Question

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​

Answer 1

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 2

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