# What are the intercepts of the line 2y=-x+1?

Oct 16, 2015

I found:
$\left(1 , 0\right)$
$\left(0 , \frac{1}{2}\right)$

#### Explanation:

x-intercept:
set $y = 0$
you get:
$0 = - x + 1$
so $x = 1$

y-intercept:
set $x = 0$
you get:
$2 y = 1$
so $y = \frac{1}{2}$

Oct 16, 2015

$\left(x , y\right) \to \left(0 , \frac{1}{2}\right) \text{ and } \left(1 , 0\right)$

#### Explanation:

The final answers are at parts ( 2 ) and ( 3 )

Before you can determine the intercepts you need to manipulate the equation so that you only have y on the left hand side of the equals sign and everything else on the other side.
To isolate y and still maintain balance multiply both sides by $\frac{1}{2}$

Step1. $\text{ } \frac{1}{2} \left(2 y\right) = \frac{1}{2} \left(- x + 1\right)$

$\frac{2}{2} y = - \frac{1}{2} x + \frac{1}{2}$

But $\frac{2}{2} = 1$ giving;

$y = - \frac{1}{2} x + \frac{1}{2}$ .........................( 1 )

Now to find the intercepts:

.*******
Step2. The graph crosses the x-axis at y=0

Substitute y=0 in (1) giving:

$0 = - \frac{1}{2} x + \frac{1}{2}$

Add $\frac{1}{2} x$ to both sides so that you may part isolate $x$

$\left(0\right) + \frac{1}{2} x = \left(- \frac{1}{2} x + \frac{1}{2}\right) + \frac{1}{2} x$

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

Multiply both sides by 2 giving:

$x = 1$
so one of the points where it crosses is at $y = 0 , x = 1$ ......( 2 )

.**********
Step3. The graph crosses the y-axis at x=0

Substituting y = 0 in equation ( 1 ) gives:

$y = \frac{1}{2}$ ....................( 3 )
so the other point where it crosses is at $y = \frac{1}{2} , x = 0$ .......( 3 )