lines mn and pq are parallel. lines rs and tv intersect them. which statements are true about these lines…

Explanation:

Step1: Recall slope - formula

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Find slope of line MN

Let two points on line MN be $M(-2,-1)$ and $N(3,3)$. Then $m_{MN}=\frac{3+1}{3 + 2}=\frac{4}{5}\neq\frac{2}{3}$.

Step3: Find slope of line PQ

Let two points on line PQ be $P(-2,-3)$ and $Q(3,1)$. Then $m_{PQ}=\frac{1 + 3}{3+2}=\frac{4}{5}$. Since $m_{PQ}$ is a real - valued number, it is not undefined.

Step4: Find slope of line RS

Let two points on line RS be $R(-1,4)$ and $S(1,1)$. Then $m_{RS}=\frac{1 - 4}{1+1}=-\frac{3}{2}$.

Step5: Check parallelism of RS and TV

Let two points on line TV be $T(-4,2)$ and $V(0,-4)$. Then $m_{TV}=\frac{-4 - 2}{0 + 4}=-\frac{3}{2}$. Since $m_{RS}=m_{TV}=-\frac{3}{2}$, lines RS and TV are parallel.

Step6: Check perpendicularity

The product of the slopes of two perpendicular lines is $- 1$. The slope of $MN$ (and $PQ$) is $\frac{4}{5}$, and the slope of $RS$ is $-\frac{3}{2}$. $\frac{4}{5}\times(-\frac{3}{2})=-\frac{6}{5}\neq - 1$, so RS is not perpendicular to MN and PQ.

Answer:

The slope of line RS is $-\frac{3}{2}$, Lines RS and TV are parallel.