NIFT UG 2018 Question Paper with solutions

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Question : 8 of 150

Marks: +1, -0

Find the area of

∆ A B C

whose vertices are

A(10,-6), B(2,5)

and

C(-1,3)

12 sq units
$\;{37}/{2}$ sq units
455 sq units
$\;{49}/{2}$ sq units

Solution:

Here,

x_{1}=10, x_{2}=2, x_{3}=-1

and

\;\; y_{1}=-6, y_{2}=5, y_{3}=3

Area of

∆\ABC=\;{1}/{2}\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\}

=\;{1}/{2}\{10(5-3)+2(3+6)-1(-6-5)\}

=\;{1}/{2}(20+18+11)

=\;{49}/{2}

sq units

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