Question

Find the are of a triangle whose vertices are (2,3) (4,0) 6,-3)​

Answer 1

Given :-

Let us consider a triangle ABC having vertices

• A (2, 3)

• B (4, 0)

• C (6, - 3)

To Find :-

• Area of triangle ABC

Solution :-

Given that vertices of triangle ABC are

• A (2, 3)

• B (4, 0)

• C (6, - 3)

We know,

Area of triangle ABC is

$\sf \ Area =\dfrac{1}{2}| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Here,

• • x₁ = 2

• • x₂ = 4

• • x₃ = 6

• • y₁ = 3

• • y₂ = 0

• • y₃ = - 3

Thus,

$\sf \ Area =\dfrac{1}{2}|2(0 + 3) + 4( - 3 - 3) + 6(3 - 0)|$

$\sf \ Area =\dfrac{1}{2}|6 - 24 + 18|$

$\sf \ Area =\dfrac{1}{2}|24 - 24 |$

$\sf \ Area =\dfrac{1}{2}|0 |$

$\rm :\implies\:\sf \ Area \: = \: 0$

Additional Information :-

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the midpoint of AB, then coordinates of C is

$\rm :\longmapsto\: \sf \: C\: = \: \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg)$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: B(x_2,y_2)$

Distance Formula :-

 

Let us consider a line segment joining the points A and B and then distance between A and B is

${\underline{\boxed{\rm{\quad Distance \: AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

Section Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the point on AB, which divides in the ratio m : n, then coordinates of C is

$\rm :\longmapsto\: \sf \: C \: = \: \bigg(\dfrac{nx_1+mx_2}{m + n} , \dfrac{ny_1+my_2}{m + n} \bigg)$