TS ICET 23 May 2018 Shift 2 Paper

Question : 140 of 200

Marks: +1, -0

The vertices of △ABC are A(4.2) , B(6,5) and C(1,4) . The median through A meets BC at D.Then

\AD^2

Solution:

A(4, 2), B(6, 5), C(1, 4)
D is the midpoint of BC

D=(7/9,9/2)

A D=(4,2),(\;{7}/{2},{9}/{2})

=√{(4-{7}/{2})^{2}+(2-{9}/{2})^{2}}=√{{1}/{4}+\;{25}/{4}}=√{{26}/{4}}

∴A D^{2}={26}/{4}=\;{13}/{2}

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