Question

Prove that $\left( {a,b + c} \right)\left( {b,c + a} \right)\left( {c,a + b} \right)$ are collinear points.

Answer 1

We use the concept of collinear points and that three points in general form a triangle and if the points are collinear then the area of the triangle formed by collinear points is equal to zero.

• Three points are said to be collinear if they lie on the same line.
• Area of a triangle formed by points $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ is given by \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|
• Determinant of a matrix \left[ {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right] = a(ei - hf) - b(di - fg) + c(dh - eg)

Complete step-by-step solution:
We are given three points $\left( {a,b + c} \right);\left( {b,c + a} \right);\left( {c,a + b} \right)$
Let us assume three points form a triangle.
Then we can write the area of triangle formed by the three points as \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} a&{b + c}&1 \\\ b&{c + a}&1 \\\ c&{a + b}&1 \end{array}} \right|
We calculate the determinant value using column transformations
$\Rightarrow$Area of triangle = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} a&{b + c}&1 \\\ b&{c + a}&1 \\\ c&{a + b}&1 \end{array}} \right|
Applying column operation ${C_1} \to {C_1} + {C_2}$ to the determinant
$\Rightarrow$Area of triangle = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} {a + b + c}&{b + c}&1 \\\ {a + b + c}&{c + a}&1 \\\ {a + b + c}&{a + b}&1 \end{array}} \right|
Since all elements of first column are same we can take out the same element
$\Rightarrow$Area of triangle = \dfrac{1}{2}(a + b + c)\left| {\begin{array}{*{20}{c}} 1&{b + c}&1 \\\ 1&{c + a}&1 \\\ 1&{a + b}&1 \end{array}} \right|
Applying column operation ${C_1} \to {C_1} - {C_3}$ to the determinant
$\Rightarrow$Area of triangle = \dfrac{1}{2}(a + b + c)\left| {\begin{array}{*{20}{c}} 0&{b + c}&1 \\\ 0&{c + a}&1 \\\ 0&{a + b}&1 \end{array}} \right|
Since one whole column of the determinant is zero, then the value of the determinant is zero.
$\Rightarrow$Area of triangle$= \dfrac{1}{2}(a + b + c) \times 0$
We know when zero multiplied by any other number is always zero
$\Rightarrow$Area of triangle$= 0$
Since the area of the triangle is zero, then the three points which are said to be vertices of the triangle are collinear points.

$\therefore$Three points $\left( {a,b + c} \right)\left( {b,c + a} \right)\left( {c,a + b} \right)$ are collinear points.

Note: Alternate method:
We can also prove the determinant equal to zero by showing the two columns identical to each other. Since we know area of triangle is given by
$\Rightarrow$Area of triangle = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} a&{b + c}&1 \\\ b&{c + a}&1 \\\ c&{a + b}&1 \end{array}} \right|
Applying column operation${C_1} \to {C_1} + {C_2}$to the determinant
$\Rightarrow$Area of triangle = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}} {a + b + c}&{b + c}&1 \\\ {a + b + c}&{c + a}&1 \\\ {a + b + c}&{a + b}&1 \end{array}} \right|
Since all elements of first column are same we can take out the same element
$\Rightarrow$Area of triangle = \dfrac{1}{2}(a + b + c)\left| {\begin{array}{*{20}{c}} 1&{b + c}&1 \\\ 1&{c + a}&1 \\\ 1&{a + b}&1 \end{array}} \right|
Here first and the third columns are identical to each other, so the value of determinant is zero
$\Rightarrow$Area of triangle$= \dfrac{1}{2}(a + b + c) \times 0$
$\Rightarrow$Area of triangle$= 0$