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How do you find the slope and intercept of $x - 2 y = 0$ $x-2y=0$ ?

1 Answer

Jan 14, 2016

Slope $= \frac{1}{2}$ $= 1/2$
y-intercept: $0$ $0$ (i.e. at $(0, 0)$ $(0,color(green)(0))$
x-intercept: $0$ $0$ (i.e. at $(0, 0)$ $(color(green)(0),0)$

Explanation:

For a linear equation in the general form:
$XXX A x + B y = C$ $color(white)("XXX")color(red)(A)x+color(blue)(B)y=C$
the slope is given by the equation:
$XXX m = - \frac{A}{B}$ $color(white)("XXX")m=-color(red)(A)/color(blue)(B)$

In this case
$XXX x - 2 y = 0 XXX \Rightarrow XXX 1 x + (- 2 y) = 0$ $color(white)("XXX")x-2y=0color(white)("XXX")rArrcolor(white)("XXX")color(red)(1)x+(color(blue)(-2)y) =0$

and therefore
$XXX m = - (\frac{1}{- 2}) = \frac{1}{2}$ $color(white)("XXX")m=-(color(red)(1)/(color(blue)(-2)))=1/2$

The y-intercept is the value of $y$ $y$ when $x = 0$ $x=color(cyan)(0)$
$XXX (0 - 2 y = 0) \to (y = 0)$ $color(white)("XXX")(color(cyan)(0)-2y=0) rarr (y=0)$

The x-intercept is the value of $x$ $x$ when $y = 0$ $y=color(orange)(0)$
$XXX (x - 2 (0) = 0) \to (x = 0)$ $color(white)("XXX")(x-2(color(orange)(0))=0) rarr (x=0)$

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