which point is on the line that passes through point h and is perpendicular to line fg? (-6, 10) (-2, -12)…

Explanation:

Step1: Find the slope of line FG

Let (F=( - 8,-8)) and (G=(8,4)). The slope formula is (m=\frac{y_2 - y_1}{x_2 - x_1}). So, (m_{FG}=\frac{4-( - 8)}{8-( - 8)}=\frac{4 + 8}{8 + 8}=\frac{12}{16}=\frac{3}{4}).

Step2: Find the slope of the perpendicular line

The slope of a line perpendicular to a line with slope (m) is (m'=-\frac{1}{m}). So the slope of the line perpendicular to (FG) is (m'=-\frac{4}{3}). Let (H=(6,-6)).

Step3: Use the point - slope form (y - y_1=m'(x - x_1))

The point - slope form of a line is (y+6=-\frac{4}{3}(x - 6)). Expand it: (y+6=-\frac{4}{3}x+8), then (y=-\frac{4}{3}x + 2).

Step4: Check each point

For point ((-6,10)): Substitute (x=-6) into (y=-\frac{4}{3}x + 2), (y=-\frac{4}{3}\times(-6)+2=8 + 2=10). For point ((-2,-12)): Substitute (x = - 2) into (y=-\frac{4}{3}x+2), (y=-\frac{4}{3}\times(-2)+2=\frac{8}{3}+2=\frac{8 + 6}{3}=\frac{14}{3}\neq-12). For point ((0,-2)): Substitute (x = 0) into (y=-\frac{4}{3}x+2), (y=2\neq-2). For point ((4,2)): Substitute (x = 4) into (y=-\frac{4}{3}x+2), (y=-\frac{4}{3}\times4+2=-\frac{16}{3}+2=-\frac{16 - 6}{3}=-\frac{10}{3}\neq2).

Answer:

((-6,10))