JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 14)
In a 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 $$\Delta $$ ABC (in sq. units) is :
12
4
5
9
Explanation
Median through C is x = 4
So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and
C(4, y) is D which lies on the median through B.
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$$ \therefore $$$$\,\,\,$$ D = $$\left( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right)$$
Now, $${{1 + 4 + 2 + y} \over 2}$$ = 5 $$ \Rightarrow $$ y = 3.
So, C $$ \equiv $$ (4, 3).
The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1).
The area of triangle $$\Delta $$ABC = 3 $$ \times $$ $$\Delta $$AGC
= 3 $$ \times $$ $${1 \over 2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9.
So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and
C(4, y) is D which lies on the median through B.
$$ \therefore $$$$\,\,\,$$ D = $$\left( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right)$$
Now, $${{1 + 4 + 2 + y} \over 2}$$ = 5 $$ \Rightarrow $$ y = 3.
So, C $$ \equiv $$ (4, 3).
The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1).
The area of triangle $$\Delta $$ABC = 3 $$ \times $$ $$\Delta $$AGC
= 3 $$ \times $$ $${1 \over 2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9.
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