Chapter 13, Problem 10CR

To determine

To prove: $QRST$ is a trapezoid.

Explanation of Solution

Given information:

Four vertices of trapezoid are,

  $Q(-1,-2)$ , $R(6,-2)$ , $S(10,1)$ and $T(7,4)$

Formula used: Slope of a line passing through two point is $\frac{y_2-y_1}{x_2-x_1}$

Proof: we have to show that two of the four line-segment are parallel.

As we know that parallel lines have equal slope let’s find slope for each line segment.

Slope of $QR$ is $\frac{y_2-y_1}{x_2-x_1}=\frac{6-(-1)}{-2-(-2)}=0$

Slope of $RS$ is $\frac{y_2-y_1}{x_2-x_1}=\frac{10-6}{1-(-2)}=\frac{4}{3}$

Slope of $ST$ is $\frac{y_2-y_1}{x_2-x_1}=\frac{10-7}{1-4}=1$

Slope of $TQ$ is $\frac{y_2-y_1}{x_2-x_1}=\frac{7-(-1)}{4-(-2)}=\frac{4}{3}$

As from the above it is obtained that the slope of $RS$ and $TQ$ are same. That means $RS$ and $TQ$ are parallel.

Hence, $QRST$ is a trapezoid.