Chapter 1.4, Problem 54AYU

In Problems 47-72, find an equation for the line with the given properties. Express your answer using either the general form or the slope-intercept form of the equation of a line, whichever you prefer.

$x\text{-intercept}=-4;y\text{-intercept}=4$

To determine

To find: The equation of the line with $x\text{-intercept }=-4$ and $y\text{-intercept }=4$

Answer to Problem 54AYU

Solution:

$\text{<mi>x</mi> }-\text{ <mi>y</mi> }+4=0$

Explanation of Solution

Given:

$x\text{-intercept }=-4$, $y\text{-intercept }=4$

Therefore the points on the line are $\left(-4,0\right)$ and $\left(0,4\right)$.

Formula used:

Slope of the line $\text{<mi>m</mi> }=\frac{y_2-y_1}{x_2-x_1}$

Point slope form of a line $\text{<mi>y</mi> }-\text{ <mi>y</mi>}₁ =\text{ <mi>m</mi>}(\text{<mi>x</mi> }-\text{ <mi>x</mi>}₁)$

Calculation:

First compute the slope of the line with $(x₁,\text{ <mi>y</mi>}₁)=\left(-4,0\right)$ and $(x₂,\text{ <mi>y</mi>}₂)=\left(0,4\right)$.

Slope $m=\frac{y_2<mtext> </mtext>-y_1}{x_2-x_1}$

$= \frac{4-0}{0-\left(-4\right)}$

$=\frac{-1}{-2}$

$=\frac{4}{4}$

$=1$

Use the point $(x₁,\text{ <mi>y</mi>}₁)=\left(-4,0\right)$ and the slope $\text{<mi>m</mi> }=1$ to get the point slope form of the equation.

$\text{<mi>y</mi> }-0=1\left(\text{<mi>x</mi> }-\left(-4\right)\right)$

$\text{<mi>y</mi> }=1\left(\text{<mi>x</mi> }+4\right)$

$\text{<mi>y</mi> }=\text{ <mi>x</mi> }+4$

$\text{<mi>y</mi> }-\text{ <mi>y</mi> }=\text{ <mi>x</mi> }+4-\text{ <mi>y</mi>}$Subtract $y$ from both sides

$0=\text{ <mi>x</mi> }-\text{ <mi>y</mi> }+4$

$\text{<mi>x</mi> }-\text{ <mi>y</mi> }+4=0$