Question

If `A(4,-6),B(3,-2)a n dC(5,2)` are the vertices of ` A B C ,` then verify the fact that a median of a triangle ABC divides it into two triangles of equal areas.

Answer 1

It is being given that A(4, -6), B(3, -2) and C(5, 2) are the vertices of a `DeltaABC` and AD is its median. Clearly, D is the midpoint of BC.
`therefore ` the coordinates of D are `((3+5)/(2), (-2+2)/(2)) = (4, 0).`
For `Delta ABD`, we have
`A(x_(1) = 4, y_(1) = -6), B(x_(2) = 3, y_(2) = -2) "and" D(x_(3) = 4, y_(3) = 0).`
`ar(Delta ABD) = (1)/(2)|x_(1)(y_(2) -y_(3)) + x_(2)(y_(3) -y_(1)) +x_(3) (y_(1) -y_(2))|`
`=(1)/(2)|4*(-2-0) + 3* (0+6) +4*(-6+2)|`
` =(1)/(2)| 4 xx (-2) + 3 xx 6 + 4 xx (-4)|`
`=(1)/(2)| -8 +18 -16 | = (1)/(2)|-6|`
` = ((1)/(2) xx 6) = 3` sq units.
For `Delta ADC`, we have
`(x_(1) = 4, y_(1) = -6), (x_(2) = 4, y_(2) = 0)" and " (x_(3) = 5, y_(3) =2).`
`therefore ar(DeltaADC) = (1)/(2)|x_(1)(y_(2)-y_(3)) + x_(2)(y_(3)-y_(1)) + x_(3) (y_(1)-y_(2))|`
`=(1)/(2)|4*(0-2) + 4*(2+6) +5*(-6-0)|`
` = (1)/(2)|4 xx (-2) + 4 xx 8 + 5 xx (-6)|`
`=(1)/(2)|-8 +32 -30|=(1)/(2)|-6|`
`=((1)/(2) xx 6) = 3` sq units.
`therefore ar (DeltaABD) = ar(DeltaADC)`
Hene, the median AD divides `DeltaABC` into two triangles of equal areas.