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

The correct option is A isosceles



Solving equation 3x + y - 4 = 0 & x + y = 0


Substituting in (ii) y = - 2
Coordinate & B = (2, -2)
Solving equation x + y = 0 & x + 3y - 4 = 0


Substituting in equation (i) we get x = - 2
Coordinate of $A\equiv (-2,2)$
Solving equation 3x + y - 4 = 0 and x + 3y - 4 = 0
$(3x+y=4)\times 3$
$x+3y=4$


Substituting in (vi) we get y = 1
coordinate of $C\equiv (1,1)$

So, we get the vertices of triangle as $A(-2,2),B(2,-2),C(1,1).$
The distance between two points $(x_1,y_1)and(x_2,y_2)$ is given by $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$AB=\sqrt{(2-(-2){)}^2+(-2-2{)}^2}$
$AB=\sqrt{(4{)}^2+(-4{)}^2}$
$AB=\sqrt{16+16}=4\sqrt{2}$ units
$BC=\sqrt{(2-1{)}^2+(-2-1{)}^2}$
$BC=\sqrt{1^2+(-3{)}^2}=\sqrt{10}$ units
$CA=\sqrt{(1-(-2){)}^2+(1-2{)}^2}$
$CA=\sqrt{(3{)}^2+(-1{)}^2}$
$CA=\sqrt{9+1}=\sqrt{10}$ units

Here the two sides are equal hence the triangle is isosceles.