# Using Standard Form

The **standard form** of writing a line is \(Ax+By=C,\) where (A,) \(B,\) and \(C\) are integers.

This form is particularly useful for determining both the \(x\)- and \(y\)-intercepts of a line. We can determine the \(x\)-intercept by substituting 0 for \(y\) and solving for \(x.\) Similarly, we can determine the \(y\)-intercept of the line by substituting 0 for \(x\) and solving for \(y.\)

## If the equation of a line is \(3x + 5y = 60,\) what are the \(x\)-intercept and \(y\)-intercept of the line?

To find the \(x\)-intercept, we substitute 0 for \(y\) and solve: \[\begin{align} 3x + 5(0) &= 60 \\ 3x &= 60 \\ x &= 20.\end{align}\]

To find the \(y\)-intercept, we substitute 0 for \(x\) and solve: \[\begin{align} 3(0) + 5y &= 60 \\ 5y &= 60 \\ x &= 12.\end{align}\]

The \(x\)-intercept is \((12,0)\) and the \(y\)-intercept is \((0,20).\)

## If the \(x\)-intercept and \(y\)-intercept of a line are \((5,0)\) and \((0,6)\), respectively, what is the equation of the line?

Dividing both sides of the standard form equation by \(C\) yields the equation \(\frac{A}{C}x+\frac{B}{C}y=1.\) Given this equation, the \(x\)-intercept is \(\left(\frac{C}{A},0\right)\) and the \(y\)-intercept is \(\left(0,\frac{B}{C}\right).\)

Since our \(x\)-intercept is 5, \(\frac{A}{C} = \frac{1}{5}.\) Since our \(y\)-intercept is 6, \(\frac{B}{C} = \frac{1}{6}.\)

Substituting our known values into the equation, we have \(\frac{1}{5}x + \frac{1}{6}y = 1.\) Multiplying both sides by \(30\) yields \(6x + 5y = 30\). \(_\square\)

**Cite as:**Using Standard Form.

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