Point Slope Form to Standard form. How to convert

Example of Converting from Point Slope to Standard Form

Convert $$ y - 3 = 5(x - 4)$$ 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.

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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.

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Practice Converting Equations

Practice 1

Convert (y - 2) = -5(x - 1) to standard form.

Step 1

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

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

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

Convert $$(y + 2) = -4(x + 3)$$ to standard form.

Step 1

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

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

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

Convert $$ (y + 3) = - \frac 3 4(x-5) $$ to standard form.

Step 1

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

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

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

Convert $$ (y + \frac 1 2 ) = - \frac 3 5 ( x + 7 ) $$ to standard form.

Step 1

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

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

Step 2

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