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Question: If the three points \[A\left( 1,6 \right),B\left( 3,-4 \right)\] and \(C\left( x,y \right)\) are col...

If the three points A(1,6),B(3,−4)A\left( 1,6 \right),B\left( 3,-4 \right) and C(x,y)C\left( x,y \right) are collinear, then the equation satisfying by xx and yy is,
(a) 5x+y−11=05x+y-11=0
(b) 5x+13y+5=05x+13y+5=0
(c) 5x−13y+5=05x-13y+5=0
(d) 13x−5y+5=013x-5y+5=0

Explanation

Solution

Hint:Since, the points are collinear then we will apply the formula which is used to find the area of a triangle and we will equate that area equal to 0. This is because the line cannot be a triangle. By the use of this formula we will find the equation satisfying x and y. The formula for the area of triangle equated to 0 is given by 12∣x1  y1  1 x2  y2  1 x3  y3  1 ∣=0\dfrac{1}{2}\left| \left. \begin{aligned} & {{x}_{1}}\,\,{{y}_{1}}\,\,1 \\\ & {{x}_{2}}\,\,{{y}_{2}}\,\,1 \\\ & {{x}_{3}}\,\,{{y}_{3}}\,\,1 \\\ \end{aligned} \right| \right.=0.

Complete step-by-step answer:
First we will use the trick which says that the area of the triangle can also be carried out to the points that are collinear. The condition which is already given is that the points are collinear.
This means that the area of the triangle is zero. So now we will use the formula for the area of the triangle. It is given by 12∣x1  y1  1 x2  y2  1 x3  y3  1 ∣=0\dfrac{1}{2}\left| \left. \begin{aligned} & {{x}_{1}}\,\,{{y}_{1}}\,\,1 \\\ & {{x}_{2}}\,\,{{y}_{2}}\,\,1 \\\ & {{x}_{3}}\,\,{{y}_{3}}\,\,1 \\\ \end{aligned} \right| \right.=0 thus, after substituting the values into the formula we get 12∣1       6    1 3  −4    1 x    y     1 ∣=0\dfrac{1}{2}\left| \left. \begin{aligned} & 1\,\,\,\,\,\,\,6\,\,\,\,1 \\\ & 3\,\,-4\,\,\,\,1 \\\ & x\,\,\,\,y\,\,\,\,\,1 \\\ \end{aligned} \right| \right.=0 therefore, after expanding we get 12[1(−4−y)−6(3−x)+1(3y+4x)]=0\dfrac{1}{2}\left[ 1\left( -4-y \right)-6\left( 3-x \right)+1\left( 3y+4x \right) \right]=0

& \Rightarrow \dfrac{1}{2}\left[ -4-y-18+6x+3y+4x \right]=0 \\\ & \Rightarrow \dfrac{1}{2}\left[ 10x+2y-22 \right]=0 \\\ & \Rightarrow 5x+y-11=0 \\\ \end{aligned}$$ Hence the correct option is (a). Note: Clearly the third point is in (x, y) form so the formula will imply an equation of any line of a triangle. The points are collinear if the slope of any pairs of two points out of three points are equal. Here it means that if we consider $$A\left( 1,6 \right)$$, $B\left( 3,-4 \right)$ and $C\left( x,y \right)$ also if we take the pairs AB, BC and CA together then Slope of AB = Slope of BC = Slope of CA. We will apply the formula given by $\dfrac{{{y}_{3}}-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{{{x}_{3}}-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. In this formula we will substitute the points of A and C. Therefore we have $\begin{aligned} & \dfrac{{{y}_{3}}-6}{y-6}=\dfrac{{{x}_{3}}-1}{x-1} \\\ & \Rightarrow \left( {{y}_{3}}-6 \right)\left( x-1 \right)=\left( {{x}_{3}}-1 \right)\left( y-6 \right) \\\ & \Rightarrow {{y}_{3}}\left( x-1 \right)-6\left( x-1 \right)={{x}_{3}}\left( y-6 \right)-1\left( y-6 \right) \\\ & \Rightarrow {{y}_{3}}x-{{y}_{3}}-6x+6={{x}_{3}}y-6{{x}_{3}}-y+6 \\\ \end{aligned}$ Since, the points are collinear therefore we will substitute point B in this equation and it will satisfy the equation ${{y}_{3}}x-{{y}_{3}}-6x+6={{x}_{3}}y-6{{x}_{3}}-y+6$. Therefore, we have $\begin{aligned} & {{y}_{3}}x-{{y}_{3}}-6x+6={{x}_{3}}y-6{{x}_{3}}-y+6 \\\ & \Rightarrow -4x-\left( -4 \right)-6x+6=3y-6\left( 3 \right)-y+6 \\\ & \Rightarrow -4x+4-6x+6=3y-18-y+6 \\\ & \Rightarrow -10x+10=2y-12 \\\ & \Rightarrow 10x+2y-12-10=0 \\\ & \Rightarrow 10x+2y-22=0 \\\ & \Rightarrow 5x+y-11=0 \\\ \end{aligned}$ Hence, the required equation is $5x+y-11=0$ and the correct option is (a).