if P(3,3);Q(0,4);R(-1,2)are vertices of mid point of triangle ABC . find co ordinates of A,B,C . also find area triangle APC.
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let A(a,b) B(c,d) C(e,f) be the vertices of the triangle
given P (3,3) is midpoint of A and B
given Q (0,4) is midpoint of B and C
given R (-1,2) is midpoint of A and C
so, a+c = 6 , b+d = 6 , c+e = 0 , d+f = 8 , a +e = -2 , b+f = 4
solving them we get,
a = 2
b = -4
c = 4
d = 5
e = -4
f = 3
vertices of triangle are A(2,-4) B (4,5) C(-4,3)
Area of triangle formed by A(a,b) B(c,d) C(e,f) is
\begin{lgathered}\frac{1}{2} (a(d - f) + c(f - b) + e(b - d)) \\\end{lgathered}
2
1
(a(d−f)+c(f−b)+e(b−d))
\begin{lgathered}\frac{1}{2}(2(5 - 3) + ( 4)(3 - ( - 4)) + ( - 4)( - 4 - 5)) \\\end{lgathered}
2
1
(2(5−3)+(4)(3−(−4))+(−4)(−4−5))
34 square units
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