Question

Give one pair of values of x and y such that (x, y) is equidistant from the points
(-1,8) and (3, 4). Justify your steps.​

Answer 1

Step-by-step explanation:

Let P(x, y) be equidistant from the given points A(-1, 8) & B(3, 4)

$\therefore PA = PB\\</p><p>\therefore PA^2 = PB^2 \\ \therefore(x + 1)^{2} + (y - 8)^{2} \\ = (x - 3)^{2} + (y - 4)^{2} \\ \therefore \: {x}^{2} + 2x + 1 + {y}^{2} - 16y + 64 \\ = {x}^{2} - 6x + 9 + {y}^{2} - 8y + 16 \\ \therefore \: 2x - 16y + 65 = - 6x - 8y + 25 \\ \therefore \: 2x - 16y + 65 + 6x + 8y - 25 = 0 \\ \therefore \: 8x - 8y + 50 = 0 \\ \therefore \: 4x - 4y + 25 = 0 \\ \therefore \:4y = 4x + 25 \\ \huge \red{\boxed{\therefore \:y = x + \frac{25}{4} }} \\ \\Now\: at \: x = 1 \\ y = \frac{29}{4} \\ \therefore y = 7.25$

Thus, (x, y) = (1, 7.25) is equidistant from the given points A(-1, 8) & B(3, 4).