Question

triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x+y =5 and x = 4. Then area of AABC (in sq. units) is (a) 12







(b) 9 (c) 4
(d) 5​

Answer 1

Answer:

We have given that, In a triangle ABC , coordinates of A are (1,2) and the equation of the medians through B and C are x+y=5 and x=4 respectively. Thus, The equation of the median through B is x+y=5.

(7,−3),(4,3)

ΔABC has vertex A(1,2)

Given that midpoint of AC is D and mid point of AB is E

Also,

equation of medium BD is x+y=5

equation of medium CE is x=4

let coordinate of B(p,q) and C(r,s)

∴ coordinate of E≡(

2p+1

, 2q+1

)

But E lies on the line segment CE (x=4)

∴

2P+1

=4

⇒P=7

And point B(p,q) lies on the line BD [x+y=4]

∴ p+q=4

⇒q=−3

∴ B≡(7,−3)

Point C(r,s) lies on CE [x=4]

∴ r=4

Coordinates of D are ≡[ 2r+1

2s+2

[Similarly point D lies on BD [x+y=5]

∴

2r+1

+ 2s+2

=5

r+s=7

⇒s=3

∴ C≡(4,3)

∴ coordinates of B & C are (7,−3) & (4,3)

solution

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