Co-ordinate Geometry MCQ Quiz - Objective Question with Answer for Co-ordinate Geometry - Download Free PDF

Given:

Vertices are (3, 5), (- 2, 0) and (6, 4)

Formula used:

Area of triangle = 1/2[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Calculations:

 Area of triangle = 1/2[3(0 – 4) - 2(4 – 5) + 6(5 – 0)]

⇒ 1/2[- 12 + 2 + 30]

⇒ 1/2 × 20

⇒ 10

⇒ Area of triangle = 10 unit2

∴ The area of the triangle is 10 unit2

Given:

The point = P (2, 3)

Formula used:

The distance between two points A (x, y) and B (a, b) = \(\sqrt {{(a\ -\ x)^2}\ +\ {(b\ -\ y)^2}}\)

Calculation:

Let us assume the distance be X

⇒ The point of the x-axis = (2, 0)

⇒ The distance between point P from point x-axis = \(\sqrt {{(2\ -\ 2)^2}\ +\ {(0\ -\ 3)^2}}\) = √(0 + 9) = √9 = 3 m

∴ The required result will be 3 m.

Given:

(x, y) divides the line segment joining (4, 3) and (2, 5) in the ratio 1 : 1 internally.

Formula Used:

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Calculation:

x = (mx₂ + nx₁)/(m + n)

⇒ x = (1 × 2 + 1 × 4)/(1 + 1)

⇒ x = (2 + 4)/2

⇒ x = 6/2

⇒ x = 3

y = (my₂ + ny₁)/(m + n)

⇒ y = (1 × 5 + 1 × 3)/(1 + 1)

⇒ y = (5 + 3)/2

⇒ y = 8/2

⇒ y = 4

∴ The value of (x, y) is (3, 4).

Given:

m : n = 2 : 5

If the point C (1, 1) divides the line segment AB internally in ratio 2 ∶ 5, where A (3, 7). 

Concept used: 

P(x, y) = \(\dfrac{mx_2 + nx_1}{m + n} ,\dfrac{my_2 + ny_1}{m + n} \)

Calculation:

⇒ (1, 1) = \(\dfrac{2x + 5(3)}{2 + 5} ,\dfrac{2y + 5(7)}{2 + 5} \)

⇒ 2x = 7 - 15

⇒ x = -4

⇒ 2y = 7 - 35

⇒ y = -14

∴ B = (-4, -14)

Given:-

Vertices of triangle = (1,2), (-4,-3), (4,1)

Formula Used:

Area of triangle = ½ [x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)]

whose vertices are (x1, y1), (x2, y2) and (x3, y3)

Calculation:

⇒ Area of triangle = (1/2) × [1(-3 – 1) + (-4) (1 – 2) + 4{2 – (-3)}]

= (1/2) × {(-4) + 4 + 20}

= 20/2

= 10 sq. units

A(4, 1), B(1, 1) and C(3, 5) are the vertices of a triangle. Then,

⇒ AB² = (1 - 1)² + (1 - 4)² = 9

⇒ BC² = (5 - 1)² + (3 - 1)² = 20

⇒ AC² = (5 - 1)² + (3 - 4)² = 17

Given:

Coordinate of the centroid = (4,8)

Coordinate of the vertex 1 = (9,7)

Coordinate of the vertex 2 = (1,4)

Concept used:

If the coordinates of the vertices of a triangle are (x1, y1), (x2, y2), (x3, y3), then the formula for the centroid of the triangle is given below:

The centroid of a triangle = ((x1 + x2+ x3)/3, (y1+ y2 + y3)/3)

 \(Area = \frac{1}{2}\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\\ {{x_2}}&{{y_2}}&1\\ {{x_3}}&{{y_3}}&1 \end{array}} \right| \)

Calculation:

Let the coordinate of the third vertex be (a,b).

According to the question,

(a + 9 + 1) ÷ 3 = 4

⇒ a = 2

(b + 7 + 4) ÷ 3 = 8

⇒ b = 13

So, the coordinate of the third vertex is (2,13)
Coordinates of three vertices of triangle are (9,7) , (2,13) & (1,4).
We can calculate the area of a triangle By the Vertex formula.

\(A = \frac{1}{2}\left| {\begin{array}{*{20}{c}} 9&7&1\\ 2&{13}&1\\ 1&4&1 \end{array}} \right|\ \)

So, the area of the triangle

A = (1/2) [9(13 - 4) + 2(4 - 7) + 1(7 - 13)] = 34.5 Unit²

∴ The area of triangle is 34.5 unit².

Given:

The distance between two points (-6, y) and (18, 6) is 26 units.

Formula used:

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

Where, 

The distance between two points (x₁, y₁) and (x₂, y₂) is D units.

Calculation:

The distance between two point = 26 units

The value of the first co-ordinate = (x₁, y₁) = (-6, y)

The value of the second co-ordinate = (x2, y2) = (18, 6)

According to the question,

⇒ D = \(\sqrt{(18-(-6))^2 + (6-y)^2} \)

⇒ 26 = \(\sqrt{(24)^2 + (6-y)^2} \)

Squaring on both sides of the equation.

⇒ 676 = 24² + (6 - y)²

⇒ 676 = 576 + (6 - y)2

⇒ 100 = (6 - y)2

⇒ 10² = (6 - y)2

Taking square root on both sides of the equation.

⇒ 10 = 6 - y

⇒ y = -4

∴ The required answer is -4.

Concept Used:

Equation of a circle with center (h, k) and radius r is

(x - h)² + (y - k)² = r²

Calculation:

Distance between the center and the end of the diameter is the radius.

\( ⇒ r = \sqrt{(2-(-2))^2+(3-5)^2} = \sqrt{4^2+2^2} = \sqrt{20}\)

Hence equation of given circle be (x - (-2))2 + (y - 5)2 = 20

⇒(x + 2)2 + (y - 5)2 = 20

Now, another end of the diameter should also satisfy the equation.
From options, there is only one point that satisfies the equation of circle i.e., (-6, 7)

⇒ (-6 + 2)2 + (7 - 5)2 =  (-4)2 + (2)2 = 16 + 4 = 20 = RHS

Hence, (-6, 7) is another end of the diameter.

Shortcut TrickCalculation:

The center of a circle lies at the mid-point of diameter.

By using mid-point theorem

⇒ (2 + x)/2 = - 2

⇒ x = (- 4 - 2) = - 6

And,

⇒ (3 + y)/2 = 5

⇒ y = (10 - 3) = 7

∴ The co-ordinate of other end of diameter = (- 6, 7).

⇒ Section formula for internal division = {[(mx₂ + nx₁)/(m + n)], [(my₂ + ny₁)/(m + n)]}

⇒ Here, x₁, y₁ = (- 1, 0) and x₂, y₂ = (2, 6). m : n = 2 : 1

⇒ [(2 × 2) + (1 × - 1)]/(2 + 1), [(2 × 6) + (1 × 0)]/(2 + 1) = (1, 4)

Given:

For Quadrilateral ABCD,

A (- 4, - 2)

B (- 3, - 5)

C (3, - 2)

D (2, 3)

Concept:

If a ΔABC is formed from points A (x₁, y₁), B (x₂, y₂), C (x₃, y₃)

then, the area of ΔABC = \(\dfrac{1}{2}\)[x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂)]

Calculation:

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔADC

Using the formula,

A (- 4, - 2), B (- 3, - 5), C (3, - 2)

Area of ΔABC = \(\dfrac{1}{2}\)[- 4 (- 5 + 2) + (- 3) (- 2 + 2) + 3 (- 2 + 5)]

= (1 / 2) × (12 + 9)

= (21 / 2) sq. unit

Now,

A (- 4, - 2), D (2, 3), C (3, - 2)

Area of ΔADC = \(\dfrac{1}{2}\)[- 4 (3 + 2) + 2 (- 2 + 2) + 3 (- 2 - 3)]

= (1 / 2) × (- 20 - 15)

= (35 / 2) sq. unit (Area can not be Negative)

Area of quadrilateral ABCD = (21 / 2) + (35 / 2) sq. unit

= 28 sq. unit

Given:

Curve 5x + 6y - 30, a linear equation

Concept:

Intercept is the shadow region on the axis by a line

OR where the line cuts the axis.

So, suppose line cuts x-axis at point(a,0), remember that y will always be zero on the entire x-axis.

Calculation:

Put the equation equals zero

⇒ 5x + 6y - 30 = 0

⇒ 5x + 6y = 30

For x-intercept, put y = 0 in the equation

⇒ 5x = 30

⇒ x = 6

∴ x-intercept of the equation = 6

Given:

Lines are 2x + 5y = 12    ----(i)

x + y = 3     ----(ii)

And x axis i.e y = 0

Calculation:

Point of intersection of equation (i) and (ii), we get 

(2x + 5y) - (2x + 2y) = 12 - 2 × 3

⇒ 3y = 6

⇒ y = 2 

Form (ii), we get 

⇒ x = 1

Point of intersection of equation (i) and (ii) is (1, 2)

Point of intersection equation (i) and x-axis is (6, 0)

Point of intersection equation (ii) and x axis is (3, 0)

From graph we have to find the area of triangle 

Base (AB) = 6 - 3 = 3 unit 

Height (CM) = Y-coordinate = 2 unit

∴ The required area = (1/2) × 3 × 2 = 3 sq. unit

As we know,

Area of triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) = 1/2 × [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)]

Area of the triangle whose vertices are A(-4, -2), B(-3, -5) and C(3, -2)

⇒ 1/2 × [{(-4) × (-3)} + {(-3) × 0} + {3 × 3}]

⇒ 1/2 × [12 + 0 + 9]

⇒ (1/2) × 21