Question

use the table to answer the question.
coordinates of line ab and cd

line	point 1	point 2
cd	(3,5)	(_,-1)

the table includes two points that fall on each of the lines, line ab and line cd. what must the value of the missing coordinate be in order to prove the lines are perpendicular?
(1 point)
○ 1
○ -1
○ 5
○ -15

Explanation:

Step1: Calculate slope of line AB

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For line $AB$ with points $(-3,6)$ and $(3,8)$, we have $m_{AB}=\frac{8 - 6}{3-(-3)}=\frac{2}{6}=\frac{1}{3}$.

Step2: Determine slope of line CD for perpendicularity

If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of line $CD$ be $m_{CD}$. Since $m_{AB}\times m_{CD}=-1$ and $m_{AB}=\frac{1}{3}$, then $m_{CD}=-3$.

Step3: Use slope formula for line CD to find missing x - coordinate

Let the missing x - coordinate of the point on line $CD$ be $x$. Using the slope formula $m_{CD}=\frac{y_2 - y_1}{x_2 - x_1}$ with points $(3,5)$ and $(x,-1)$ and $m_{CD}=-3$. We have $-3=\frac{-1 - 5}{x - 3}$. Cross - multiply: $-3(x - 3)=-6$. Expand: $-3x+9=-6$. Subtract 9 from both sides: $-3x=-15$. Divide by $-3$: $x = 5$.

Answer:

$5$