Question

Find the area of a triangle with vertices (-2, -3), (3, 2) and (-1,-8).​

Answer 1

$\sf{Answer}$

Given :-

Vertices of triangle (-2 , -3 ) ,( 3, 2 ) and ( -1 , -8 )

To find :-

• Area of triangle

Formula to know :-

Area of triangle =

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

Solution :-

Vertices (-2 , -3) , (3 , 2) and (-1 , -8 )

${x_1 = -2}$

${x_2 = 3}$

${x_3 = -1}$

${y_1 = -3}$

${y_2 = 2}$

${y_3 = -8}$

Substituting values in above formula

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

$\frac{1}{2} | - 2(2 - ( - 8)) + 3( - 8 - ( - 3)) - 1( - 3 -2) |$

$\frac{1}{2} | - 2(2 + 8) + 3( - 8 + 3) - 1( - 5)|$

$\frac{1}{2} | - 2(10) + 3(-5) + 5|$

$\frac{1}{2} | - 20 -15+ 5|$

$\frac{1}{2} | - 30|$

$\frac{30}{2}$

${15 sq.units}$

So, area of triangle is 15 sq.units

__________________

Know more :-

Distance formula:-

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Centroid formula:-

$\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}$

Section formula Internal

division:-

$\dfrac{mx_2+nx_1}{m+n}, \dfrac{my_2+ny_1}{m+n}$

Section formula External division:-

$\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}$

Answer 2

Answer:

दुफ्फुद्ज युक्सफीय इफूप अच्छी