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

Explanation:

Step1: Find slope of line FG

Let (F(x_1,y_1)=(- 8,-8)) and (G(x_2,y_2)=(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{12}{16}=\frac{3}{4}).

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is - 1. Let the slope of the line perpendicular to FG be (m_1). Then (m_{FG}\times m_1=-1), so (m_1=-\frac{4}{3}).

Step3: Assume point - slope form of line passing through H

Let (H(x_0,y_0)=(6,-6)). The point - slope form of a line is (y - y_0=m_1(x - x_0)). So the equation of the line is (y+6=-\frac{4}{3}(x - 6)).

Step4: Check each point

For point ((-6,10))

Substitute (x=-6) and (y = 10) into (y+6=-\frac{4}{3}(x - 6)). Left - hand side: (10 + 6=16). Right - hand side: (-\frac{4}{3}(-6 - 6)=-\frac{4}{3}\times(-12)=16).

For point ((-2,-12))

Substitute (x=-2) and (y=-12) into (y + 6=-\frac{4}{3}(x - 6)). Left - hand side: (-12+6=-6). Right - hand side: (-\frac{4}{3}(-2 - 6)=-\frac{4}{3}\times(-8)=\frac{32}{3}\neq-6).

For point ((0,-2))

Substitute (x = 0) and (y=-2) into (y+6=-\frac{4}{3}(x - 6)). Left - hand side: (-2 + 6=4). Right - hand side: (-\frac{4}{3}(0 - 6)=-\frac{4}{3}\times(-6)=8\neq4).

For point ((4,2))

Substitute (x = 4) and (y = 2) into (y+6=-\frac{4}{3}(x - 6)). Left - hand side: (2+6=8). Right - hand side: (-\frac{4}{3}(4 - 6)=-\frac{4}{3}\times(-2)=\frac{8}{3}\neq8).

Answer:

((-6,10))