Find the equation of a line with a slope of 13/2
that passes through the point (−2, — 10).
1 answer:
Answer: ![y=\frac{13}{2} x+3](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B13%7D%7B2%7D%20x%2B3)
Step-by-step explanation:
Remember the point-slope equation which is
where
is your <u>point</u> and
is your <u>slope.</u>
<u />
Given that we substitute what you have:
![y-(-10) =\frac{13}{2} (x - (-2))](https://tex.z-dn.net/?f=y-%28-10%29%20%3D%5Cfrac%7B13%7D%7B2%7D%20%28x%20-%20%28-2%29%29)
Minus and minus give us positive:
![y+10 =\frac{13}{2} (x +2)](https://tex.z-dn.net/?f=y%2B10%20%3D%5Cfrac%7B13%7D%7B2%7D%20%28x%20%2B2%29)
Multiply slope into the parenthesis:
![y+10= \frac{13}{2}x + \frac{13}{2}*2\\](https://tex.z-dn.net/?f=y%2B10%3D%20%5Cfrac%7B13%7D%7B2%7Dx%20%2B%20%5Cfrac%7B13%7D%7B2%7D%2A2%5C%5C)
Calculate it:
![y+10= \frac{13}{2}x +13](https://tex.z-dn.net/?f=y%2B10%3D%20%5Cfrac%7B13%7D%7B2%7Dx%20%2B13)
Isolate the
by subtracting 10 from both sides:
![y=\frac{13}{2} x + 13 -10](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B13%7D%7B2%7D%20x%20%2B%2013%20-10)
Your final equation is:
![y=\frac{13}{2} x +3](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B13%7D%7B2%7D%20x%20%2B3)
Hope this makes sense!
And here's the graph to prove that the line actually goes through the point
:
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