﻿ Point Slope Form to Standard form. How to convert

# Point Slope Form to Standard Form

How to convert?

It is a common task to convert equation of line from point slope form to standard form.

### Example of Converting from Point Slope to Standard Form

Step 1

Distribute $$\red m$$, which represents the slope to $$\blue{x-x1}$$.

$$y-y_1 = \red m \blue{ ( x - x_1)} \\ y-3 = \red 5 \blue{( x - 4) } \\ y-3 = \red 5 \cdot \blue x -\red 5 \cdot \blue 4 \\ y-3 = 5x -20 \\$$
Step 2

Move y1 to the other side by adding its opposite to both sides of the equation and simplify.

Step 3

"Move" the x term to the other side by adding its opposite to both sides.

### Use our Calculator

You can use the calculator below to find the equation of a line from any two points. Just type numbers into the boxes below and the calculator (which has its own page here) will automatically calculate the equation of line in point slope and standard forms.

$$(x_1, y_1)$$
,
$$(x_2, y_2)$$
,



### Practice Converting Equations

##### Practice 1
Step 1

Distribute $$\red m$$, which represents the slope to $$\blue{x-x1}$$.

$$y-y_1 = \red m \blue{ ( x - x_1)} \\ y-2 = \red {-5} \cdot \blue x - \red { -5} \cdot \blue 1 \\ y-2 = -5x + 5$$
Step 2

Move y1 to the other side by adding its opposite to both sides of the equation and simplify.

Step 3

"Move" the x term to the other side by adding its opposite to both sides.

##### Practice 2
Step 1

Distribute $$\red m$$, which represents the slope to $$\blue{x-x1}$$.

$$y-y_1 = \red m \blue{ ( x - x_1)} \\ y + 2 = \red {-4} \blue{ ( x +3 )} \\ y + 2 = \red {-4} \cdot \blue x +\red {-4} \cdot \blue 3 \\ y + 2 = -4x + -12 \\ y + 2 = -4x-12$$
Step 2

Move y1 to the other side by adding its opposite to both sides of the equation and simplify.

Step 3

"Move" the x term to the other side by adding its opposite to both sides.

##### Practice 3
Step 1

Distribute $$\red m$$, which represents the slope to $$\blue{x-x1}$$.

$$y-y_1 = \red m \blue{ ( x - x_1)} \\ y + 3 = \red{-\frac{3}{4} } \blue{ ( x - 5)} \\ y + 3 = \red{-\frac{3}{4} } \cdot \blue x - \red{-\frac{3}{4} }\cdot \blue 5 \\ y + 3 =-\frac{3}{4} x - -\frac{15}{4} \\ y + 3 =-\frac{3}{4} x +\frac{15}{4}$$
Step 2

Move y1 to the other side by adding its opposite to both sides of the equation and simplify.

Step 3

"Move" the x term to the other side by adding its opposite to both sides.

Step 4

An optional step, but if you want integer coefficients, then multiply by the least common multipleof the fractions in the equation.

##### Practice 4
Step 1

Distribute $$\red m$$, which represents the slope to $$\blue{x-x1}$$.

$$y-y_1 = \red m \blue{ ( x - x_1)} \\ (y + \frac 1 2 ) = \red {- \frac 3 5} \blue{ ( x + 7 ) } \\ \\ (y + \frac 1 2 ) = \red {- \frac 3 5} \cdot \blue x + \red { - \frac 3 5} \cdot \blue 7 \\ (y + \frac 1 2 ) = -\frac 3 5 x - \frac {21}{5} \\ (y + \frac 1 2 ) = -\frac 3 5 x - \frac {21}{5}$$
Step 2

Move y1 to the other side by adding its opposite to both sides of the equation and simplify.

Step 3

"Move" the x term to the other side by adding its opposite to both sides.

Step 4

An optional step, but if you want integer coefficients, then multiply by the least common multipleof the fractions in the equation.