Question

The coordinates of point A are (-6, -10) and the coordinates of point B are (x,-4). The distance between A to point B is 10 units. Find the x-coordinate of point B if B lies in the fourth quadrant.

Answer 1

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-10})\qquad B(\stackrel{x_2}{x}~,~\stackrel{y_2}{-4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ 10=\sqrt{[x-(-6)]^2+[-4-(-10)]^2}\implies 10=\sqrt{(x+6)^2+(-4+10)^2} \\\\\\ 10^2=(x+6)^2+(6)^2\implies 100=x^2+12x+36+36 \\\\\\ 100=x^2+12x+72\implies 0=x^2+12x-28 \\\\\\ 0=(x+14)(x-2)\implies x= \begin{cases} -14\\ \boxed{2} \end{cases}$

because B is on the IV Quadrant, the x-coordinate must be positive.