Calculating Triangle Area with Vertices in Geometric Progression

Let the vertices of the triangle be $A = (a, b)$, $B = (x_1, y_1)$, and $C = (x_2, y_2)$.

We are given that the x-coordinates $a, x_1, x_2$ are in geometric progression (GP) with the common ratio $r$. This means:

• $x_1 = ar$
• $x_2 = ar^2$

Similarly, the y-coordinates $b, y_1, y_2$ are in geometric progression (GP) with the common ratio $s$. This means:

• $y_1 = bs$
• $y_2 = bs^2$

So, the vertices of the triangle are:

• $A = (a, b)$
• $B = (ar, bs)$
• $C = (ar^2, bs^2)$

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

\(\text{Area} = \dfrac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)

Let's substitute the coordinates of our triangle vertices into this formula:

\(x_1 = a, y_1 = b\)

\(x_2 = ar, y_2 = bs\)

\(x_3 = ar^2, y_3 = bs^2\)

\(\text{Area} = \dfrac{1}{2} |a(bs^2 - bs^2) + ar(bs^2 - b) + ar^2(b - bs)|\)

Let's simplify the expression inside the absolute value:

\(a(bs^2 - bs^2) = a \cdot 0 = 0\)

Ah, there was a typo in the initial substitution. Let's correct it using the formula $\text{Area} = \dfrac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ with $(x_1, y_1) = (a, b)$, $(x_2, y_2) = (ar, bs)$, and $(x_3, y_3) = (ar^2, bs^2)$.

\(\text{Area} = \dfrac{1}{2} |a(bs - bs^2) + ar(bs^2 - b) + ar^2(b - bs)|\)

Now, factor out common terms:

\(\text{Area} = \dfrac{1}{2} |ab(s - s^2) + abr(s^2 - 1) + abr^2(1 - s)|\)

Factor out $ab$:

\(\text{Area} = \dfrac{1}{2} |ab [ (s - s^2) + r(s^2 - 1) + r^2(1 - s) ]|\)

Rewrite terms to factor $(s-1)$ or $(1-s)$:

\(\text{Area} = \dfrac{1}{2} |ab [ s(1 - s) + r(s - 1)(s + 1) + r^2(1 - s) ]|\)

Replace $(1-s)$ with $-(s-1)$:

\(\text{Area} = \dfrac{1}{2} |ab [ -s(s - 1) + r(s - 1)(s + 1) - r^2(s - 1) ]|\)

Factor out $(s-1)$ from the terms inside the square brackets:

\(\text{Area} = \dfrac{1}{2} |ab (s - 1) [ -s + r(s + 1) - r^2 ]|\)

Simplify the expression inside the square brackets:

\(-s + r(s + 1) - r^2 = -s + rs + r - r^2\)

Rearrange and factor the terms:

\(rs + r - s - r^2 = s(r - 1) - r(r - 1)\)

Factor out $(r-1)$:

\(s(r - 1) - r(r - 1) = (r - 1)(s - r)\)

Substitute this back into the area expression:

\(\text{Area} = \dfrac{1}{2} |ab (s - 1) (r - 1) (s - r)|\)

This can also be written as:

\(\text{Area} = \dfrac{1}{2} |ab (r - 1) (s - 1) (s - r)|\)

Given the options, the expression is presented without the absolute value sign. We match our derived expression with the options.

The formula for the area of the triangle is $\dfrac{1}{2}ab(r-1)(s-1)(s-r)$.

Revision Table: Triangle Area with GP Coordinates

Additional Information: Geometric Progression and Triangle Area

A geometric progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In this problem, the x-coordinates $a, ar, ar^2, ...$ form a GP with common ratio $r$, and the y-coordinates $b, bs, bs^2, ...$ form a GP with common ratio $s$.

The formula for the area of a triangle using coordinates is derived from the determinant method or vector cross product, and it gives the signed area. Taking the absolute value ensures the area is non-negative.

When the vertices are given as $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area can also be calculated as:

\(\text{Area} = \dfrac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|\)

Substituting the given coordinates:

\(\text{Area} = \dfrac{1}{2} \left| \det \begin{pmatrix} a & b & 1 \\ ar & bs & 1 \\ ar^2 & bs^2 & 1 \end{pmatrix} \right|\)

Expanding the determinant:

\(\text{Area} = \dfrac{1}{2} |a(bs - bs^2) - b(ar - ar^2) + 1(ar \cdot bs^2 - ar^2 \cdot bs)|\)

\(\text{Area} = \dfrac{1}{2} |abs(1 - s) - abr(1 - r) + abr s(s - r)|\)

\(\text{Area} = \dfrac{1}{2} |ab [s(1 - s) - r(1 - r) + rs(s - r)]|\)

\(\text{Area} = \dfrac{1}{2} |ab [s - s^2 - r + r^2 + rs^2 - r^2s]|\)

This expression looks different but should simplify to the same result. Let's check the previous algebraic simplification, which seems more direct.

The previous simplification was:

\(\dfrac{1}{2} |ab [ -s(s - 1) + r(s - 1)(s + 1) - r^2(s - 1) ]|\)

\(\dfrac{1}{2} |ab (s - 1) [ -s + r(s + 1) - r^2 ]|\)

\(\dfrac{1}{2} |ab (s - 1) [ -s + rs + r - r^2 ]|\)

\(\dfrac{1}{2} |ab (s - 1) [ s(r - 1) - r(r - 1) ]|\)

\(\dfrac{1}{2} |ab (s - 1) (r - 1) (s - r)|\)

This confirms the earlier derivation using the first area formula form.