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Sagot :
Recall that the general equation for a line is y=mx+b where m is the slope and b is the y-intercept.
First, let's find the slope by finding
(y2-y1)/(x2-x1):
(-8-0)/(-5-3)
-8/-8
1
Now we know the equation is y=1x+b, or y=x+b.
By plugging in one of the two points we know is on the line, we can solve for b.
0=3+b
b=-3
So the equation is:
y=x-3
First, let's find the slope by finding
(y2-y1)/(x2-x1):
(-8-0)/(-5-3)
-8/-8
1
Now we know the equation is y=1x+b, or y=x+b.
By plugging in one of the two points we know is on the line, we can solve for b.
0=3+b
b=-3
So the equation is:
y=x-3
- (8,0)
- (3,-5)
Slope:-
[tex]\\ \rm\hookrightarrow m=\dfrac{-5}{3-8}=-5/-5=1[/tex]
Equation of line in point slope form
[tex]\\ \rm\hookrightarrow y-y_1=m(x-x_1)[/tex]
[tex]\\ \rm\hookrightarrow y=x-8[/tex]
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