Question

Prove that the distance between the origin and the point (–6, –8) is twice the distance between the points (4, 0) and (0, 3).

Answer 1

Given: O ≡ (0, 0); A ≡ (–6, –8); B ≡ (4, 0); C ≡ (0, 3).

To Prove: Distance OA = 2 · distance BC.

Proof [Step-wise]:

1. Use the distance formula: distance between (x₁, y₁) and (x₂, y₂) is `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`.

2. Compute OA:

OA = Distance between (0, 0) and (–6, –8)

= `sqrt((-6 - 0)^2 + (-8 - 0)^2`

= `sqrt(36 + 64)`

= `sqrt(100)`

= 10

3. Compute BC:

BC = Distance between (4, 0) and (0, 3)

= `sqrt((0 - 4)^2 + (3 - 0)^2` 

= `sqrt(16 + 9)`

= `sqrt(25)`

= 5

4. Compare:

OA = 10 

= 2 × 5 

= 2 · BC

Therefore, the distance from the origin to (–6, –8) is twice the distance between (4, 0) and (0, 3).