Chapter 7: Coordinate Geometry

True.

Justification:

Using distance formula, d = √ ((x₂ – x₁)² + (y₂ – y₁)²)

We can find,

Hence, triangle ABC and DEF are similar.

True.

Justification:

Plotting the points P(- 4, 2), A (- 4, 6) and B (- 4, – 6) on a graph paper and connecting the points we get the graph,

Hence, from the graph it is clear that, point P ( – 4, 2) lies on the line segment joining the points A ( – 4, 6) and B ( – 4, – 6),

False

Justification:

The points are collinear if area of a triangle formed by its points is equals to the zero.

Given,

x₁ = 0, x₂ = 0, x₃ = 3 and

y₁ = 5, y₂ = – 9, y₃ = 6

From the above equation, it is clear that the points are not collinear.

False

Justification:

We know that, the points lying on perpendicular bisector of the line segment joining the two points is equidistant from the two points.

i.e., PA should be equals to the PB.

Using distance formula,

d = √ ((x₂ – x₁)² + (y₂ – y₁)²)

True.

Justification:

Coordinates of A = (x₁, y₁) = (3, 1)

Coordinates of B = (x₂, y₂) = (12, – 2)

Coordinates of C = (x₃, y₃) = (0, 2)

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

Δ = ½ [3 – (2 – 2) + 12(2 – 1) + 0{1 – (- 2)}]

Δ = ½ [3(- 4) + 12(1) + 0]

Δ = ½ (- 12 + 12) =0

Area of ΔABC = 0

Since, the points A (3, 1), B (12, – 2) and C (0, 2) are collinear.

Therefore, the points A (3, 1), B (12, – 2) and C (0, 2) can’t be the vertices of a triangle.

False

Justification:

The given points are A (4, 3), B (6, 4), C (5, –6) and D (–3, 5)

Finding the distance between A and B

Finding the distance between D and A

Since the distances are different, we can conclude that the points are not the vertices of a parallelogram.