Hi all, currently teaching a sec 3 NA student ,but really dun quite understand how to derive the answer .I need help to get the detailed workings .Thanks
There are 2 points P(-2,1) and Q(3,4) in the coordinate plane .A line with equation y=3x+7 passes through P . R is a point on the line such that gradient of RQ = [-1/3]
Find
(a)
y = 3x+7 ––– (1)
The equation of the line thru Q with gradient -â…“ is given by
y-4 = -â…“(x-3)
line QR: 3y+x = 15 ––– (2)
Since both lines intersect at R, sub (1) into (2):
3(3x+7)+x = 15
10x = -6
∴ x = -0.6, y = 5.2
∴ R = (-0.6, 5.2)