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 formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. So, $m_{MN}=\frac{2 - 0}{4-(-4)}=\frac{2}{8}=\frac{1}{4}$.

Step2: Find slope of perpendicular line

The slope of a line perpendicular to a line with slope $m$ is $m'=-\frac{1}{m}$. So, slope of line perpendicular to MN is $m'=- 4$.

Step3: Use point - slope form with point P

Assume $P(2,-4)$. Point - slope form is $y - y_1=m'(x - x_1)$. Substituting values: $y+4=-4(x - 2)$.

Step4: Find x - intercept

Set $y = 0$ in $y+4=-4(x - 2)$. Then $0 + 4=-4(x - 2)$. Expand: $4=-4x + 8$. Rearrange: $4x=4$, so $x = 1$.

Answer:

C. $(1,0)$