Question

the straight line x+y=0,3x+y-4=0,x+3y-4=0 from a triangle which is

Answer 1

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step by step explaination

$x + y = 0 \: \: \: \: \: - (i)$
$3x + y - 4 = 0 \: \: \: \: - (ii)$
$x + 3y - 4 = 0 \: \: \: - (iii)$

by solving eq. (i and (ii
$x = 2 \: \: \: and \: \: \: y = - 2$
solution set - A(2,-2)

by solving eq. (i & (iii
$x = - 2 \: \: \: and \: \: \: y = 2$
solution set - B(-2,2)

by solving eq. (ii &(iii
$3(4 - 3y) + y - 4 = 0$
$we \: get \: \: x = 1 \: \: \: and \: y = 1$
solution set - C(1,1)

distance b/w
A & B
$= \sqrt{ {(2 + 2)}^{2} + {( - 2 - 2)}^{2} }$
$= 4 \sqrt{2}$
distance b/w
A & C
$= \sqrt{ {(2 - 1)}^{2} + {( - 2 - 1)}^{2} }$
$= \sqrt{10}$
distance b/w
B & C
$\sqrt{ {( - 2 - 1)}^{2} + {(2 - 1)}^{2} }$
$= \sqrt{10}$
clearly
AC = BC
so, it's a isosceles triangle

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