Math, asked by aishwaryanetha8, 6 months ago

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

Answered by asrithagorantla
5

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)|

Answered by akansha804
0

Answer:

The area of the triangle is,

A. |ab(r^{2}-1)|.

Step-by-step explanation:

Consider a triangle with vertices: (x_{1},\ y_{1}),\ (x_{2},\ y_{2}),\ \text{and}\ (x_{3},\ y_{3}).

Then the equation for the area of a triangle is:

\text{Area}=\frac{1}{2}|y_{1}(x_{2}-x_{3})+y_{2}(x_{3}-x_{1})+y_{3}(x_{1}-x_{2})|

For the provided scenario, the vertices of the triangle are:

(x_{1},\ y_{1})=(a,b)\\\\(x_{2},\ y_{2})=(x_{1},\ y_{1})\\\\(x_{3},\ y_{3})=(x_{2},\ y_{2})

The area of the triangle with the above vertices is,

\text{Area}=\frac{1}{2}|b(x_{1}-x_{2})+y_{1}(x_{2}-a)+y_{2}(a-x_{1})|

Now, consider the facts:

  • The values a,\ x_{1}\ \text{and}\ x_{2} are in GP having a recurrent ratio r, and
  • The values b,\ y_{1}\ \text{and}\ y_{2} are in GP having a recurrent common ratio s
  • And the relationship between these two recurrent ratios is,

        s-r=2 or s=r+2.

For the GP a,\ x_{1}\ \text{and}\ x_{2}, the relation between the values is expressed as follows:

\frac{x_{1}}{a}=\frac{x_{2}}{x_{1}}=r\\\\\Rightarrow\ x_{1}=ar\ \text{and}\ x_{2}=x_{1}r\\\\\Rightarrow\ x_{2}=(ar)\cdot r\\\\\Rightarrow\ x_{2}=ar^{2}...(i)

For the GP b,\ y_{1}\ \text{and}\ y_{2} , the relation between the values is expressed as follows:

\frac{y_{1}}{b}=\frac{y_{2}}{y_{1}}=s\\\\\Rightarrow\ y_{1}=bs\ \text{and}\ y_{2}=y_{1}s\\\\\Rightarrow\ y_{2}=(bs)\cdot s\\\\\Rightarrow\ y_{2}=bs^{2}...(ii)

Substitute the values of x_{1},\ x_{2},\ y_{1}\ \text{and}\ y_{2} in the equation for "Area" and solve as follows:

\text{Area}=\frac{1}{2}\cdot |b(x_{1}-x_{2})+y_{1}(x_{2}-a)+y_{2}(a-x_{1})|\\\\=\frac{1}{2}\cdot |b(ar-ar^{2})+bs(ar^{2}-a)+bs^{2}(a-ar)|\\\\=\frac{1}{2}\cdot |ba[(r-r^{2})+s(r^{2}-1)+s^{2}(1-r)]|\\\\=\frac{1}{2}\cdot |ba[r(1-r)+s(r-1)(r+1)+s^{2}(1-r)]|\\\\=\frac{1}{2}\cdot |ba(1-r)[r-s(1+r)+s^{2}]|\\\\=\frac{1}{2}\cdot |ba(1-r)(s-r)(s-1)|

Now substitute s=r+2 in the equation for area of the triangle, and simplify as follows:

\text{Area}=\frac{1}{2}\cdot |ba(1-r)(s-r)(s-1)|\\\\=\frac{1}{2}\cdot |ba(1-r)(r+2-r)(r+2-1)|\\\\=\frac{1}{2}\cdot |2\cdot ba\cdot (1-r)(r+1)|\\\\=|ba\cdot (1-r^{2})|\\\\\sim |ab(r^{2}-1|

Thus, the area of the triangle with vertices (a, b),\ (x_{1},\ y_{1})\ \text{and}\ (x_{2},\ y_{2}) is |ab(r^{2}-1)|.

Correct option: A

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