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Question
A(6, –3), В(0, 5) and C(–2, 1) are vertices of ΔАВС. Points P(3, 1) and Q(2, –1) lie on sides AB and AC respectively. Check whether `(AP)/(PB) = (AQ)/(QC)`.
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Solution

Given: The vertices of ΔABC are A(6, –3), В(0, 5) and C(–2, 1).
Points P(3, 1) and Q(2, –1) lies on sides AВ and AC respectively.
To check: `(AP)/(PB) = (AQ)/(QC)`
By using distance formula,
AP = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt((3 - 6)^2 + [1 - (-3)]^2`
= `sqrt((-3)^2 + (1 + 3)^2`
= `sqrt(9 + (4)^2`
= `sqrt(9 + 16)`
= `sqrt(25)`
= 5 ...(1)
PB = `sqrt((0 - 3)^2 + (5 - 1)^2`
= `sqrt((-3)^2 + (4)^2`
= `sqrt(9 + 16)`
= `sqrt(25)`
= 5 ...(2)
AQ = `sqrt((2 - 6)^2 + [-1 - (-3)]^2`
= `sqrt((-4)^2 + (-1 + 3)^2`
= `sqrt(16 + (2)^2`
= `sqrt(4 + 16)`
= `sqrt(20)`
= `2sqrt(5)` ...(3)
QC = `sqrt([(-2) - 2]^2 + [1 - (-1)]^2`
= `sqrt((-4)^2 + (2)^2`
= `sqrt(16 + 4)`
= `sqrt(20)`
= `2sqrt(5)` ...(4)
Putting the value of AP, PB, AQ and QC is
`(AP)/(PB) = (AQ)/(QC)`
⇒ `5/5 = (2sqrt(5))/(2sqrt(5))`
⇒ 1 = 1
So, in the triangle ABС:
`(AP)/(PB) = (AQ)/(QC)`
Hence Proved.
