Question

if A(0,0) B(1,3) C(k,0) are the vertices of triangle ABC whose area is 3 square unit, then value of k is ​

Answer 1

Given

Three vertices of triangle ABC

• A (0,0) , B (1,3) , C (k,0)

Area of the triangle is 3 square units.

To find

• Value of k

Formula required

• Formula to calculate the area of a triangle with vertices $\sf{(x_1,y_1),\;(x_2,y_2),\;(x_3,y_3)}$ is given by

$\purple{\bigstar}\boxed{\sf{Ar(\triangle) = \dfrac{1}{2}\mid [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\mid}}$

Solution

On comparing

• $\sf{x_1}$ = 0 , $\sf{y_1}$ = 0
• $\sf{x_2}$ = 1 , $\sf{y_2}$ = 3
• $\sf{x_3}$ = k , $\sf{y_3}$ = 0

Using formula for area of triangle

$\implies\sf{3 = \dfrac{1}{2}\mid [0(3-0) + 1(0-0)+k(0-3)]\mid }$

$\implies\sf{3 = \dfrac{1}{2}\mid [0+ 0-3k]\mid }$

$\implies\sf{3 = \dfrac{1}{2}\times(3k)}$

$\implies\sf{3k=6}$

$\implies\boxed{\boxed{\sf{\;k=2\;}}}\;\;\purple{\bigstar}$

Therefore,

• Value of k is 2 .

Answer 2

Answer:

Given :

• If the area of triangle with vertices (1 ,3) (0, 0) and (K ,0) is 3 square units

To Find : -

• whose area is 3 square unit, then value of k is

Solution : -

Area = 1/2[x_1(y_2 - y_3) + x2(y_3 - y_1) + x3(y1 - y2) ]

Substituting the values

x_1 = 1 y_ 1 = 3

x_2 = 0 y_2 = 0

x_3 = K y_3 = 0

Then we get

A = 1/2[1( 0 - 0) +0(0 - 3) + K(3 - 0) ]

As Area is 3square units then

3 = 3K /2

K = 6/3

K = 2

Answer is k = 2