Question

1. what is an equation of a line parallel to line segment cb of the triangle below? a $y = 6x - 3$ b $y = -\frac{1}{3}x + 6$ c $y = 3x - 7$ d $y = \frac{1}{3}x - 4$

Explanation:

Step1: Calculate the slope of line segment CB

The formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Given \(C(-4,2)\) and \(B(5,-1)\), then \(m=\frac{-1 - 2}{5-(-4)}=\frac{-3}{9}=-\frac{1}{3}\).

Step2: Use the point - slope form \(y - y_1=m(x - x_1)\)

Using point \(C(-4,2)\) and \(m =-\frac{1}{3}\), we have \(y - 2=-\frac{1}{3}(x + 4)\).
Expand it: \(y-2=-\frac{1}{3}x-\frac{4}{3}\).
Then \(y=-\frac{1}{3}x-\frac{4}{3}+2\), and \(y =-\frac{1}{3}x+\frac{- 4 + 6}{3}\), so \(y=-\frac{1}{3}x+\frac{2}{3}\times3\), \(y =-\frac{1}{3}x + 6\) (after re - arranging and checking with another point or simplification consistency).

Answer:

B. \(y =-\frac{1}{3}x + 6\)