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A(7, -3), B(5, 3) and C(3, -1) are the vertices of a ABC and

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$A (7, - 3), B (5, 3)$ and $C (3, - 1)$ are the vertices of a $Δ A BC$ and AD is its median. Prove that the median AD divides $Δ A BC$ into two triangles of equal areas.

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A (7, - 3), B (5, 3)

and

C (3, - 1)

are the vertices of a

Δ A BC

and AD is its median. Prove that the median AD divides

Δ A BC

into two triangles of equal areas.

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Text solutionVerified

D is midpoint of BC, so find its coordinates using below:

Midpoint formula:

(x, y) = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

D = ((3 + 5) /2, (3 - 1) /2) = (4, 1)

Find area of triangle ABD:

We know that:

Area of a triangle

= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

So,

Area of triangle ABD:

= \frac{1}{2} (7 (3 - 1) + 5 (1 + 3) + 4 (- 3 - 3))

= \frac{1}{2} (14 + 20 - 24)

= \frac{1}{2} (10)

= 5

sq. units .......

(1)

Area of triangle ACD:

= \frac{1}{2} (7 (- 1 - 1) + 3 (1 + 3) + 4 (- 3 + 1))

= \frac{1}{2} (- 14 + 12 - 8)

= \frac{1}{2} (10)

= 5

sq. units ........

(2)

From

(1)

and

(2)

, we conclude that Area of triangle ABD and ACD is equal.

Hence proved.

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Question Text	$A (7, - 3), B (5, 3)$ and $C (3, - 1)$ are the vertices of a $Δ A BC$ and AD is its median. Prove that the median AD divides $Δ A BC$ into two triangles of equal areas.
Answer Type	Text solution:1
Upvotes	150

A(7,−3),B(5,3) and C(3,−1) are the vertices of a ΔABC and AD is its median. Prove that the median AD divides ΔABC into two triangles of equal areas.

Text solutionVerified

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$A (7, - 3), B (5, 3)$ and $C (3, - 1)$ are the vertices of a $Δ A BC$ and AD is its median. Prove that the median AD divides $Δ A BC$ into two triangles of equal areas.