Question

The area of the triangle formed by the points (7,9), (3,
-7) and (-3,3) is (in square units)
72
58
64
68​

Answer 1

$Let \: A(7,9) = (x_{1},y_{1}) ,$

$\: B(3,-7) = (x_{2},y_{2}) \:and$

$C(-3,3)) = (x_{3},y_{3}) \:are \: vertices \: of$

$triangle \: ABC$

$Area \: \triangle \: ABC$

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

$= \frac{1}{2}|7(-7-3)+3(3-9)+(-3)(9+7)|$

$= \frac{1}{2}|7(-10)+3(-6)+(-3)(16)|$

$= \frac{1}{2}| -70-18-48|$

$= \frac{1}{2}|-136|$

$= \frac{136}{2}$

$= 68$

Therefore.,

$\red{ Area \: of \: the \: triangle }$

$\green { = 68 \: square \:units }$

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