Question

consider line segments pq, ts and sw represented below in the cartesian plane.

• the equation associated with line segment pq is $y = \frac{3}{5}x + 61$.
• line segment ts is parallel to line segment pq.
• the $y$-coordinate of point s is 56.

what is the equation associated with line segment sw?

Explanation:

Step1: Find slope of TS

TS || PQ, so slope of TS = slope of PQ = $\frac{3}{5}$.

Step2: Find equation of TS

TS passes through T(0,20), so equation: $y = \frac{3}{5}x + 20$.

Step3: Find coordinates of S

y-coordinate of S is 56. Substitute into TS equation:
$56 = \frac{3}{5}x + 20$
$\frac{3}{5}x = 36$
$x = 60$. So S(60,56).

Step4: Calculate slope of SW

Points S(60,56) and W(82,45). Slope $m = \frac{45-56}{82-60} = \frac{-11}{22} = -\frac{1}{2}$.

Step5: Find equation of SW

Use point S(60,56) and slope $-\frac{1}{2}$:
$y - 56 = -\frac{1}{2}(x - 60)$
$y = -\frac{1}{2}x + 30 + 56$
$y = -\frac{1}{2}x + 86$.

Answer:

$y = -\frac{1}{2}x + 86$