Question

The variable line drawn through the point $(1,3)$ meets the $x-axis$ at $A$ and $y-axis$ at $B$. If the rectangle $OAPB$ is completed. Where $O$ is the origin, then locus of $P$ is?
A. ${\displaystyle \frac{1}{y}}+{\displaystyle \frac{3}{x}}=1$.
B. $x+3y=1$
C. ${\displaystyle \frac{1}{x}}+{\displaystyle \frac{3}{y}}=1$
D. $3x+y=1$

Answer 1

Hint: In this problem we need to find the locus of the point $P$ according to the given conditions. Given that the line passes through the point $(1,3)$ which is assumed to be $C$ meets the $x-axis$ at $A$ and $y-axis$ at $B$. So, we will assume the coordinates of the points $A$ and $B$ as $(h,0)$, $(0,k)$ respectively. Now we will calculate the slope of the line $BC$, $AC$. Use the geometry rule which is the slope of the lines $BC$, $AC$ are equal because they both are single lines. Now simplify the equation and replace $h$, $k$ with $x$, $y$ respectively to get the required result.

Complete step-by-step answer:
Given data, The variable line drawn through the point $(1,3)$ meets the $x-axis$ at $A$ and $y-axis$ at $B$ and the rectangle $OAPB$ is completed. Where $O$ is the origin. The diagrammatic representation of the above data is given by

Let the coordinates of the points $A$ and $B$ are assumed to be $(h,0)$, $(0,k)$ respectively.
Now the slope of the line $BC$ will be given by
$m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$
Substituting the values $(x_1,y_1)=(0,k)$, $(x_2,y_2)=(1,3)$ in the above equation, then we will get
$\begin{array}{rl}& m={\displaystyle \frac{3-k}{1-0}}\\ & \Rightarrow m={\displaystyle \frac{3-k}{1}}\end{array}$
Now the slope of the line $AC$ will be calculated by substituting $(x_1,y_1)=(h,0)$, $(x_2,y_2)=(1,3)$ in $m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}$, then we will get
$\begin{array}{rl}& m={\displaystyle \frac{3-0}{1-h}}\\ & \Rightarrow m={\displaystyle \frac{3}{1-h}}\end{array}$
Now the line $BC$, $AC$ represents the same line, so the slopes of the two lines should be equal, then we will have
${\displaystyle \frac{3-k}{1}}={\displaystyle \frac{3}{1-h}}$
Doing cross multiplication in the above equation, then we will get
$(3-k)(1-h)=3$
Using distribution law of multiplication in the above equation, then we will have
$3-3h-k+hk=3$
Cancelling the term $3$ which is on both sides of the above equation and divide whole equation with $hk$, then we will get
${\displaystyle \frac{-3h-k+hk}{hk}}=0$
Simplifying the above equation by using mathematical operations, then we will have
$\begin{array}{rl}& {\displaystyle \frac{-3h}{hk}}-{\displaystyle \frac{k}{hk}}+{\displaystyle \frac{hk}{hk}}=0\\ & \Rightarrow -{\displaystyle \frac{3}{k}}-{\displaystyle \frac{1}{h}}+1=0\\ & \Rightarrow {\displaystyle \frac{1}{h}}+{\displaystyle \frac{3}{k}}=1\end{array}$
Replace the terms $h$, $k$ with $x$, $y$ respectively in the above equation, then we will get
${\displaystyle \frac{1}{x}}+{\displaystyle \frac{3}{y}}=1$

So, the correct answer is “Option c”.

Note: We can also use the collinear property form the points $A$, $C$, $B$ and simplify the determinant obtained to get the required result. Here we need to solve the determinant $\left|\begin{array}{ccc}h& 0& 1\\ 1& 3& 1\\ 0& k& 1\end{array}\right|=0$ which shows that the points are collinear.