# How do you graph the line with slope -1/2 passing through point (-3,-5)?

Apr 25, 2017

See explanation

#### Explanation:

$\textcolor{b r o w n}{\text{The slope (gradient) is stated as a single value so this is a straight line graph}}$

$\textcolor{b l u e}{\text{Determine the equation of the line}}$

Gradient is $m = \left(\text{change in y")/("change in x reading left to right}\right)$

$m = \left(\text{change in y")/("change in x reading left to right}\right) = \frac{- 1}{2}$

So the y value becomes less as you read left to right. The slope is down.

Let the given point be ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(- 3 , - 5\right)$

Using the standardised equation format $y = m x + c$

We have by substituting the values for ${P}_{1}$

${y}_{1} = m {x}_{1} + c \text{ "->" } - 5 = \left(- \frac{1}{2}\right) \left(- 3\right) + c$

$\text{ } - 5 = + \frac{3}{2} + c$

Subtract $\frac{3}{2}$ from both sides to get $c$ on its own.

$\text{ } - 5 - \frac{3}{2} = 0 + c$

$\text{ } - \frac{13}{2} = c$

So the equation of the line that passes through the point (-3,-5) is:
$y = - \frac{1}{2} x - \frac{13}{2}$
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$\textcolor{b l u e}{\text{Determine the value of the y-intercept}}$

The y-intercept is the value of the constant $c = - \frac{13}{2}$

ie: set $x = 0$
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$\textcolor{b l u e}{\text{Determine the value of the x-intercept}}$

Set $y = 0$

$y = - \frac{1}{2} x - \frac{13}{2} \text{ "->" } 0 = - \frac{1}{2} x - \frac{13}{2}$

$\frac{1}{2} x = - \frac{13}{2}$

$x = - \frac{26}{2} = - 13$
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Mark your points on the x and y axis and draw a straight line through them. Also show the given point.