Question

The line segment joining the points A(2,1) and B(5,-8) is trisected at the points p and q such that p is nearer to A. If p also lies on the line given by 2x-y+k=0, find the value of k

Answer 1

Answer:

Step-by-step explanation:

• A(2,1)

2x-y+k=0    k=?

2*2-1+k=0

4-1+k=0

3+k=0

k=-3

• B(5,-8)

2x-y+k=0

2*5-(-8)+k=0

10+8+k=0

18+k=0

k=-18

Answer 2

Answer:

  k = -8

Step-by-step explanation:

The location of P is ...

  P = (2A+B)/3 = (2·2+5, 2·1-8)/3 = (3, -2)

Putting this point into the equation for the line, we have ...

  2(3) -(-2) +k = 0

  8 +k = 0

  k = -8