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How do you find the slope and y intercept for: $x + 8 y = 0$ $x +8y=0$ ?

1 Answer

Nov 12, 2015

Slope: $(- \frac{1}{8})$ $(-1/8)$
y-intercept: $0$ $0$

Explanation:

Method 1
A linear equation in the form:
$XXX A x + B y = C$ $color(white)("XXX")Ax+By=C$
the slope is $(- \frac{A}{B})$ $(-A/B)$
and
the y-intercept can be evaluated by setting $x$ $x$ to $0$ $0$ and solving for $y$ $y$

For the given equation
$XXX x + 8 y = 0$ $color(white)("XXX")x+8y=0$
the slope is $(- \frac{(1)}{8}) = - \frac{1}{8}$ $(-((1))/8) = -1/8$
and with $x = 0$ $x=0$
$XXX 0 + 8 y = 0 \Rightarrow y = 0$ $color(white)("XXX")0+8y=0 rArr y=0$

Method 2
Convert the equation in its given form into slope-intercept form.

$x + 8 y = 0$ $x+8y=0$

$\to 8 y = - x$ $rarr 8y=-x$

$\to y = - \frac{1}{8} x$ $rarr y = -1/8x$

$\to y = (- \frac{1}{8}) x + 0$ $rarr y = (-1/8)x + 0$
$XXX$ $color(white)("XXX")$ which is the slope-intercept form for a linear equation
with slope $(- \frac{1}{8})$ $(-1/8)$ and y-intercept $0$ $0$

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