Question

which point is on the line that passes through point r and is perpendicular to line pq? (-6, 10) (-4, -8) (0, -1) (2, 4)

Explanation:

Step1: Find slope of line PQ

Let $P(x_1,y_1)=(-6,4)$ and $Q(x_2,y_2)=(4, - 4)$. The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. So, $m_{PQ}=\frac{-4 - 4}{4+6}=\frac{-8}{10}=-\frac{4}{5}$.

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 PQ be $m_1$. Then $m_{PQ}\times m_1=-1$. So, $-\frac{4}{5}\times m_1=-1$, and $m_1 = \frac{5}{4}$.

Step3: Use point - slope form with point R

Let point $R(4,2)$. The point - slope form of a line is $y - y_0=m(x - x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope. So the equation of the line passing through $R$ with slope $\frac{5}{4}$ is $y - 2=\frac{5}{4}(x - 4)$.

Step4: Check each point

For point $(-6,10)$:
Substitute $x=-6$ and $y = 10$ into $y - 2=\frac{5}{4}(x - 4)$.
Left - hand side: $10 - 2=8$.
Right - hand side: $\frac{5}{4}(-6 - 4)=\frac{5}{4}\times(-10)=-\frac{25}{2}
eq8$.
For point $(-4,-8)$:
Substitute $x=-4$ and $y=-8$ into $y - 2=\frac{5}{4}(x - 4)$.
Left - hand side: $-8 - 2=-10$.
Right - hand side: $\frac{5}{4}(-4 - 4)=\frac{5}{4}\times(-8)=-10$.
For point $(0,-1)$:
Substitute $x = 0$ and $y=-1$ into $y - 2=\frac{5}{4}(x - 4)$.
Left - hand side: $-1 - 2=-3$.
Right - hand side: $\frac{5}{4}(0 - 4)=\frac{5}{4}\times(-4)=-5
eq-3$.
For point $(2,4)$:
Substitute $x = 2$ and $y = 4$ into $y - 2=\frac{5}{4}(x - 4)$.
Left - hand side: $4 - 2=2$.
Right - hand side: $\frac{5}{4}(2 - 4)=\frac{5}{4}\times(-2)=-\frac{5}{2}
eq2$.

Answer:

B. $(-4,-8)$