Question

If \[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&1&1\\{{b_1}}&{{b_2}}&{{b_3}}\\{{a_1}}&{{a_2}}&{{a_3}}\end{array}} \right|\] then find the nature of two triangles whose coordinates are \[({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})\]and \[({a_1},{b_1}),({a_2},{b_2}),({a_3},{b_3})\].
A. Congruent
B. Similar
C. Equal in areas
D. Right angled triangles

Answer 1

Hints Write the formula of the area of two triangles. Then use the given equation and properties of the determinant to obtain the relation between two triangles.

Formula used
The area of a triangle with vertices \[({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})\] is
\[\dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|\].
The value of determinant A and its transpose is always equal. That is,
\[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{x_1}}&{{x_2}}&{{x_3}}\\{{y_1}}&{{y_2}}&{{y_3}}\\1&1&1\end{array}} \right|\]
Also,
\[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{x_2}}&{{x_3}}\\{{y_1}}&{{y_2}}&{{y_3}}\\1&1&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&1&1\\{{x_1}}&{{x_2}}&{{x_3}}\\{{y_1}}&{{y_2}}&{{y_3}}\end{array}} \right|\].

Complete step by step solution
The given equation is,
\[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&1&1\\{{b_1}}&{{b_2}}&{{b_3}}\\{{a_1}}&{{a_2}}&{{a_3}}\end{array}} \right|\]
Now we will transpose the row and column of the right side determinat:
\[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}1&{{b_1}}&{{a_1}}\\1&{{b_2}}&{{a_2}}\\1&{{b_3}}&{{a_3}}\end{array}} \right|\]
Now interchange the column 1 and column 2 of the right side determinant:
\[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = -\left| {\begin{array}{*{20}{c}}{{b_1}}&1&{{a_1}}\\{{b_2}}&1&{{a_2}}\\{{b_3}}&1&{{a_3}}\end{array}} \right|\]
Now interchange column 2 and column 3 of the right side determinant:
\[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = (-1)^2\left| {\begin{array}{*{20}{c}}{{b_1}}&{{a_1}}&1\\{{b_2}}&{{a_2}}&1\\{{b_3}}&{{a_3}}&1\end{array}} \right|\]
Now interchange column 1 and column 2 of the right side determinant:
\[\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = (-1)^3\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
Taking absolute signs on both sides:
\[\left| \left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| \right| = \left| (-1)^3\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right| \right| \]
Multiply both sides by \[\dfrac{1}{2}\]
\[\left| \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| \right| = \left| \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right| \right|\]
Area of triangle whose vertices are \[({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})\] = Area of triangle whose vertices are \[({a_1},{b_1}),({a_2},{b_2}),({a_3},{b_3})\]
Therefore, the area of the two triangles is the same.

The correct option is “C”.

Additional information The area of a triangle is always a positive quantity. The area formed by three collinear points of a triangle is zero. The general formula of the area of a triangle is base multiplied by height divided by 2. But in coordinate geometry when three vertices of the triangle are given we use the determinant form to obtain the area of the given triangle.

Note Students often do mistakes when they interchange the rows or columns of a determinant. They forgot to multiply -1 when they interchange the rows. When we interchange two consecutive rows or columns then we multiply -1.