Question

Point A and B have co-ordinates (7, −3) and (1, 9) respectively. Find:

1. the slope of AB.
2. the equation of perpendicular bisector of the line segment AB.
3. the value of ‘p’ of (−2, p) lies on it.

Answer 1

Coordinates of A are (7, −3) of B = (1, 9)

i. ∴ Slope (m) = `(y_2 - y_1)/(x_2 - x_1)`

= `(9 - (-3))/(1 - 7)`

= `(9 + 3)/(1 - 7)`

= `12/(-6)`

= −2

ii. Let PQ is the perpendicular bisector of AB intersecting it at M.

∴ Co-ordinates of M will be

= `((x_1 + x_2)/2, (y_1 + y_2)/2)`

= `((7 + 1)/2, (-3 + 9)/2)`

= `(8/2, 6/2)`

= (4, 3)

∴ Slope of PQ = `1/2`  ...(m₁m₂ = −1)

∴ Equation of PQ is given by

y − y₁ = m(x − x₁)

`\implies y - 3 = 1/2 (x - 4)`

`\implies` 2y − 6 = x − 4

`\implies` x − 2y + 6 − 4 = 0

`\implies` x − 2y + 2 = 0   ...(i)

iii. ∵ Point (−2, p) lies on equation (i); we get

∴ −2 − 2p + 2 = 0

`\implies` −2p + 0 = 0

`\implies` −2p = 0

∴ p = 0