## How to Graph Lines Using Standard Form

Read,5 minutes

There are lots of ways to express the equation of a line. There's slope-intercept form, point-slope form, and additionally the topic of this page. Each of them expresses the equation of a line, as well as each of them, has its own **advantages** and **disadvantages.** For example, the point-slope form makes it simple for finding the line's equation whenever one just knows the **slope** as well as a **single point** on the line. Standard form additionally has a few distinct uses, however, we will cover more on it later.

## Define Standard Form?

Standard Form is shown as: \(Ax \ + \ Bx \ = \ C \ , \ A \ ≠ \ 0 \ , \ B \ ≠ \ 0\)

Where \(A\) and \(B\) are **coefficients** and \(C\) is a **constant.**

**Examples:**

- \(2x \ + \ 4y \ = \ 8\)
- \(5x \ - \ 7y \ = \ 12\)
- \(3x \ - \ 9y \ = \ -18\)

If the equation is given in standard form, one can’t identify the slope and \(y\)-intercept needed for graphing.

**Therefore... what is done?**

There are really \(2\) different **methods** that can be used to graph linear equations written in **standard form.** One can utilize either one, so here’s both of them and you can decide which one you like best.

Here is an example solved by \(2\) methods.

### Method one: Rewriting an Equation in Slope Intercept Form

For this lesson, we will find out the way to **convert** a standard form equation into an equation written in slope-intercept form.

So, let’s begin! If you’ve got a standard form equation, it can be rewritten in slope-intercept form.

The objective whenever you rewrite a standard form equation in slope intercept form is to redo the equation so it reads as \(y \ = \ mx \ + \ b\). You’ll have to utilize your knowledge of **solving equations.** Don’t forget to utilize **opposite** operations and whatever is done to one side of an equation, **must** be done to the other side.

So we can redo \(6x \ - \ 2y \ = \ 4\) in slope-intercept form so we can graph the equation easily.

**Step1:** Isolate \(y\) for \(x\) by Subtracting \(6x\) from both sides.

\(6x \ - \ 2y \ \color{red}{-6x} \ = \ 4 \ \color{red}{-6x}\)

**Step2:** Rewrite the right side of the equation.

\(-2y \ = \ 4 \ \color{red}{-6x}\)

**Step3:** Divide all the terms by the coefficient of \(y\).

\(\frac{-2y}{\color{red}{-2}} \ = \ \frac{4}{\color{red}{-2}} \ - \ \frac{6x}{\color{red}{-2}}\)

**Step4:** If possible, simplify.

\(y \ = \ 3x \ - \ 2\)

**Step5:** Graph the line.

\(y\)-intercept \(= \ -2\) , \(Slop \ = \ 3\):

See the way making a graph is fairly simple after it gets written in a slope-intercept form. It’s always **possible** to redo standard form equations in a slope-intercept form. Be certain you solve the equation for \(y\), and then you are done!

### Method two: Utilizing X and Y Intercept for Graphing Linear Equations

Now you know one method of graphing standard form equations – via converting it into a slope-intercept form. But there’s also an additional method of graphing standard form equations, which is to **locate** its \(x\) and \(y\) **intercepts.** So, let’s go over what the word intercepts means. Intercept is where your line **crosses** an **axis.** You have an \(x\), as well as a \(y\) intercept.

The point where a line touches the \(x\)-axis is known as an \(x\)**-intercept.** The point where the line touches the \(y\)-axis is known as a \(y\)**-intercept.** If you can locate the points where a line crosses an \(x\) and \(y\)-axis, after that you get 2 points and can **draw** a line.

Whenever equations are presented in standard form, it’s fairly simple to locate intercepts.

For graphing a linear equation utilizing \(X\) and \(Y\) Intercept you **must** do the following steps:

- Locate the \(x\)-intercept of the line via using
**zero**for \(y\). - Locate the \(y\)- intercept of the line via using a
**zero**for the \(x\). **Connect**the \(2\) points.

**Now let's see another Example**

Draw a graph of \(6x \ - \ 3y \ = \ -6\).

**Answer:**

- Firstly
**isolate**\(y\) for \(x\): \(6x \ - \ 3y \ = \ -6 \ → \ -3y \ = \ -6x \ -6 \ → \ y \ = \ 2x \ + \ 2\) - Locate the \(x\)−
**intercept**of the line via using a**zero**for \(y\). \(2x \ + \ 2 \ = \ y \ → \ 2x \ + \ 2 \ = \ 0 \ → \ 𝑥 \ = \ \frac{-2}{2} \ = -1\) - Locate the \(y\)−
**intercept**of the line via using a**zero**for the \(x\). \(2x \ + \ 2 \ = \ y \ → \ 2(0) \ + \ 2 \ = \ y \ → \ y \ = \ 2\) - After that: \(x\)−intercept: \((−1,0)\) and \(y\)−intercept: \((0,2)\)

### Exercises for Graphing Lines Using Standard Form

**1) **Sketch the graph of the line: \(3y \ - \ 6x \ = \ 3\)

**2) **Sketch the graph of the line: \(\frac{1}{3} \ y \ - \ x \ = \ -\frac{2}{3}\)

**3) **Sketch the graph of the line: \(-6y \ + \ 6x \ = \ 30\)

**4) **Sketch the graph of the line: \(-5y \ + \ \frac{5}{2} \ x \ = \ -15\)

**5) **Sketch the graph of the line: \(11y \ + \ 22x \ = \ 11\)

**6) **Sketch the graph of the line: \(2y \ + \ 3x \ = \ -4\)

**7) **Sketch the graph of the line: \(-7y \ - \ 7x \ = \ 14\)

**8) **Sketch the graph of the line: \(3y \ + \ 9x \ = \ 9\)

**9) **Sketch the graph of the line: \(-6y \ - \ 4x \ = \ -12\)

**10) **Sketch the graph of the line: \(-15y \ - \ 6x \ = \ 30\)

**1) **Sketch the graph of the line: \(3y \ - \ 6x \ = \ 3\)

\(\color{red}{3y \ - \ 6x \ = \ 3 \ ⇒ \ y \ = \ 2x \ + \ 1 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{-\frac{1}{2}}\) |

\(\color{red}{1}\) | \(\color{red}{0}\) |

**2) **Sketch the graph of the line: \(\frac{1}{3} \ y \ - \ x \ = \ -\frac{2}{3}\)

\(\color{red}{\frac{1}{3} \ y \ - \ x \ = \ -\frac{2}{3} \ ⇒ \ y \ = \ 3x \ - \ 2 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{\frac{2}{3}}\) |

\(\color{red}{-2}\) | \(\color{red}{0}\) |

**3) **Sketch the graph of the line: \(-6y \ + \ 6x \ = \ 30\)

\(\color{red}{-6y \ + \ 6x \ = \ 30 \ ⇒ \ y \ = \ x \ - \ 5 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{5}\) |

\(\color{red}{-5}\) | \(\color{red}{0}\) |

**4) **Sketch the graph of the line: \(-5y \ + \ \frac{5}{2} \ x \ = \ -15\)

\(\color{red}{-5y \ + \ \frac{5}{2} \ x \ = \ -15 \ ⇒ \ y \ = \ \frac{1}{2} \ x \ + \ 3 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{-6}\) |

\(\color{red}{3}\) | \(\color{red}{0}\) |

**5) **Sketch the graph of the line: \(11y \ + \ 22x \ = \ 11\)

\(\color{red}{11y \ + \ 22x \ = \ 11 \ ⇒ \ y \ = \ -2x \ + \ 1 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{\frac{1}{2}}\) |

\(\color{red}{1}\) | \(\color{red}{0}\) |

**6) **Sketch the graph of the line: \(2y \ + \ 3x \ = \ -4\)

\(\color{red}{2y \ + \ 3x \ = \ -4 \ ⇒ \ y \ = \ -\frac{3}{2} \ x \ - \ 2 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{-1 \ \frac{1}{3}}\) |

\(\color{red}{-2}\) | \(\color{red}{0}\) |

**7) **Sketch the graph of the line: \(-7y \ - \ 7x \ = \ 14\)

\(\color{red}{-7y \ - \ 7x \ = \ 14 \ ⇒ \ y \ = \ -x \ - \ 2 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{-2}\) |

\(\color{red}{-2}\) | \(\color{red}{0}\) |

**8) **Sketch the graph of the line: \(3y \ + \ 9x \ = \ 9\)

\(\color{red}{3y \ + \ 9x \ = \ 9 \ ⇒ \ y \ = \ -3x \ + \ 3 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{1}\) |

\(\color{red}{3}\) | \(\color{red}{0}\) |

**9) **Sketch the graph of the line: \(-6y \ - \ 4x \ = \ -12\)

\(\color{red}{-6y \ - \ 4x \ = \ -12 \ ⇒ \ y \ = \ -\frac{2}{3} \ x \ + \ 2 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{3}\) |

\(\color{red}{2}\) | \(\color{red}{0}\) |

**10) **Sketch the graph of the line: \(-15y \ - \ 6x \ = \ 30\)

\(\color{red}{-15y \ - \ 6x \ = \ 30 \ ⇒ \ y \ = \ -\frac{2}{5} \ x \ - \ 2 \ ⇒}\)

\(\color{red}{y}\) | \(\color{red}{x}\) |

\(\color{red}{0}\) | \(\color{red}{-5}\) |

\(\color{red}{-2}\) | \(\color{red}{0}\) |