Question

Given points A(2,1,3) and B(0,5,0), find a point C in the xy-plane that creates an isosceles right triangle such that:_______

a. AB and BC are the legs of the right triangle
b. AB is the hypotenuse of the right triangle (for part ii, just doing all the set-up work is sufficient; you do not need to solve and find the points)

Answer 1

a.)Point C will be either (4.82, 7.41) or ( -4.82, 2.59)

b.) -x² + 2x + 6y - y² - 5= 0 , 8y - 4x = 11

Explanation:

As given A(2,1,3) and B(0,5,0)

a.)

We have to find C in the xy-plane

⇒ C is of the form (x, y, 0) ( Because in xy plane , z coordinate is zero)

Now, given AB and BC are the legs of the right triangle .

AB = (0, 5, 0) - (2, 1, 3) = ( -2, 4, -3)

BC = (x, y, 0) - (0, 5, 0) = ( x, y-5, 0)

As AB and BC are legs of triangle

⇒AB is perpendicular to BC

⇒(AB).(BC) = 0

⇒( -2, 4, -3).( x, y-5, 0) = 0

⇒ -2x + 4(y-5) -0 = 0

⇒ -2x + 4y- 20 = 0

⇒ -x + 2y- 10 = 0

⇒ 2y = 10 + x

⇒ y = 5 +
`(x)/(2)` ......(1)

As ABC is making an isosceles right triangle

It means two sides of the triangle are equal

Now,

|AB| = √(-2)² + (4)² + (-3)² = √4+16+9 = √29

|BC| = √(x)² + (y-5)² + 0² = √(x)² + (y-5)²

Now,

AS AB = BC

⇒|AB| = |BC|

⇒|AB|² = |BC|²

⇒29 = (x)² + (y-5)²

⇒29 = (x)² + (5 +
`(x)/(2)` -5)²

⇒29 = (x)² + (
`(x)/(2)` )²

⇒29 = x²( 1 +
`(1)/(4)` ) = x²(
`(5)/(4)` )

⇒ x² =
`(116)/(5)` = 23.2

⇒ x = 4.82 , -4.82

⇒y = 7.41 , 2.59

so, Point C will be either (4.82, 7.41) or ( -4.82, 2.59)

b.)

As given AB is the hypotenuse of the right triangle

S, AC and CB will perpendicular to each other

Now,

AC = (x, y, 0) - ( 2, 1, 3) = ( x-2, y-1, -3)

CB = (0, 5, 0) - (x, y, 0) = (-x, 5-y, 0)

As, AC and CB is perpendicular

⇒(AC).(CB) = 0

⇒( x-2, y-1, -3).(-x, 5-y, 0) = 0

⇒(x-2)(-x) + (y-1)(5-y) - 0 = 0

⇒-x² + 2x + 5y - y² - 5 + y = 0

⇒-x² + 2x + 6y - y² - 5= 0 ..........(2)

As ABC is making an isosceles right triangle

It means two sides of the triangle are equal

Now,

|AC| = √(x-2)² + (y-1)² + (0-3)² = √(x-2)² + (y-1)² + 9

|BC| = √(x)² + (y-5)² + 0² = √(x)² + (y-5)²

Now,

AS AC = BC

⇒|AC| = |BC|

⇒|AC|² = |BC|²

⇒(x-2)² + (y-1)² + 9 = (x)² + (y-5)²

⇒x² + 4 - 4x + y² + 1 - 2y + 9 = x² + y² + 25 - 10y

⇒ 4 - 4x + 1 - 2y + 9 = 25 - 10y

⇒14 - 4x - 2y = 25 - 10y

⇒10y - 2y - 4x = 25 - 14

⇒8y - 4x = 11 ........(3)

Now, we have 2 equations

-x² + 2x + 6y - y² - 5= 0 , 8y - 4x = 11

Solve these equations to find x, y