Question

Find the value of x. if the area the triangle formed by the vertices
-2,-2).(-1.-3) and (0) is 3 sq units.​

Answer 1

$Let \: A(x_{1},y_{1}) = (-2,-2) , \\ B(x_{2},y_{2}) = (-1,-3) , \: and \\ C(x_{3},y_{3}) = (x,0) , are \: three \\veticies \: of \: a \: triangle$

$Area \: of \: \triangle ABC = 3\:square \:units$

$\implies \frac{1}{2}|x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2}) |= 3$

$\implies |-2(-3-0)+(-1)[0-(-2)]+x[-2-(-3)] = 6$

$\implies 6 -2 + x = 6$

$\implies 4 + x = 6$

$\implies x = 6 - 4$

$\implies x = 2$

Therefore.,

$\red { Value \: of \: x } \green { = 2 }$

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