Question

The straight lines x+y = 0, 3x+y-4 = 0, and x+3y-4 = 0 form a triangle which is
a. isosceles
b. equilateral
c. right angled
d. none of these.

Answer 1

Dear student
x+y = 0 … (1) slope = -1
3x+y = 4 … (2) slope = -3/1
x+3y = 4 … (3) slope = -1/3

So no two lines are perpendicular to each other as m1xm2 is not equal to -1 for any two lines.

From (1) and (2) intersection point is

A(2,-2)

From (2) and (3) intersection point is

B(1,1)

From (3) and (1) intersection point is
C(-2,2)

AC² = [(2-(-2)] ² + [-2-2] ² = 32
BC² = [1-(-2)] ²+[1-2] ² = 10
AB²= [2-1] ² + [-2-1] ² = 10
As BC² = AB² which means BC = AB and AC is different, the triangle is a isosceles triangle.

Regards