if a x1 x2 are in gp with common ratio r and b y1 y2 are in gp with common ratio s where s-r=2 then the area of the triangle with vertices (a,b),(x1,y1),(x2,y2) is
A)|ab(r^2-1)|
B)ab(r^2-s^2)
C)ab(s^2-1)
D)abrs
Answers
Answer:
A)|ab(r²2-1)
Step-by-step explanation:
∆=1/2|a b 1|
|ar bs 1|
|ar² bs²1|
=ab/2|(r-s)(s-1)(1-r)|
=|ab(r²2-1)|
Answer:
The area of the triangle is,
A. .
Step-by-step explanation:
Consider a triangle with vertices: .
Then the equation for the area of a triangle is:
For the provided scenario, the vertices of the triangle are:
The area of the triangle with the above vertices is,
Now, consider the facts:
- The values are in GP having a recurrent ratio , and
- The values are in GP having a recurrent common ratio
- And the relationship between these two recurrent ratios is,
or .
For the GP , the relation between the values is expressed as follows:
For the GP , the relation between the values is expressed as follows:
Substitute the values of in the equation for "Area" and solve as follows:
Now substitute in the equation for area of the triangle, and simplify as follows:
Thus, the area of the triangle with vertices is .
Correct option: A