# The gradient is m = &#x2212;<!-- − --> 4 / 3 , and the point given is ( 3

The gradient is $m=-4/3$, and the point given is $\left(3/2,3\right)$
Let $c$ be the $y$-intercept
$y=mx+c$
$y=\left(-4/3\right)x+c$
If I substitute $x$ and $y$ from the point given, $3=\left(-4/3\right)\left(3/2\right)+c$ and look for $c$, I get $c=5$
But according to my answer sheet, $c=33/8$ by using $y-{y}_{1}=m\left(x-{x}_{1}\right)$
Why are the two methods giving different answers?
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Marquis Matthews
The answer sheet is wrong. Using $y-{y}_{1}=m\left(x-{x}_{1}\right)$, we get $y-3=\left(-4/3\right)\left(x-3/2\right)$. Distributing the $-4/3$, we get $y-3=\left(-4/3\right)x+4/2=\left(-4/3\right)x+2$. Now add three to both sides, and we get $y=\left(-4/3\right)x+5$, so $c=5$ again.
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Alaina Holt
It looks like someone flipped the gradient at some point. If you use $-\frac{3}{4}$ as the gradient, you end up with $c=\frac{33}{8}$.