Question

Find Y so that the points A(0 , 0), B(2 , 2) and C(-4 , Y) are the vertices of a right triangle with hypotenuse AC.

Answer 1

The points are A(0 , 0), B(2 , 2) and C(-4 , Y)

Let A (x1, y1) = (0, 0) and B (x2, y2) = (2, 2)

Lenth of the Line Segment =    √ [(x2 - x1)^2 + (y2 - y1)^2] -------------> (1)

Substitute A (x1, y1) = (0, 0) and B (x2, y2) = (2, 2) in Equation (1)

AB  =  √ [(2 - 0)^2 + (2 - 0)^2]

       =  √ [(2 - 0)^2 + (2 - 0)^2]

       =  √ [4 + 4]

AB  =  √8

Substitute B (x1, y1) = (2, 2) and C (x2, y2) = (-4, Y) in Equation (1)

BC  =  √ [(-4 - 2)^2 + (Y - 2)^2]

       =  √ [(-6)^2 + (Y^2 - 4Y + 4)]

       =  √ [36 + Y^2 - 4Y + 4]

BC   =  √ [Y^2 - 4Y + 40]

Substitute C (x1, y1) = (-4, Y) and A (x2, y2) = (0, 0) in Equation (1)

CA  =  √ [(0 + 4)^2 + (0 - Y)^2]

CA  =  √ [16 + Y^2]

All the right anular triangles follows Pythagorean Law

Hence,

(AC)^2  =   (AB)^2 + (BC)^2 ----------------> (2)

Substitute the values of AB, BC and AC in Eq (2)

[ √ (16 + Y^2) ]^2  =  [ √8 ]^2  +  [ √ (Y^2 - 4Y + 40) ]^2

16 + Y^2  =  8 +  y^2 - 4y + 40

8 +  Y^2 - 4y + 40 - (16 + Y^2)  =  0

8 +  Y^2 - 4Y + 40 - 16 - Y^2  =  0

8 - 4Y + 40 - 16  =  0

48 - 4Y - 16 =  0

32 - 4Y  =  0

4Y  =  32

Y  =  32/4

Y  =  8

Answer :

The Value of Y  is  8.