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Q14.

Expert-verified
Found in: Page 244

### Algebra 1

Book edition Student Edition
Author(s) Carter, Cuevas, Day, Holiday, Luchin
Pages 801 pages
ISBN 9780078884801

# Write an equation in point-slope form for the line that passes through each point with the given slope.$\left(8,3\right),m=-2$

The equation of the straight line for the given condition is $\mathbit{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbit{x}\mathbf{+}\mathbf{19}$.

See the step by step solution

## Step 1. State the concept of a straight line.

The standard form of slope and intercept form of the straight line is $y=mx+c$.

Here, $m$ is the slope of the straight line and $c$ is the $y$-intercept of the equation.

## Step 2. State the equation of a line slope-point form.

The standard form of slope and point form of the straight line is $y-{y}_{1}=m\left(x-{x}_{1}\right)$

Here $m$ is the slope of the straight line and is the point from which the straight line passes.

$\left({x}_{1},{y}_{1}\right)=\left(8,3\right)\phantom{\rule{0ex}{0ex}}m=-2$

## Step 3. Substitute the values to get the required equation.

Substitute the above values in the equation $y-{y}_{1}=m\left(x-{x}_{1}\right)$

$y-{y}_{1}=m\left(x-{x}_{1}\right)\phantom{\rule{0ex}{0ex}}y-3=-2\left(x-8\right)\phantom{\rule{0ex}{0ex}}y-3=-2x+16\phantom{\rule{0ex}{0ex}}y=-2x+19$

The equation of the straight line is $y=-2x+19$.