Question

$\underline{ \Huge{\color{red}{\mathbb{QUESTION}}}}$
Find the area of triangle formed by the points P(-1.5 , 3), Q(6 , -2) and R(-3 , 4).

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Answer 1

$\huge\boxed{\fcolorbox{black}{pink}{Question}}$

Find the area of triangle formed by the points P(-1.5 , 3), Q(6 , -2) and R(-3 , 4).

$\huge\boxed{\fcolorbox{black}{pink}{Answer}}$

Let ;

$~~~~~~~~~~~⟹ P (x_{1} , y_{1}) => (-1.5 , 3)$

$~~~~~~~~~~~⟹ Q (x_{2} , y_{2}) => (6 , -2)$

$~~~~~~~~~~~⟹ R (x_{3} , y_{3}) => (-3 , 4)$

Area of ∆PQR ;

$\boxed{⇢\frac{1}{2} |[x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})]|}$

$~~~~~~~~~~~~~~~$

${:⟹ \frac{1}{2} |[1.5(-2 - (4)) + 6(4 - (3)) + (-3)(3 -(-2))]|}$

$~~~~~~~~~~~~~~~$

${:⟹ \frac{1}{2}|[1.5(- 2 - 4 ) + 6(4 - 3) + (-3)(3 + 2)]|}$

$~~~~~~~~~~~~~~~$

$:⟹ {\frac{1}{2}|[1.5(-6) + 6(1) + (-3)(5)]|}$

$~~~~~~~~~~~~~~~$

$:⟹ \frac{1}{2}[9 + 6 - 15]$

$~~~~~~~~~~~~~~~$

$:⟹ \frac{1}{2}[9 + 6 - 15]$

$~~~~~~~~~~~~~~~$

$:⟹ \frac{1}{2}[15 - 15]$

$~~~~~~~~~~~~~~~$

$:⟹ \frac{1}{2}[\cancel{15} \cancel{-15}]$

$~~~~~~~~~~~~~~~$

$:⟹ \frac{1}{2}(0)$

The rule is that anything multiplied by 0 is equal to 0.

$:⟹ 0$

If the area of triangle is 0square unties , then it's vertices will be collinear