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:

 A(-1, 1) B, (5, 1), C(5, 6) and D(-1, 6)

Formula used:

Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃)

=  \(\dfrac{1}{2}\) |x₁(y₂-y₃) +x₂(y₃-y₁) +x₃(y₁-y₂) |

Calculations:

Area of ∆ABC =  \(\dfrac{1}{2}\)  | − 1[1 − 6)] + (5) (6 - 1) + 5 [1 - 1)] |

=  \(\dfrac{1}{2}\)  | 5 + 25 |

= 15 sq. units

Area of ∆ACD =  \(\dfrac{1}{2}\) |− 1(6 – 6) + 5(6 – 1) + (-1)( 1 − 6) |

=  \(\dfrac{1}{2}\) | 25 + 5 |

= 15 sq. units

Area of the quadrilateral ABCD = 15 + 15

= 30 sq. units

∴ the area of the quadrilateral is 30 sq. units.

∴ The answer is 30 sq.units.

Given:

Points: P(-2, 1), Q(α, β), R(4, -1)

α - β = -3

Formula used:

For collinear points, slope of PQ = slope of QR

Slope formula: (y₂ - y₁) / (x₂ - x₁)

Calculation:

Slope of PQ = (β - 1) / (α + 2)

Slope of QR = (-1 - β) / (4 - α)

Since P, Q, R are collinear:

⇒ (β - 1) / (α + 2) = (-1 - β) / (4 - α)

Cross-multiplying:

⇒ (β - 1)(4 - α) = (-1 - β)(α + 2)

⇒ 4β - β α - 4 + α = -α -2 - β α - 2β

⇒ 4β + α + α + 2β = -2 + 4

⇒ 3β + α = 1

Given that α - β = -3:

From α - β = -3 ⇒ α = β - 3

Substitute in 3β + α = 1:

⇒ 3β + (β - 3) = 1

⇒ 4β - 3 = 1

⇒ β = 1

So, α = 1 - 3 = -2

α + β = -2 + 1 = -1

∴ The correct answer is -1.

Given:

x + 4 = 0

Calculations:

x = – 4

The equation on the graph is as shown below:

∴ The above equation is parallel to y-axis and passes through (-4, 0)

Concept:

1. The slope of a line passing through the distinct points (x1, y1) and (x2, y2) is

m =  \(\rm { (y_2 - y_1)}\over{(x_2 - x_1 )}\)

2. If two lines having slopes m₁ and m₂ are perpendicular to each other then m₁m₂ = - 1

Calculation:

Let M(h, k) is required point on the line

x + y = 3          ------(1)

On comparing it from y = mx + c, we will get

The slope of a given line m = -1  

Let the slope of line PM is m'

∵ line PM is perpendicular to a given line, therefore

mm' = -1 ⇒ m' = 1

So the slope of the perpendicular line PM is 1.

Also the slope of PM = \(\frac{k\ -\ 3}{h\ -\ 2}\ =\ 1\)

⇒  k – 3 = h – 2

⇒  h – k = - 1       ------(2)

Since point M lying on the given line, therefore it will satisfy the equation (1)

⇒ h + k = 3       ------(3)

Solving (2) and (3), we get

h = 1, k = 2

So the coordinates of M are (1, 2).

Hence option (2) is correct.

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.

Formula:

Slope of the line = (y₂ – y₁)/(x₂ – x₁)

Given:

y₂ = 2,    y₁ = -4,    x₂ = 5,    x₁ =3

Calculation:

⇒ {2 – (- 4)}/{5 – 3}

⇒ (6)/(2)

⇒ 3

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.

Let the reflected point be (x, y).

Then the midpoint of (-1, 3) and (x, y) i.e. [(x-1)/2, (y+3)/2] must lie on the line x = -4.

⇒ (x – 1) / 2 = -4

⇒ x = -8 + 1 = -7

Since the line joining the original and the reflected point in perpendicular to x = -4, hence its slope will be equal to 0.

⇒ [y – 3]/[(-7) – (-1)] = 0

⇒ y – 3 = 0

⇒ y = 3

Given:

x₁ = 4, x₂ = 3, y₁  = 3, y₂ = - 2

Formula:

Distance between two points = √[(x₁ – x₂)² + (y₁ – y₂)²]

Calculation:

√[(4 – 3)2 + (3 – {-2})2]

⇒ √[(1)² + (5)²]

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)

Concept:

Consider expression: a₁x + b₁y + c = 0 and a₂x + b₂y + c = 0

Condition for the system has unique solution if a1/a2 ≠ b1/b2

Condition for the system has infinite solution if a1/a2 = b1/b2 = c₁/c₂

Condition for the system has no solution if a1/a2 = b1/b2 ≠ c1/c2

Calculation:

The equation is x – Ky = 2, 3x + 2y = 5

Here, a1 = 1, b1 = -k, a2 = 3, b2 = 2

1/3 ≠ -k/2

k ≠ -2/3

∴ The required value of K should not be equal to -2/3