Chapter 1, Problem 22RE

$y\text{-intercept }=-2$; containing the point $\left(5,-3\right)$

To determine

To find: The equation of a line having $y\text{-intercept}$ $-2$ and containing the point $\left(5,-3\right)$. To also sketch a graph of the equation.

Answer to Problem 22RE

Solution:

The general form of equation of the line is $x+5y=-10$

Explanation of Solution

Given:

The $y\text{-intercept}$ $-2$ and containing the point $\left(5,-3\right)$. The $y\text{-intercept}$ $-2$ means the line contains the point $\left(0,-2\right)$.

Formula used:

Point slope form of an equation of a line.

$(y-y₁)=m(x-x₁)$

$m=\frac{y_2-y_1}{x_2-x_1}$

Calculation:

To find the slope, let $\text{A}=(x₁,y₁)=\left(0,-2\right)$ and $\text{B}=(x₂,y₂)=\left(5,-3\right)$.

$m=\frac{-3-\left(-2\right)}{5-0}$

$=\frac{-3+2}{5}$

$=\frac{-1}{5}$

Now using this slope $m=\frac{-1}{5}$ and the point $\text{A}\left(0,-2\right)$, the equation can be found.

$y-\left(-2\right)=\frac{-1}{5}\left(x-0\right)$

$5\left(y+2\right)=-x$   Multiply 5 to both sides.

$5y+10=-x$

$5y+10+x=-x+x$   Add $x$ to both sides.

$x+5y+10-10=-10$   Subtract 10 from both sides.

$x+5y=-10$

The general form of equation of the line is $x+5y=-10$.