Question

given n(1,6), o(-7,-2), p(-5,-8), and q(-7,y). find y such that $overline{no} parallel overline{pq}$.

Explanation:

Step1: Find slope of $\overline{NO}$

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points $N(1,6)$ and $O(-7,-2)$, we have $m_{NO}=\frac{-2 - 6}{-7 - 1}=\frac{-8}{-8}=1$.

Step2: Find slope of $\overline{PQ}$

For points $P(-5,-8)$ and $Q(-7,y)$, the slope $m_{PQ}=\frac{y+8}{-7 + 5}=\frac{y + 8}{-2}$.

Step3: Set slopes equal

Since $\overline{NO}\parallel\overline{PQ}$, their slopes are equal. So $1=\frac{y + 8}{-2}$.

Step4: Solve for $y$

Cross - multiply: $-2=y + 8$. Then subtract 8 from both sides: $y=-2-8=-10$.

Answer:

$y=-10$