Question

Find a point on the x-axis , which is equidistant from the points (7,6) and (3,4).

• A. (15/2 , 0)
• B. (5, 0)
• C. (10, 0)
• D. (2, 0)

Answer 1

Answer:

A. (15/2 , 0)

let P(a,0) be the point on the x-axis that is equidistant from the points A(7,6) and B (3,4)

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$d_{PA}=\sqrt{(a-7)^2+(0-6)^2}$

$d_{PB}=\sqrt{(a-3)^2+(0-4)^2}$

${(a-7)^2+(0-6)^2}={(a-3)^2+(0-4)^2}$

$a^2-14a+49+36=a^2-6a+9+16$

$-8a+60=0$

$-8a=-60$

$a=60/8=15/2$

$=(15/2,0)$