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
2 answers:
Answer:
Step-by-step explanation:
2x-y+k=0 k=?
2*2-1+k=0
4-1+k=0
3+k=0
k=-3
2x-y+k=0
2*5-(-8)+k=0
10+8+k=0
18+k=0
k=-18
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
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Answer:
x = -2 —> y = 4
x= 0 —> y = 1
x= 3 —> y = 1/8
Step-by-step explanation:
<h3>X= –2</h3>
<h3>X= 0</h3>
<h3>X= 3</h3>
I hope I helped you^_^