if points are A (5,1), B(3,2) and C(1,4) prove that AC=AB+BC
Answers
Given: vertices of ΔABC = A( -5 , -1 ) , B( 3 , -5 ) & C( 5 , 2 )
To Prove: area of ΔABC = 4 times are of triangle formed by joining mid points of ΔABC
Let say the triangle formed by joining the vertices is ΔPQR
Formulas use are the following,
Area\:of\:Triangle=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|AreaofTriangle=
2
1 ∣x 1
(y 2
−y 3 )+x 2
(y 3 −y 1 )+x 3
(y 1 −y 2 )∣
Coordinates\:of\:Mid-Point=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})CoordinatesofMid−Point=(
2
x
1
+x
2
,
2
y
1
+y
2
)
Now,
Area\:of\:\Delta\,ABC=\frac{1}{2}\left|-5(-5-2)+3(2+1)+5(-1+5)\right|AreaofΔABC=
2
1
∣−5(−5−2)+3(2+1)+5(−1+5)∣
=\frac{1}{2}\left|-5(-7)+3(3)+5(4)\right|=
2
1
∣−5(−7)+3(3)+5(4)∣
=\frac{1}{2}\left|35+9+20\right|=
2
1
∣35+9+20∣
=\frac{1}{2}\times64=
2
1
×64
=32\:unit^2=32unit
2
Mid point of AB = P , Mid Point of CB = Q & Mid Point of AC = R
Coordinate of P = (\frac{-5+3}{2},\frac{-1-5}{2})(
2
−5+3
,
2
−1−5
)
= (\frac{-2}{2},\frac{-6}{2})(
2
−2
,
2
−6
)
= (-1,-3)(−1,−3)
Coordinate of Q = (\frac{5+3}{2},\frac{2-5}{2})(
2
5+3
,
2
2−5
)
= (\frac{8}{2},\frac{-3}{2})(
2
8
,
2
−3
)
= (4,\frac{-3}{2})(4,
2
−3
)
Coordinate of R = (\frac{-5+5}{2},\frac{-1+2}{2})(
2
−5+5
,
2
−1+2
)
= (\frac{0}{2},\frac{1}{2})(
2
0
,
2
1
)
= (0,\frac{1}{2})(0,
2
1
)
text\Area\:of\:\Delta\,PQR=\frac{1}{2}\left|-1(\frac{-3}{2}-\frac{1}{2})+4(\frac{1}{2}+3)+0(-3+\frac{-3}{2})\right|textAreaofΔPQR=
2
1
∣
∣
∣
−1(
2
−3
−
2
1
)+4(
2
1
+3)+0(−3+
2
−3
)
∣
∣
∣
=text\frac{1}{2}\left|-1(\frac{-4}{2})+4(\frac{1+6}{2})+0\right|text=
2
1
∣
∣
∣
−1(
2
−4
)+4(
2
1+6
)+0
∣
∣
∣
=text\frac{1}{2}\left|-1(-2)+2(7)+0\right|text=
2
1
∣−1(−2)+2(7)+0∣
=text\frac{1}{2}\left|2+14+0\right|text=
2
1
∣2+14+0∣
=8\:unit^2=8unit
2
Therefore, ar ΔABC = 4 × ΔPQR
Hence Proved,