Question

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

• A. Isoceles
• B. Equilateral
• C. Right angle
• D. None of these

Answer 1

Answer: (A)

Isoceles

$x+y=4$
and $3x+y=4$
$\therefore x=0,andy=4$
$x+y=4$
and $x+3y=4$
$\therefore x=4$
and $y=0$
Similarly when
$3x+y=4$
and $x+3y=4$
$\therefore thenx=1andy=1$
$\therefore A=(0,4),B=(1,1)andC=(4,0)$
$\therefore AB=\sqrt{{(0-1)}^2+{(4-1)}^2}$
$=\sqrt{10}$
and $BC=\sqrt{{(1-4)}^2+{(1-0)}^2}$
$=\sqrt{10}$
$\therefore AB=BC$
$\therefore \Delta$ is isoceles.