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

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 \(m_{FG}\times m_1=-1\). Since \(m_{FG}=\frac{3}{4}\), we have \(\frac{3}{4}m_1=-1\), so \(m_1 =-\frac{4}{3}\).

Step3: Assume point \(H(x_0,y_0)\) and use point - slope form

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

\((4,2)\)