Question

find the area of triangle whose vertices are P(6,-2), Q(-3,5), R(-1,-2)​

Answer 1

correct explanation

hope you understand

Answer 2

GIVEN :-

• Co-ordinates of triangle are P(6,-2) , Q(-3,5) , R(-1,-2).

TO FIND :-

• Area of triangle.

TO KNOW :-

$\\ \small \boxed{\sf \: area \: of \: triangle = \dfrac{1}{2} \{ x_{1}( y_{2} - y_{3}) + x_{2}(y_{3} - y_{1} )+ x_{3}(y_{1} - y_{2}) \} }$

SOLUTION :-

Here ,

• P(x1,y1) = (6,-2)
• Q(x2,y2) = (-3,5)
• R(x3,y3) = (-1,-2)

Putting values in formula ,

Area of triangle ,

$\\ \sf \implies \: \dfrac{1}{2} \{6(5 - ( - 2)) + ( - 3)( - 2 - ( - 2)) + ( - 1)( - 2 - 5) \} \\ \\ \implies \sf \: \dfrac{1}{2} \{6(5 + 2) - 3( - 2 + 2) - 1( - 2 - 5) \} \\ \\ \implies \sf \: \dfrac{1}{2} \{6(7) - 3(0) - 1( - 7) \} \\ \\ \implies \sf \: \dfrac{1}{2} (42 - 0 + 7) \\ \\ \implies \sf \: \dfrac{1}{2} (49)$

Hence , area of triangle is 49/2 unit².