Question

Lines $x+y=4$, $3x+y=4$, $x+3y=4$ from a triangle which is

• A. Right-angled
• B. obtuse-angled isosceles
• C. acute-angled isosceles
• D. none

Answer 1

The correct option is C acute-angled isosceles

Given

$x+y=4$........(1)

$3x+y=4$.......(2)

$x+3y=4$........(3)

solving (1) and (2)

$3x+y=4x+y=4\_\_\_\_\_\_\_2x=0$

$\therefore x=0y=4A=(0,4)$

solving (1) and (3)

$x+3y=4x+y=4\_\_\_\_\_\_\_2y=0$

$\therefore y=0x=4B=(4,0)$

solving (2) and (3)

$3x+y=43x+9y=12\_\_\_\_\_\_\_-8y=-8$

$\therefore y=1x=1C=(1,1)$

Distance between $A$ & $B$

$=\sqrt{(0-4{)}^2+(4-0{)}^2}$

$=\sqrt{4^2+4^2}$

$=\sqrt{32}=4\sqrt{2}$

= Distance between $A$ & $C$

$\sqrt{(0-1{)}^2+(4-1{)}^2}$

$=\sqrt{1^2+3^2}$

$=\sqrt{10}$

Distance between $B$ & $C$

$\sqrt{(4-1{)}^2+(0-1{)}^2}$

$=\sqrt{10}$

As $AC=BC$

So, $△ABC$, is isosceles triangle

using pythagoras theorem,

$A{C}^2+B{C}^2<A{B}^2$

as $10+10=20<(4\sqrt{2}{)}^2$

$20<32$

So $△ABC$ is acute $\angle$ ed isosceles triangle.