Examples on Area of a Triangle Formula

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Example – 3

Find the area of the triangle, the co-ordinates of whose vertices are \(\begin{align}\left( ap,\,\,\frac{a}{p} \right),\left( aq,\,\,\frac{a}{q} \right)\,\,\text{and}\,\,\left( ar,\,\,\frac{a}{r} \right)\end{align}\).

Solution: Using the result obtained is Art - 3, we have,

\[\begin{align}&\; \Delta =\left| \,\frac{1}{2}\,\,\,\left| \begin{matrix} ap & \frac{a}{p} & 1  \\ aq & \frac{a}{q} & 1  \\   ar & \frac{a}{r} & 1  \\
\end{matrix} \right|\, \right| \\  &  \\  & \quad=\frac{1}{2}\left| ap\left( \frac{a}{q}-\frac{a}{r} \right)+aq\left( \frac{a}{r}-\frac{a}{p} \right)+ar\left( \frac{a}{p}-\frac{a}{q} \right)\, \right| \\ &\quad =\frac{{{a}^{2}}}{2}\left| \frac{p\left( r-q \right)}{qr}+\frac{q\left( p-r \right)}{pr}+\frac{r\left( q-p \right)}{pq} \right| \\  &\quad =\frac{{{a}^{2}}\left| {{p}^{2}}\left( q-r \right)+{{q}^{2}}\left( r-p \right)+{{r}^{2}}\left( p-q \right) \right|}{2\left| pqr \right|} \\ 
\end{align}\]

Example – 4

Assume two fixed points in the co-ordinate plane: \(A\left( a,\,0 \right)\,\,\text{and}\,\,B\left( -a,\,0 \right)\) . A variable point C(x, y) moves in the plane in such a way that CA + CB is a constant k. Use the distance formula to evaluate the condition that the co-ordinates of C must satisfy.

Solution: We have,

\[\begin{align} & \qquad CA=\sqrt{{{\left( x-a \right)}^{2}}+{{y}^{2}}} \\  & \text{and}\,\,CB=\sqrt{{{\left( x+a \right)}^{2}}+{{y}^{2}}} \end{align}\]

From the constraint specified in the question, we have

\[\begin{align}& \qquad \qquad CA+CB=k \\ & \Rightarrow\qquad C{{A}^{2}}+C{{B}^{2}}+2\,CA\cdot CB={{k}^{2}} \\ & \Rightarrow\qquad {{\left( x-a \right)}^{2}}+{{y}^{2}}+{{\left( x+a \right)}^{2}}+{{y}^{2}}+2\sqrt{\left( {{\left( x-a \right)}^{2}}+{{y}^{2}} \right)\left( {{\left( x+a \right)}^{2}}+{{y}^{2}} \right)}={{k}^{2}} \\ & \Rightarrow \qquad2\sqrt{{{\left( {{x}^{2}}-{{a}^{2}} \right)}^{2}}+{{y}^{4}}+2{{y}^{2}}\left( {{x}^{2}}+{{a}^{2}} \right)\,}=\,\,{{k}^{2}}-2\left( {{x}^{2}}+{{y}^{2}}+{{a}^{2}} \right) \\ & \Rightarrow\qquad 2\sqrt{{{x}^{4}}+{{y}^{4}}+{{a}^{4}}+2{{x}^{2}}{{y}^{2}}-2{{a}^{2}}{{x}^{2}}+2{{a}^{2}}{{y}^{2}}}\,\,=\,\,{{k}^{2}}-2\left( {{x}^{2}}+{{y}^{2}}+{{a}^{2}} \right) \\ 
\end{align}\]

Squaring both sides and cancelling out the common terms on both sides, we obtain

\[\begin{align} &\qquad\quad -8{{a}^{2}}{{x}^{2}}={{k}^{4}}+8{{a}^{2}}{{x}^{2}}-4{{k}^{2}}\left( {{x}^{2}}+{{y}^{2}}+{{a}^{2}} \right) \\ & \Rightarrow \qquad4{{k}^{2}}{{x}^{2}}-16{{a}^{2}}{{x}^{2}}+4{{k}^{2}}{{y}^{2}}={{k}^{4}}-4{{k}^{2}}{{a}^{2}} \\  & \Rightarrow \qquad 4{{x}^{2}}\left( {{k}^{2}}-4{{a}^{2}} \right)+4{{k}^{2}}{{y}^{2}}={{k}^{2}}\left( {{k}^{2}}-4{{a}^{2}} \right) \\ 
\end{align}\]

This is the relation that the co-ordinates of the variable point C (x, y) must satisfy. All the pairs (x, y) which satisfy this equation, when plotted on the co-ordinate plane, will trace out the path on which the variable point C is constrained to move. In other words, this equation specifies the locus of the point C.

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