Triangle formed by lines x + y - 4 = 0,, 3x + y = 4, x + 3y = 4 is

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9.Straight Line

easy

The triangle formed by the lines $x + y - 4 = 0,\,$ $3x + y = 4,$ $x + 3y = 4$ is

A

Isosceles

B

Equilateral

C

Right-angled

D

None of these

(IIT-1983)

Solution

(a) The vertices of triangle are the intersection points of these given lines. The vertices of

$\Delta $ are $A(0,\,4),$ $B(1,2),$$C(4,0)$

Now, $AB = \sqrt {{{(0 – 1)}^2} + {{(4 – 1)}^2}} = \sqrt {10} $

$BC = \sqrt {{{(1 – 4)}^2} + {{(0 – 1)}^2}} = \sqrt {10} $

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

$\because AB = BC$; $\bigtriangleup$ is isosceles.

Standard 11

Mathematics

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