Question

Find the coordinates of the point(s) on x-axis which are at distance of 13 units from the point A(4, 5).

• A. (1, 0) and (-8, 0)
• B. (16, 0) and (-8, 0)
• C. (8, 0) and (-1, 0)
• D. (-6, 0) and (8, 0)

Answer 1

The correct option is B (16, 0) and (-8, 0)

Given: A(4, 5), distance = 13 units
Let the point on the x-axis is P(x, 0).
So, PA = 13 units

The distance between two points $(x_1,y_1)\text{and}(x_2,y_2)$ is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$

PA = $\sqrt{(x-4{)}^2+(0-5{)}^2}$
13 = $\sqrt{(x-4{)}^2+25}$
169 = $(x-4{)}^2+25$
$(x-4{)}^2$ = 144
$x-4=\pm 12$
$x-4=12$ and $x-4=-12$
x = 16 and x = -8
Hence, points are (16, 0) and (-8, 0).