﻿ Standard form to Point Slope form. How to convert forms. Examples and practice problems showing the steps . Plus a video tutorial # Standard Form to Point Slope Form

How to convert from Ax + By = C to y - y1 = m(x-x1)

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

### Example Problem

Step 1

Isolate the $$\red y$$ term.

$3x + \red { 5 y} = 15$
Step 2

Multiply all terms by the multiplicative inverse of the coefficient of y.

$$5y = -3x + 15 \\ \red{\frac 1 5} \cdot 5y = \red{\frac 1 5} \cdot -3x + \red{\frac 1 5 } \cdot 15 \\ \frac{5y}{5} = \frac{-3x}{5} + \frac{15}{5}$$
Step 3

Simplify.

$$\frac{5y}{5} = \frac{-3x}{5} + \frac{15}{5} \\ y = - \frac 3 5 x + 3$$
Step 4

Substitute a convenient value for x into your equation and then solve for y.

You're doing this to get the values of $$( \blue{x_1}, \blue{y_1})$$ for the point slope formula : $$y - \blue y_1 = m(x - \blue x_1)$$

Remember that you can pick any value that you want. You're just choosing a value for $$x$$ and then finding its associated $$y$$ value.

Let's choose $$x = \blue 5$$. Yes, you could choose x = 0 and make your life really easy! After you solve for the y value, then .

$$y = - \frac 3 5 x + 3 \\ \text{ Let's choose } x = \blue 5 \\ y = -\frac 3 5 \cdot \blue 5 + 3 \\ y = -3 + 3 \\ y = \blue 0 \\ \text{ Or, using } x = \blue 0 \\ y = - \frac 3 5 x + 3 \\ y = -\frac 3 5 \cdot \blue 0 + 3 \\ y = \blue 3$$
Step 5

Substitute the x value you picked and y value you solved for into the general form of point slope formula.

$$\text{ Using } x = \blue 5 \\ y -y_1 = m(x - x_1) \\ \boxed { y - \blue 0 = -\frac 3 5 (x - \blue 5) } \\ \text{ Or, using } x = \blue 0 \\ y -y_1 = m(x - x_1) \\ \boxed { y - \blue 3 = -\frac 3 5 (x - \blue 0) } \\$$

Note that both of the above equations are equivalent. They both are valid. Neither equation is 'better'.

### Practice Converting Equations

Step 1

Isolate the $$\red y$$ term.

$$\red { 3y} - 2x = -12$$
Step 2

Multiply all terms by the multiplicative inverse of the coefficient of y.

$$3y = 2x -12 \\ \red { \frac 1 3 } \cdot 3y = \red { \frac 1 3 } \cdot 2x - \red { \frac 1 3 }\cdot 12 \\ \frac{ 3y}{3} = \frac{2x}{3} - \frac {12}{3}$$
Step 3

Simplify.

$$\frac{ 3y}{3} = \frac{2x}{3} - \frac {12}{3} \\ y = \frac 2 3 x - 4$$
Step 4

Substitute a convenient value for x into your equation and then solve for y.

You're doing this to get the values of $$( \blue{x_1}, \blue{y_1})$$ for the point slope formula : $$y - \blue y_1 = m(x - \blue x_1)$$

Remember that you can pick any value that you want. You're just choosing a value for $$x$$ and then finding its associated $$y$$ value.

Let's choose $$x = \blue 3$$. Yes, you could choose x = 0 and make your life really easy! After you solve for the y value, then .

$$y = \frac 2 3 x - 4 \\ \text{ Using } x = \blue 3 \\ y = \frac 2 3 \cdot \blue 3 - 4 \\ y = 2 -4 \\ y = - 2 \\ \text{ Or, using } x = \blue 0 \\ y = \frac 2 3 \cdot \blue 0 - 4 \\ y = -4$$
Step 5

Substitute the x value you picked and y value you solved for into the general form of point slope formula.

$$\text{ Using } x = \blue 3 \\ y -y_1 = m(x - x_1) \\ \boxed { y + \blue 2 = \frac 2 3 (x - \blue 3) } \\ \text{ Or, using } x = \blue 0 \\ y -y_1 = m(x - x_1) \\ \boxed { y + \blue 4 = \frac 2 3 (x - \blue 0) } \\$$

Note that both of the above equations are equivalent. They both are valid. Neither equation is 'better'.

Step 1

Isolate the $$\red y$$ term.

$$\red {4y} - 5x = 20$$
Step 2

Multiply all terms by the multiplicative inverse of the coefficient of y.

$$4y - 5x = 20 \\ \red{ \frac 1 4 }\cdot 4y = \red{ \frac 1 4 } \cdot 5x + \red{ \frac 1 4 } \cdot 20 \\ \frac{4y}{4} = \frac{5x}{4} + \frac{20}{4}$$
Step 3

Simplify.

Step 4

Substitute a convenient value of x into your equation and solve for y (You're doing this to get the point x1, y1). Let's choose x = 3. Yes, you could choose x = 0 and make your life really easy!Then substitute the x and y values that found into the point slope formula.

Step 1

Isolate the 'y' term.

Step 2

Multiply all terms by the multiplicative inverse of the coefficient of y.

Step 3

Simplify.

Step 4

Substitute a convenient value of x into your equation and solve for y (You're doing this to get the point x1, y1). Let's choose x = 3. Yes, you could choose x = 0 and make your life really easy!Then substitute the x and y values that found into the point slope formula.

### More challenging problems

Step 1

Isolate the 'y' term.

Step 2

Multiply all terms by the multiplicative inverse of the coefficient of y.

Step 3

Simplify.

Step 4

Substitute a convenient value of x into your equation and solve for y (You're doing this to get the point x1, y1). Let's choose x = 3. Yes, you could choose x = 0 and make your life really easy!Then substitute the x and y values that found into the point slope formula.

Step 1

Isolate the 'y' term.

Step 2

Multiply all terms by the multiplicative inverse of the coefficient of y.

Step 3

Simplify.

Step 4

Substitute a convenient value of x into your equation and solve for y (You're doing this to get the point x1, y1). Let's choose x = 3. Yes, you could choose x = 0 and make your life really easy!Then substitute the x and y values that found into the point slope formula.