This question asks us to find the cartesian equation of l.
First we must find the points P and Q. To do this we substitute t with 1/4 to find P and substitute t with 2 to find Q.Doing this we get the coordinates for P and Q.For P: x= 4(1/4) = 1 y= 4/(1/4) = 16 P(1,16)For Q: x = 4(2) = 8 y = 4/2 = 2 Q(8,2)
The equation of line l is found by using the standard method y - y* = m(x - x*) where y* and x* are points on the line and m is the gradient.We must find the gradient of PQ before we find the gradient of l. To do this we simply use dy/dx: (Gradient PQ) = (16-2)/(1-8) = -(14/7) = -2
As PQ is perpendicular to l, we follow this formula. (Gradient of l) * (Gradient PQ) = -1: Gradient of l must be 1/2 as gradient of l = -1/(Gradient PQ)
We can now use y - y* = m(x - x*). We know that l passes through the origin, and therefore x* = 0 and y* = 0: y - 0 = 1/2(x - 0) y = 1/2x (this is the equation of l, as required)
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