Write the equation of a line that passes through (-10,3) and (-2,-5).
What I have: y= -3/13x +b
After doing y2 - y1/x2 - x, I got a slope of -3/13, but I'm confused on how to get the y intercept.
huh?
(y2-y1) = -5-3 = -8
(x2-x1) = -2 +10 = 8
so slope = -8/8 = -1
so y = -1 x + b
use either point now
3 = -1(-10) + b
b = -7
so I get
y = -x -7
================
check that with the other point
-5 = 2 -7 sure enough
use either point in your equation:
-3/13(-10)+b = 3
30 + 13b = 39
13b = 9
-3/13(-2)+b = -5
6+13b = -65
13b = -71
Hmmm. They should come out the same. Let's check your slope:
m = (-5-3)/(-2+10) = -1
So, y = x+b
-(-10)+b = 3
10+b = 3
b = -7
-(-2)+b = -5
2+b = -5
b = -7
So, now we know b.
You can forget about finding b if you just use the two-point or point-slope form. We already have the slope, so you can use either point to get the equation of the line
y-3 = -(x+10)
y = -x-7
y+5 = -(x+2)
y = -x-7
Thank you guys.
To find the y-intercept, we can substitute the coordinates of one of the given points into the equation of the line and solve for the y-intercept (b). Let's use the point (-10, 3) since both points are valid.
Start with the equation you have:
y = -3/13x + b
Substitute the coordinates of the point (-10, 3) into the equation:
3 = (-3/13)(-10) + b
Now simplify and solve for b:
3 = 30/13 + b
3 - 30/13 = b
(39/13) - (30/13) = b
9/13 = b
So, the y-intercept (b) is 9/13.