Two lines in the xy-plane are drawn through the point (3,4) and the two trisection points of the line segment joining the points (-4,5) and (5,-1). Which of the following choices is the equation of one of these lines?

- $3x – 2y – 1 = 0$
- $4x – 5y + 8 = 0$
- $5x + 2y – 23 = 0$
- $x + 7y – 31 = 0$
- $x – 4y + 13 = 0$

## Solution

Between A & B, \triangle x = 9 and \triangle y = 6. Consult the figure below.

So, T1 = ( 4 + 3, 5 – 2) = (-1,3), and

T2 = (-4 + 6, 5 – 4) = (2, 1)

We find the slopes of T1P and T2P and see if they match the given equations’ slopes.

Slope T1P:

m = (4 – 3)/(3 – ( 1)) = 1/4 (e?)

Slope T2P:

m = (4 – 1)/(3 – 2) = 3/1 = 3 (nope)

So it seems to be (e).

Check: e): x – 4y + 13 = 0 Does it contain T1? -1 – 4(3) + 13 = 0 OK.

Does it have P? 3 – 4(4) + 13 OK.

It is, in fact, e).