the line L goes through the points A (3,1) and B(4,-2). Find the equation for L.
First we need to consider what a straight line looks like:
y = mx + c (if we get something different we've gone wrong somewhere)one way to get the equation is using a point (general point is (x1 , y1) ), the gradient (m) and the following equation:
(y - y1) = m(x - x1)
so we can find the gradient using rise over run:
m = (y1-y2)/(x1-x2)
so using the points A and B:
m = (-2-1) / (4-3)= -3
so now we have the gradient we can chose either A or B to put into the equation here we use A:
(y - 1) = -3(x - 3)
first we expand the brackets on the RHS:
y - 1 = -3x + 9
then take the -1 over from the RHS to the LHS
y = -3x + 10 (under line your answer so the examiner can find your answer easily)
check does this look like what we expect?yes takes y = mx + c form.
do the two points both work?
for A (3,1) x= 3 and y=1:
1 = -33 + 10= -9 +10 = 1so is consistent
for B (4, -2) x= 4 and y= -2:
-2 = -34 +10 = -12 + 10 = -2is also consistent
now we know it is correct
