)

2. (2, 1)

3. (-1, 2)

4. (1, 2)

Answer

Substituting (1, -2) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(1) - 5(-2) + 7

⇒ 3 + 10 + 7

⇒ 20

Since, L.H.S. ≠ R.H.S., (1, -2) does not satisfy the equation.

Substituting (2, 1) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(2) - 5(1) + 7

⇒ 6 - 5 + 7

⇒ 8

Since, L.H.S. ≠ R.H.S., (2, 1) does not satisfy the equation.

Substituting (-1, 2) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(-1) - 5(2) + 7

⇒ -3 - 10 + 7

⇒ -6

Since, L.H.S. ≠ R.H.S., (-1, 2) does not satisfy the equation.

Substituting (1, 2) in L.H.S. of the equation 3x - 5y + 7 = 0, we get :

⇒ 3x - 5y + 7

⇒ 3(1) - 5(2) + 7

⇒ 3 - 10 + 7

⇒ 0

Since, L.H.S. = R.H.S., (1, 2) satisfies the equation.

Hence, Option 4 is the correct option.

The pair of equation x = a and y = b graphically represents lines which are

1. parallel

2. intersecting at (b, a)

3. coincident

4. intersecting at (a, b)

Answer

From graph,

x = a and y = b intersects at (a, b).

Hence, Option 4 is the correct option.

The distance of the point P(2, 3) from the x-axis is

1. 2 units

2. 3 units

3. 1 unit

4. 5 units

Answer

The distance between the point P(2, 3) and x-axis can be determined by assuming a point A(2, 0) on x-axis.

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow AP = \sqrt{(2 - 2)^2 + (3 - 0)^2} \\[1em] \Rightarrow AP = \sqrt{0 + 3^2} \\[1em] \Rightarrow AP = \sqrt{9} \\[1em] \Rightarrow AP = 3 \text{ units}.$

Hence, Option 2 is the correct option.

The distance of the point P(-4, 3) from the y-axis is

1. 5 units

2. -4 units

3. 4 units

4. 3 units

Answer

The distance between the point P(-4, 3) and y-axis can be determined by assuming a point A(0, 3) on y-axis.

By distance formula,

$\Rightarrow d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \Rightarrow AP = \sqrt{(-4 - 0)^2 + (3 - 3)^2} \\[1em] \Rightarrow AP = \sqrt{(-4)^2 + 0} \\[1em] \Rightarrow AP = \sqrt{16} \\[1em] \Rightarrow AP = 4 \text{ units}.$

Hence, Option 3 is the correct option.

The distance of the point P(-6, 8) from the origin is

1. 8 units

2. $2\sqrt{7}$ units

3. 10 units

4. 6 units

Answer

Origin (O) = (0, 0) and P = (-6, 8).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore OP = \sqrt{(-6 - 0)^2 + (8 - 0)^2} \\[1em] = \sqrt{(-6)^2 + 8^2} \\[1em] = \sqrt{36 + 64} \\[1em] = \sqrt{100} \\[1em] = 10 \text{ units}.$

Hence, Option 3 is the correct option.

The distance between the points A(0, 6) and B(0, -2) is

1. 6 units

2. 8 units

3. 4 units

4. 2 units

Answer

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(0 - 0)^2 + (-2 - 6)^2} \\[1em] = \sqrt{0 + (-8)^2} \\[1em] = \sqrt{64} \\[1em] = 8 \text{ units}.$

Hence, Option 2 is the correct option.

The distance between the points (0, 5) and (-5, 0) is

1. 5 units

2. $5\sqrt{2}$ units

3. $2\sqrt{5}$ units

4. 10 units

Answer

Let, A = (0, 5) and B = (-5, 0).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(-5 - 0)^2 + (0 - 5)^2} \\[1em] = \sqrt{(-5)^2 + (-5)^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2}\text{ units}.$

Hence, Option 2 is the correct option.

AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of its diagonal is

1. 5 units

2. 3 units

3. $\sqrt{34}$ units

4. 4 units

Answer

Since, AOBC is the rectangle.

So, AB will be the diagonal.

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(5 - 0)^2 + (0 - 3)^2} \\[1em] = \sqrt{(5)^2 + (-3)^2} \\[1em] = \sqrt{25 + 9} \\[1em] = \sqrt{34} \text{ units}.$

Hence, Option 3 is the correct option.

If the distance between the points (2, -2) and (-1, x) is 5 units, then one of the value of x is

1. -2

2. 2

3. -1

4. 1

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

Distance between (2, -2) and (-1, x) is 5 units.

$\therefore 5 = \sqrt{(-1 - 2)^2 + [x - (-2)]^2} \\[1em] \Rightarrow 5 = \sqrt{(-3)^2 + [x + 2]^2} \\[1em] \Rightarrow 5 = \sqrt{9 + x^2 + 4 + 4x} \\[1em] \Rightarrow 5 = \sqrt{x^2 + 4x + 13} \\[1em] \Rightarrow x^2 + 4x + 13 = 5^2 \text{ [On squaring both sides]} \\[1em] \Rightarrow x^2 + 4x + 13 = 25 \\[1em] \Rightarrow x^2 + 4x - 12 = 0 \\[1em] \Rightarrow x^2 + 6x - 2x - 12 = 0 \\[1em] \Rightarrow x(x + 6) - 2(x + 6) = 0 \\[1em] \Rightarrow (x - 2)(x + 6) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -6.$

As distance cannot be negative,

∴ x = 2

Hence, Option 2 is the correct option.

The distance between the points (4, p) and (1, 0) is 5 units, then the value of p is

1. 4 only

2. -4 only

3. ±4

4. 0

Answer

By distance formula,

d = $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Given,

Distance between (4, p) and (1, 0) is 5 units.

$\therefore 5 = \sqrt{(1 - 4)^2 + (0 - p)^2} \\[1em] \Rightarrow 5 = \sqrt{(-3)^2 + (-p)^2} \\[1em] \Rightarrow 5 = \sqrt{9 + p^2} \\[1em] \Rightarrow p^2 + 9 = 5^2 \text{ [On squaring both sides]} \\[1em] \Rightarrow p^2 + 9 = 25 \\[1em] \Rightarrow p^2 = 16 \\[1em] \Rightarrow p = \sqrt{16} \\[1em] \Rightarrow p = \pm 4.$

Hence, Option 3 is the correct option.

The points (-4, 0), (4, 0) and (0, 3) are the vertices of a

1. right triangle

2. isosceles triangle

3. equilateral triangle

4. scalene triangle

Answer

Let A = (-4, 0), B = (4, 0) and C = (0, 3).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{[4 - (-4)]^2 + (0 - 0)^2} \\[1em] = \sqrt{[4 + 4]^2 + 0^2} \\[1em] = \sqrt{8^2} \\[1em] = \sqrt{64} \\[1em] = 8 \text{ units} \\[1em] \therefore BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} \\[1em] = \sqrt{(-4)^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[1em] \therefore AC = \sqrt{[0 - (-4)]^2 + (3 - 0)^2} \\[1em] = \sqrt{4^2 + 3^2} \\[1em] = \sqrt{16 + 9} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units} \\[1em]$

Since, AC = BC.

∴ ABC is an isosceles triangle.

Hence, Option 2 is the correct option.

The area of a square whose vertices are A(0, -2), B(3, 1), C(0, 4) and D(-3, 1) is

1. 18 sq. units

2. 15 sq. units

3. $\sqrt{18}$ units

4. $\sqrt{15}$ units

Answer

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(3 - 0)^2 + [1 - (-2)]^2} \\[1em] = \sqrt{3^2 + [1 + 2]^2} \\[1em] = \sqrt{9 + 3^2} \\[1em] = \sqrt{9 + 9} \\[1em] = \sqrt{18} \\[1em] = 3\sqrt{2}.$

Area of square = (side)² = AB²

= $(3\sqrt{2})^2$ = 18 sq. units.

Hence, Option 1 is the correct option.

In the adjoining figure, the area of triangle ABC is

1. 15 sq. units

2. 10 sq. units

3. 7.5 sq. units

4. 2.5 sq. units

Answer

Draw a perpendicular AD from A on x-axis.

From graph,

D = (1, 0).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AD = \sqrt{(1 - 1)^2 + (0 - 3)^2} \\[1em] = \sqrt{0 + (-3)^2} \\[1em] = \sqrt{9} \\[1em] = 3 \text{ units}. \\[1em] \therefore BC = \sqrt{[4 - (-1)]^2 + (0 - 0)^2} \\[1em] = \sqrt{[4 + 1]^2 + 0} \\[1em] = \sqrt{5^2} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Area of triangle = $\dfrac{1}{2} \times$ base × height

= $\dfrac{1}{2} \times 3 \times 5$

= $\dfrac{15}{2}$ = 7.5 sq. units

Hence, Option 3 is the correct option.

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is

1. 5 units

2. 12 units

3. 11 units

4. 7 + $\sqrt{5}$ units

Answer

Let A(0, 4), B(0, 0) and C(3, 0) be the vertices of the triangle.

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(0 - 0)^2 + (0 - 4)^2} \\[1em] = \sqrt{0 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] \therefore BC = \sqrt{(3 - 0)^2 + (0 - 0)^2} \\[1em] = \sqrt{3^2 + 0} \\[1em] = \sqrt{9} \\[1em] = 3 \text{ units} \\[1em] \therefore AC = \sqrt{(3 - 0)^2 + (0 - 4)^2} \\[1em] = \sqrt{3^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = 5 \text{ units}.$

Perimeter = AB + BC + AC = 4 + 3 + 5 = 12 units.

Hence, Option 2 is the correct option.

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

1. 8 units

2. 5 units

3. 3 units

4. 25 units

Answer

Since, A is a point on y-axis, so x-coordinate = 0.

A = (0, 5).

B = (-3, 1).

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] = \sqrt{(-3 - 0)^2 + (1 - 5)^2} \\[1em] = \sqrt{(-3)^2 + (-4)^2} \\[1em] = \sqrt{9 + 16} \\[1em] = \sqrt{25} \\[1em] = 5 \text{ units}.$

Hence, Option 2 is the correct option.

The points A(9, 0), B(9, 6), C(-9, 6) and D(-9, 0) are the vertices of a

1. rectangle

2. square

3. rhombus

4. trapezium

Answer

By distance formula,

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\[1em] \therefore AB = \sqrt{(9 - 9)^2 + (6 - 0)^2} \\[1em] = \sqrt{0 + 6^2} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] \therefore BC = \sqrt{(-9 - 9)^2 + (6 - 6)^2} \\[1em] = \sqrt{(-18)^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units}. \\[1em] \therefore CD = \sqrt{[-9 - (-9)]^2 + (0 - 6)^2} \\[1em] = \sqrt{[-9 + 9]^2 + (-6)^2} \\[1em] = \sqrt{0 + 36} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ units}. \\[1em] \therefore AD = \sqrt{(-9 - 9)^2 + (0 - 0)^2} \\[1em] = \sqrt{(-18)^2 + 0} \\[1em] = \sqrt{324} \\[1em] = 18 \text{ units}.$

Since, AB = CD and BC = AD.

∴ ABCD is a rectangle.

Hence, Option 1 is the correct option.

Consider the following two statements:

Statement 1: The point (x², y) lies on the y - axis. Then the value of x is zero.

Statement 2: The abscissa of every point on y-axis is zero.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

Any point that lies on the y-axis has its x-coordinate equal to zero.

The given point is (x², y).

For this point to lie on the y-axis, its x-coordinate, which is x², must be equal to zero.

So, x² = 0.

If x² = 0, then x must be 0.

∴ Statement 1 is true.

The abscissa is the x-coordinate of a point in a Cartesian coordinate system.

All points on the y-axis are of the form (0, y), where y can be any real number.

For any point on the y-axis, its x-coordinate (abscissa) is 0.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.