Question

find the area of a triangle whose vertices are (8, 1), (3, -4),and (2, -5).​

Answer 1

Answer:

ans is 0

Step-by-step explanation:

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Answer 2

Solution:

The coordinate of the vertices of the given triangle are

$\rm \: A(x_1 = 8,y_1 = 1),B(x_2 = 3,y_2 = - 4) \: and \: C(x_3 = 2,y_3 = - 5)$

Formula

$\boxed{ \rm \: ar(</em></strong><strong><em>\</em></strong><strong><em>Delta</em></strong><strong><em> </em></strong><strong><em>\</em></strong><strong><em>:</em></strong><strong><em> </em></strong><strong><em>ABC) = \dfrac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 -y _2)| }$

by putting the value on formula

$\rm \: ar(∆ABC) = \dfrac{1}{2} |8( - 4 -( - 5) )+ 3( - 5 - 1) + 2(1-( - 4))|$

$\rm= \dfrac{1}{2} |8( - 4 + 5)+ 3( - 6) + 2(1 + 4)|$

$\rm= \dfrac{1}{2} |8( 1)+ 3( - 6) + 2(5)|$

$\rm= \dfrac{1}{2} |8 - 18 + 10|$

$\rm= \dfrac{1}{2} |18 - 18 |$

$= 0$

Its mean given point are collinear