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 \(F(x_1,y_1)=(- 8,-8)\) 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 the equation of the line passing through \(H(4,-6)\)

The point - slope form of a line is \(y - y_0=m(x - x_0)\), where \((x_0,y_0)=(4,-6)\) and \(m =-\frac{4}{3}\). So the equation is \(y+6=-\frac{4}{3}(x - 4)\), which simplifies to \(y+6=-\frac{4}{3}x+\frac{16}{3}\), and then \(y=-\frac{4}{3}x+\frac{16}{3}-6=-\frac{4}{3}x-\frac{2}{3}\).

Step4: Check each point

For point \((-6,10)\): Substitute \(x=-6\) into \(y =-\frac{4}{3}x-\frac{2}{3}\), we get \(y=-\frac{4}{3}\times(-6)-\frac{2}{3}=8-\frac{2}{3}=\frac{24 - 2}{3}=\frac{22}{3}
eq10\).
For point \((-2,-12)\): Substitute \(x = - 2\) into \(y=-\frac{4}{3}x-\frac{2}{3}\), we get \(y=-\frac{4}{3}\times(-2)-\frac{2}{3}=\frac{8}{3}-\frac{2}{3}=2
eq - 12\).
For point \((0,-2)\): Substitute \(x = 0\) into \(y=-\frac{4}{3}x-\frac{2}{3}\), we get \(y=-\frac{2}{3}
eq - 2\).
For point \((4,2)\): Substitute \(x = 4\) into \(y=-\frac{4}{3}x-\frac{2}{3}\), we get \(y=-\frac{4}{3}\times4-\frac{2}{3}=\frac{-16 - 2}{3}=-6\).

Answer:

\((4,2)\)