Answer 1

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