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Question
In Figure 2, P (5, −3) and Q (3, y) are the points of trisection of the line segment joining A (7, −2) and B (1, −5). Then y equals
Options
A. 2
B. 4
C. −4
D. `-5/2`
Solution
It is given that P and Q are the points of trisection of line segment AB.
∴ AP = PQ = QB
Now, AP = QB
`therefore sqrt((5-7)^2+(-3+2)^2)=sqrt((1-3)^2+(-5-y)^2)`
`rArrsqrt((-2)^2+(-1)^2)=sqrt((-2)+(5+y)^2)`
`rArrsqrt(4+1)=sqrt(4+25+y^2+10y)`
`rArrsqrt5=sqrt(y^2+10y+29)`
Squaring on both sides, we get
5 = y2 + 10y + 29
∴ y2 + 10y + 24 = 0
⇒ y2 + 6y + 4y + 24 = 0
⇒ y(y + 6) + 4 (y + 6) = 0
⇒ (y + 4) (y + 6) = 0
⇒ y + 4 = 0 or y + 6 = 0
⇒ y = −4 or y = − 6
From the obtained values of y, − 4 matches with the option C.
Hence, the correct answer is C.
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