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प्रश्न
Point A and B have co-ordinates (7, −3) and (1, 9) respectively. Find:
- the slope of AB.
- the equation of perpendicular bisector of the line segment AB.
- the value of ‘p’ of (−2, p) lies on it.
उत्तर
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
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