Question

which point is on the line that passes through point h and is perpendicular to line fg? (-6, 10) (-2, -12) (0, -2) (4, 2)

Explanation:

Step1: Find slope of line FG

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

Step2: Find slope of 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 \(m'=-\frac{4}{3}\).

Step3: Assume point \(H=(6,-6)\) and use point - slope form

The point - slope form of a line is \(y - y_1=m'(x - x_1)\). Substituting \(m'=-\frac{4}{3}\) and \((x_1,y_1)=(6,-6)\), we get \(y+6=-\frac{4}{3}(x - 6)\).

Step4: Check each point

For point \((-6,10)\):
Left - hand side of the line equation: \(10 + 6=16\).
Right - hand side: \(-\frac{4}{3}(-6 - 6)=-\frac{4}{3}\times(-12)=16\).
For point \((-2,-12)\):
Left - hand side: \(-12 + 6=-6\).
Right - hand side: \(-\frac{4}{3}(-2 - 6)=-\frac{4}{3}\times(-8)=\frac{32}{3}
eq-6\).
For point \((0,-2)\):
Left - hand side: \(-2 + 6 = 4\).
Right - hand side: \(-\frac{4}{3}(0 - 6)=-\frac{4}{3}\times(-6)=8
eq4\).
For point \((4,2)\):
Left - hand side: \(2+6 = 8\).
Right - hand side: \(-\frac{4}{3}(4 - 6)=-\frac{4}{3}\times(-2)=\frac{8}{3}
eq8\).

Answer:

\((-6,10)\)