Question

which point on the x - axis lies on the line that passes through point p and is perpendicular to line mn? (0, 1) (0, 4) (1, 0) (4, 0)

Explanation:

Step1: Find slope of line MN

Let \(M(-4,0)\) and \(N(4,2)\). Slope \(m_{MN}=\frac{y_N - y_M}{x_N - x_M}=\frac{2 - 0}{4-(-4)}=\frac{2}{8}=\frac{1}{4}\).

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 \(MN\) be \(m\). Then \(m\times\frac{1}{4}=-1\), so \(m = - 4\).

Step3: Assume point - slope form of line passing through \(P(2,-4)\)

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(2,-4)\) and \(m=-4\). So \(y+4=-4(x - 2)\).

Step4: Find x - intercept

Set \(y = 0\) in the equation \(y+4=-4(x - 2)\). Then \(0 + 4=-4(x - 2)\), \(4=-4x + 8\), \(4x=4\), \(x = 1\).

Answer:

C. \((1,0)\)