Question

the area of the triangle formed by the points A(2,0) , b(6,0) and c(4,6)​

Answer 1

Answer:

The given points are (2,0), (6,0) and (4,6). = 1/2 [ 6(6 - 2) ], expanding along the 2nd column and taking the positive value. = 12 square units.

Answer 2

Given :-

• A(2,0)
• B(6,0)
• C(4,6)

where,

$\sf \: x_1 = 2 \: \: \: \: \: \: \: \: x_2 = 6 \: \: \: \: \: \: \: \: \: \: \: \: \: x_3 = 4 \\ \\ \sf \:</p><p>y_1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: </p><p>y_2 = 0 \: \: \: \: \: \: \: \: \: \: </p><p>y_3 = 6$

Formula used :-

$\sf \: area \: of \: triangle = \\ \sf \: \frac{1}{2} \bigg[x_1 \big(y_2 -y_3 \big) + x_2 \big(y_3 - y_2 \big) + x_3 \big(y_1 - y_2 \big)\bigg]$

Solution :-

$\sf \: area \: of \: triangle = \\ \sf \: \dfrac{1}{2} \bigg[x_1 \big(y_2 -y_3 \big) + x_2 \big(y_3 - y_2 \big) + x_3 \big(y_1 - y_2 \big)\bigg] \\ \\ \sf \: area \: of \: triangle = \\ \sf \: \dfrac{1}{2} \bigg[2(0 - 6) + 6(6 - 0) + 4(0 - 0)\bigg] \\ \\\sf \: area \: of \: triangle = \dfrac{1}{2} \bigg[2 \times ( - 6) + 6 \times 6 + 4 \times 0 \bigg] \\ \\ \sf \implies \: \dfrac{1}{2} ( - 12 + 36 + 0) \\ \\ \sf \implies \: \dfrac{1}{2} \times 24 \\ \\ \sf \implies \: \dfrac{1}{\cancel2} \times \cancel{ 24} \\ \\ \sf \: area \: of \: triangle = \: 12 \: {unit}^{2}$

thus, area of triangle is 12 unit ² .