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?
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
Hi there! I can help you with that.
First, let's find the gradient (m) of the line using the given points (4,3) and (10,0). The gradient formula is (change in y)/(change in x):
m = (0 - 3) / (10 - 4)
m = -3 / 6
m = -0.5
So, you were correct in finding the gradient as -0.5.
Now, let's proceed with finding the equation of the line in the form y = mx + c.
We can choose any of the given points to substitute into the equation to find the value of c.
Using the point (4,3):
3 = -0.5(4) + c
3 = -2 + c
c = 3 + 2
c = 5
So, the equation of the line in the form y = mx + c is:
y = -0.5x + 5
Now, let's substitute the other given point (10,0) into the equation to make sure it satisfies the equation:
0 = -0.5(10) + 5
0 = -5 + 5
0 = 0
As you can see, the point (10,0) does satisfy the equation of the line.
Therefore, the equation of the line that passes through the points (4,3) and (10,0) is: y = -0.5x + 5.