Advertisements
Advertisements
Question
Find a point which is equidistant from the points A(–5, 4) and B(–1, 6)? How many such points are there?
Advertisements
Solution
Let P(h, k) be the point which is equidistant from the points A(–5, 4) and B(–1, 6).
∴ PA = PB ...`[∵ "By distance formula, distance" = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)]`
⇒ (PA)2 = (PB)2
⇒ (– 5 – h)2 + (4 – k)2 = (– 1 – h)2 + (6 – k)2
⇒ 25 + h2 + 10h + 16 + k2 – 8k = 1 + h2 + 2h + 36 + k2 – 12k
⇒ 25 + 10h + 16 – 8k = 1 + 2h + 36 – 12k
⇒ 8h + 4k + 41 – 37 = 0
⇒ 8h + 4k + 4 = 0
⇒ 2h + k + 1 = 0 ...(i)
Mid-point of AB = `((-5 - 1)/2, (4 + 6)/2)` = (– 3, 5) ...`[∵ "Mid-point" = ((x_1 + x_2)/2, (y_1 + y_2)/2)]`
At point (– 3, 5), from equation (i),
2h + k = 2(– 3) + 5
= – 6 + 5
= – 1
⇒ 2h + k + 1 = 0
So, the mid-point of AB satisfy the equation (i).
Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the points A and B.
Replacing h, k by x, y in above equation, we have 2x + y + 1 = 0
