Question

What is the area of ​​a triangle with the vertices (3, 5), (- 2, 0) and (6, 4):​

Answer 1

Step-by-step explanation:

ar of ∆=1/2[X1(y2-y3)+X2(y3-y1)+X3(y2-y1)]

1/2[3(0-4)-2(4-5)+6(0-5)]

1/2 [3(-4)-2(-1)+6(-5)]

1/2[-12+2-30]

1/2[-12-28]

1/2[-40]

= -20 units

Answer 2

Given :

What is the area of a triangle with the vertices (3, 5), (- 2, 0) and (6, 4)

To find :

Find the Area of triangle

Solution :

Applying Formula of area of triangle

$\implies\sf \frac{1}{2}(x_1(y_2-y_3)+x_2(y_3-y_2)+x_3(y_1-y_2)$

$\implies\sf \frac{1}{2}(3(0-4)+(-2)(4-0)+6(5-0)$

$\implies\sf \frac{1}{2}(3\times{-4}+(-2)\times{4}+6\times{5}$

$\implies\sf \frac{1}{2}(-12-8+30)$

$\implies\sf \frac{1}{2}\times{10}=5sq.unit$

Additional information :

★ same sign = addition

★ opposite sign = subtraction

★ Distance formula = $\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

★ Mid point Formula = $\sf \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}$

★ Section formula = $\sf \frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}$