Question

Find the point on X-axis which is equidistant from P ( -3 ,4) and Q (1 ,9).​

Answer 1

Answer:

Let the point on x axis be R(x, 0)

According to statement

R(x, 0) equidistant from P ( -3, 4) and Q (1, 9).

RP = RQ

${RP}^{2} = {RQ</h3><h3>}^{2} \\ {(x + 3)}^{2} + {( - 4)}^{2} = {(x - 1)}^{2} + {( - 9)}^{2} \\ {x}^{2} + 9 + 6x + 16 = {x}^{2} + 1 - 2x + 81 \\ 6x + 25 = - 2x + 82 \\ 8x = 57 \\ x = \frac{57}{8}$

Hence, the point on x axis be R(57/8, 0)

Answer 2

Answer:

pimu's answer is always correct......