Hi, need a bit of help on this one.
A straight line passes through the points (4,3) and (10,0), I need to write down the equation of the line in the form y=mx+c, I have tried but I cant seem to get the second coordinates to satisfy the equation. Here is what I have done so far.
Found the gradient = -.5
Equation= y=-.5x+c
Substituted points 4,3 into equation= 3=-.5x4+c = 3=-2+c
Added 2 to both sides to give 5=c
Equation =y=-2+5
If I substitute points 10,0 into the equation I end up with y=-.5x10+5 = -15!
Where am I going wrong please?
2 answers
m = (0-3)/10-4) = 3/6 = POSITIVE 0.5
m = (0-3)/10-4) = -3/6 = - 0.5
3 = -.5(4) + b
3 = -2 + b
b = 5
so
y = -0.5 x + 5
try (10,0)
0 = -0.5(10) + 5
0 = -5 + 5 Yes, checks
3 = -.5(4) + b
3 = -2 + b
b = 5
so
y = -0.5 x + 5
try (10,0)
0 = -0.5(10) + 5
0 = -5 + 5 Yes, checks
2 answers