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  • If D (-1 by 2, 5 by 2) , E (7, 3) and F (7 by 2 , 7 by 2)are the midpoints of sides of traingle ABC, find the area of the traingleABC.

\text{If } D\left ( \frac{-1}{2},\frac{5}{2} \right ) , E (7, 3) \ and F\left ( \frac{7}{2},\frac{7}{2} \right ) Are the midpoints of sides of  \Delta ABC find the area of the \Delta ABC

Answers (1)

 
D is the mid-point of AB using the mid-point formula we get:
-\frac{1}{2}= \frac{x_{1}+x_{2}}{2}           \frac{5}{2}= \frac{y_{1}+y_{2}}{2}

x1 + x2 = –1     ……(1)                        5 = y1 + y2       .…(2)

E is the mid-point of BC using the mid-point formula we get:
\frac{x_{2}+x_{3}}{2}= 7            \frac{y_{2}+y_{3}}{2}= 3

x2 + x3 = 14     …..(3)              y2 + y3 = 6       ….(4)
F is the mid-point of AB using the mid-point formula we get:
\frac{x_{1}+x_{3}}{2}= \frac{7}{2}               \frac{y_{1}+y_{3}}{2}= \frac{7}{2}

x1 + x3 = 7       ….(5)               y1 + y3 = 7       …..(6)
Simplifying the above equations for values of x1, y1, x2, y2, x3 and y3
x1 + x2 = –1
x3 + x3 = 14               {using (1) and (3)}
–  –   –   
x1 – x3 = –15   ….(7)
Using (5) and (7) we get
x1 + x3 = 7
x1– x3 = –15
2x1 = –8
x1 = –4
put x1 = 4 in (1) we get
–4 + x2 = –1
X2 = –1 + 4 = 3
Punt x2 = 3 in (3) we get
3 + x3 = 14
x3 = 11
Using equations (2) and (4) we get
y1 + y2 = 5
y3 + y3 = 6
-  -   –
y1 - y3 = –1      ….(8)
Adding equations (6) and (8) we get
y1 + y3 = 7
y1 – y3 = –1
-------------------
2y1 = 6
y1 = 3
Put y1 = 3 in eqn. (2)
5 – 3 = y2
2 = y2
Put y2 = 2 in eqn. (4)
2 + y3 = 6
y3 = 6 – 2
y3 = 4
Hence, x1 = –4             y1 = 3
  x2 = 3               y2 = 2
  x3 = 11             y3 = 4
A = (x1, y1) = (–4, 3), B = (x2, y2) = (3, 2), C = (x3, y3) = (11, 4)
Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]
= \frac{1}{2}\left [ \left ( -4 \right )\left ( 2-4 \right )+\left ( 3 \right )\left ( 4-3 \right ) +11\left ( 3-2 \right )\right ]
= \frac{1}{2}\left [ \left ( -4 \right )\left ( -2 \right )+3\left ( 1 \right )+11\left ( 1 \right ) \right ]
= \frac{1}{2}\left [ 8+3+11 \right ]
= \frac{1}{2}\left [ 22 \right ]
= 11 sq. units

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