Question

Find the area of triangle ABC whose vertices are A (-5, 7), B (-4, -5) and C (4, 5).

Answer 1

GIVEN :–

• Vertices of triangle are A(-5 , 7) , B(-4 , -5) & C(4 , 5).

TO FIND :–

• Area of triangle = ?

SOLUTION :–

• If vertices are $\:\: \bf A(x_1 , y_1 ) \:\:$ , $\:\: \bf B(x_2 , y_2 ) \:\:$ & $\:\: \bf C(x_3 , y_3 ) \:\:$ , then Area –

$\\ \implies \bf \: Area = \left|\begin{array}{ccc}\bf\:x_1 & \bf y_1 & \bf1 \\ \\ \bf x_2 & \bf\:y_2 & \bf 1 \\ \\ \bf x_3 & \bf y_3 & \bf1 \end{array}\right|$

• Here –

$\\ \bf { \huge{.}} \:\:\:\: x_1 = -5$

$\\ \bf { \huge{.}} \:\:\:\: x_2 = -4$

$\\ \bf { \huge{.}} \:\:\:\: x_3 = 4$

$\\ \bf { \huge{.}} \:\:\:\: y_1 = 7$

$\\ \bf { \huge{.}} \:\:\:\: y_2 = -5$

$\\ \bf { \huge{.}} \:\:\:\: y_3 = 5$

• Now put the values –

$\\ \implies \bf \: Area = \left|\begin{array}{ccc}\bf\: - 5 & \bf 7 & \bf1 \\ \\ \bf - 4 & \bf\: - 5 & \bf 1 \\ \\ \bf 4 & \bf 5 & \bf1 \end{array}\right|$

$\\ \implies \bf \: Area = - 5( - 5 - 5) - 7( - 4 - 4) + 1( - 20 + 20)$

$\\ \implies \bf \: Area = - 5( - 10) - 7( - 8) +0$

$\\ \implies \bf \: Area = 50 + 56$

$\\ \implies \large{ \boxed{ \bf Area = 106 \: \: unit}}$

▪︎ Hence , The area of triangle is 106 unit.

Answer 2

ANSWER:-

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Area of ΔABC whose vertices are is

$(x_1,y_1),(x_2,y_2) \: and \: (x_3,y_3)are$

$Area \: of ΔABC = \frac{1}{2} [x_1(y_2,y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

In ΔABC, vertices are A(−5,7),B(−4,−5) and C(4,5)

$Area \: of \: triangle \: = \frac{1}{2} (- 5 (- 5 - 5 ) - 4(5 - 7) + 4(7 + 5))$

$= \frac{1}{2} ( - 50 + 8 + 48)$

$= 5 \: sq. \: units$

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