Question

Find the area of triangle abc whose vertices are a(1,2) b(-2,3) c(-3,-4)

Answer 1

Area of ∆ABC is 11 sq-unit.

Given:

• Vertices of triangle ABC.
• A(1,2), B(-2,3), and C(-3,-4).

To find:

• Find the area of ∆ABC.

Solution:

Formula to be used:

If vertices of triangle are $A(x_1,y_1),\:B(x_2,y_2),\:and\:C(x_3,y_3)$

$\bf Ar.(\triangle\:ABC)= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Put the values of vertices in the formula.

A(1,2), B(-2,3), and C(-3,-4).

$Ar.(\triangle\:ABC)= \frac{1}{2} |1(3 - ( - 4)) + ( - 2)( -4 - 2) + ( - 3)(2 - 3)|$

or

$= \frac{1}{2} |1(3 + 4) - 2( -6) - 3( - 1)|$

or

$= \frac{1}{2} |7 + 12 + 3|$

or

$= \frac{1}{2} |22|$

or

Thus,

Area of ∆ABC= 11 sq-unit.

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