Question

The area of ABC is 3 sq.units, where A(1,3), B(0,0)and C(K,O) are the vertices. The set values
of K is

Answer 1

Answer:

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Answer 2

Answer:The value of K=2

Step-by-step explanation:

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

Then

Area= 1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2) ]

Substituting the values of

x1= 1 y1=3

x2=0 y2=0

x3= K y3=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

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