To determine the $x -$ and $y -$ intercepts of a line, we need to substitute $0$ for one variable, solve, then repeat for the other variable. In this case, we haven't been given an equation, so we will need to use the given points to write one.

Writing the Equation

The quickest way to write an equation, given two points, is by using the point-slope form. An equation in point-slope form follows a specific format.

y - y_{1} = m (x - x_{1})

In this form,

m

is the slope and the point

(x_{1}, y_{1})

lies on the graph of the line. To create our equation, we can use either of the given points for

x_{1}

and

y_{1}

as well as the slope. To calculate the slope, we will use the Slope Formula.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstitutePoints

Substitute $(- 5, - 5)$ & $(4, - 2)$

m = \frac{- 2 - ( - 5 )}{4 - ( - 5 )}

▼

Simplify

SubNeg

$a - (- b) = a + b$

m = \frac{- 2 + 5}{4 + 5}

AddTerms

Add terms

m = \frac{3}{9}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

m = \frac{1}{3}

Using the point

(- 5, - 5)

as our point and the slope

\frac{1}{3},

we have everything needed to form a point-slope form equation.

y - (- 5) = \frac{1}{3} (x - (- 5)) ⇕ y + 5 = \frac{1}{3} (x + 5)

Note that there are infinitely many correct ways to write this equation in point-slope form, as long as we use a point that lies on the line.

Finding the $x -$ intercept

Think of the point where the graph of an equation crosses the

x -

axis. The

y -

value of that

(x, y)

coordinate pair is

0,

and the

x -

value is the

x -

intercept. To find the

x -

intercept of the equation, we should substitute

0

for

y

and solve for

x .

y + 5 = \frac{1}{3} (x + 5)

Substitute

$y = 0$

0 + 5 = \frac{1}{3} (x + 5)

▼

Solve for

x

AddTerms

Add terms

5 = \frac{1}{3} (x + 5)

Distr

Distribute $\frac{1}{3}$

5 = \frac{1}{3} x + \frac{5}{3}

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

15 = x + 5

SubEqn

$LHS - 5 = RHS - 5$

10 = x

RearrangeEqn

Rearrange equation

x = 10

An

x -

intercept of

10

means that the graph passes through the

x -

axis at the point

(10, 0) .

Finding the $y -$ intercept

Let's use the same concept to find the

y -

intercept. Consider the point where the graph of the equation crosses the

y -

axis. The

x -

value of the

(x, y)

coordinate pair at the

y -

intercept is

0 .

Therefore, substituting

0

for

x

will give us the

y -

intercept.

y + 5 = \frac{1}{3} (x + 5)

Substitute

$x = 0$

y + 5 = \frac{1}{3} (0 + 5)

▼

Solve for

y

AddTerms

Add terms

y + 5 = \frac{1}{3} (5)

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

y + 5 = \frac{5}{3}

SubEqn

$LHS - 5 = RHS - 5$

y = \frac{5}{3} - 5

NumberToFrac

$a = \frac{3 \cdot a}{3}$

y = \frac{5}{3} - \frac{1 5}{3}

SubFrac

Subtract fractions

y = \frac{- 1 0}{3}

MoveNegNumToFrac

Put minus sign in front of fraction

y = - \frac{1 0}{3}

A

y -

intercept of

- \frac{1 0}{3}

means that the graph passes through the

y -

axis at the point

(0, - \frac{1 0}{3}) .

Writing the Equation

Finding the x-intercept

Finding the y-intercept

Finding the $x -$ intercept

Finding the $y -$ intercept

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