Question

The straight lines $x+y-4=0,3x+y-4=0,x+3y-4=0$ forms a triangle which is

• A. isosceles
• B. right angled
• C. equilateral
• D. None of these

Answer 1

The correct option is A isosceles

Straight lines $x+y-4=0,3x+y-4=0,x+3y-4=0$ form a triangle.

Point of interaction of line $x+y-4=0$ and $3x+y-4=0$ can be calculated by solving and we will get point of interaction $A=(2,-2)$

Point of interaction of line $3x+y-4=0$ and $x+3y-4=0$ can be calculated by solving and we will get point of interaction $B=(1,1)$

Point of interaction of line $x+3y-4=0$ and $x+y-4=0$ can be calculated by solving and we will get point of interaction $c=(-2,2)$

Hence we get a triangle $△ABC$

Using distance formula, we can calculate length of sides $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$ between points $(x_1,y_1)$ and$(x_2,y_2)$

length of AB $=\sqrt{(1-2{)}^2+(1+2{)}^2}=\sqrt{1+9}=\sqrt{10}$units

length of BC $=\sqrt{(-2-1{)}^2+(2-1{)}^2}=\sqrt{9+1}=\sqrt{10}$units

length of CA $=\sqrt{(-2-2{)}^2+(2+2{)}^2}=\sqrt{16+16}=\sqrt{32}$units

As we can observe, AB=BC, the triangle is isosceles.