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Question
If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
Solution
Given that A(4, 3), B(-1, y) and C(3, 4) are the vertices of the ΔABC.
ΔABC is a right triangle at A.
Hence by applying the Pythagoras Theorem, we have,
AB2 + AC2 = BC2 ....(1)
Let us find the distances, AB, BC and CA using the
distance formula.
`AB=sqrt((-1-4)^2+(y-3)^2)`
`BC=sqrt((3+1)^2+(4-y)^2)`
`CA=sqrt((3-4)^2+(4-3)^2)=sqrt2`
Squaring both the sides, we have
`AB^2=25+y^2+9-6y`
`BC^2=4+16+y^2-8y`
`AC^2=2`
Therefore, from equation (1), we have,
`25+y^2+9-6y+2=4+16+y^2-8y`
`36+y^2-6y=20+y^2-8y`
16-6y=-8y
16=-8y+6y
-y=16/2
y=-8
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