Question

Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Answer 1

The given lines are,

y−x=0 x+y=0 x−k=0

The point of intersection of the lines y−x=0 and x+y=0 is given by,

x=0, y=0

The point of intersection of the lines x+y=0 and x−k=0 is given by,

x=k, y=−k

The point of intersection of the lines y−x=0 and x−k=0 is given by,

x=k, y=k

Thus, the vertices of the triangle formed by the given three lines are ( 0,0 ), ( k,−k ), and ( k,k ).

The area A of the triangle with vertices ( x 1 , y 1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ) is given by,

A= 1 2 | x 1 ( y 2 − y 3 )+ x 2 ( y 3 − y 1 )+ x 3 ( y 1 − y 2 ) |(1)

Substitute the values of ( x 1 , y 1 ), ( x 2 , y 2 ) and ( x 3 , y 3 ) as ( 0,0 ), ( k,−k ), and ( k,k ) in equation (1).

A= 1 2 | 0( −k−k )+k( k−0 )+k( 0+k ) | = 1 2 | k 2 + k 2 | = 1 2 | 2 k 2 | = k 2  square units

Thus, the area of the triangle passing through the lines y−x=0, x+y=0, x−k=0 is given by k 2  square units.