Question

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​.​

Answer 1

Answer:

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

hope it helps

keep following me

Answer 2

Answer:

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