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

1.	Find the area of triangle formed by the points P(-1.5 , 3), Q(6 , -2) and R(-3 , 4).✅Don't spam or give incorrect answer.✅ Otherwise it will be reported.✅Not free points.
Answer» $\huge\boxed{\fcolorbox{black}{pink}{Question}}$ 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