Question

1. a line passes through the points (0, 8) and (6, 0).

a. create a graph of the line.
b. circle the equation that could not be used to represent the line.
i. y = -\frac{4}{3}x + 8
ii. y - 4 = \frac{4}{3}(x - 3)
iii. 3x + 4y = 24

Explanation:

Step1: Find the slope of the line

The slope \(m\) of a line passing through two points \((x_1,y_1)=(0,8)\) and \((x_2,y_2)=(6,0)\) is given by the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
Substitute the values: \(m=\frac{0 - 8}{6-0}=\frac{-8}{6}=-\frac{4}{3}\).

Step2: Check each equation

• Equation i: \(y =-\frac{4}{3}x + 8\)

This is in the slope - intercept form \(y=mx + b\) (\(m =-\frac{4}{3}\), \(b = 8\)). When \(x = 0\), \(y=8\) (y - intercept). When \(y = 0\), \(0=-\frac{4}{3}x+8\), then \(\frac{4}{3}x=8\), \(x = 6\). So this equation represents the line.

• Equation ii: \(y - 4=-\frac{4}{3}(x - 3)\)

Expand using the point - slope form \(y-y_0=m(x - x_0)\) (\(m =-\frac{4}{3}\), \((x_0,y_0)=(3,4)\)).
\(y-4=-\frac{4}{3}x + 4\), then \(y=-\frac{4}{3}x+8\). When \(x = 0\), \(y=8\); when \(y = 0\), \(0=-\frac{4}{3}x + 8\), \(x = 6\). So this equation represents the line.

• Equation iii: \(3x+4y=24\)

Solve for \(y\): \(4y=-3x + 24\), \(y=-\frac{3}{4}x+6\). The slope \(m =-\frac{3}{4}
eq-\frac{4}{3}\). When \(x = 0\), \(y = 6
eq8\). So this equation does not represent the line.

Answer:

The equation that could NOT be used to represent the line is iii. \(3x + 4y=24\).