# How do you write an equation in standard form given a line that passes through (4,-3) with m=2?

Jun 4, 2015

The equation in standard form is $2 x - y = 5$.

The general equation for the standard form of a line is $\text{A"x+"B"y="C}$, where $\text{A}$ and $\text{B}$ are integers, and $\text{A} \ne 0$, and $\text{B} \ne 0$.
http://www.mathwarehouse.com/algebra/linear_equation/standard-form-equation-of-a-line.php

Start with the equation to find the point-slope form of a line: $y - {y}_{1} = m \left(x - {x}_{1}\right)$, where $m$ is the slope, and ${x}_{1}$ and ${y}_{1}$ are the known point.

$m = 2$
${x}_{1} = 4$
${y}_{1} = - 3$

Solution:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

Substitute known values into the equation.

$\left(y - \left(- 3\right)\right) = 2 \left(x - 4\right)$ =

$y + 3 = 2 \left(x - 4\right)$ = Point-slope form.

Continue to the Standard Form.

Distribute the $2$.

$y + 3 = 2 x - 8$

Subtract $y$ from both sides.

$3 = 2 x - y - 8$

Add $8$ to both sides.

$11 = 2 x - y$

Flip the equation.

$2 x - y = 11$ = Standard Form

graph{2x-y=11 [-32.48, 32.43, -16.23, 16.24]}