
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) $\dfrac{1}{y} + \dfrac{3}{x} = 1$
(B) $x + 3y = 1$
(C) $\dfrac{1}{x} + \dfrac{3}{y} = 1$
(D) $3x + y = 1$
Answer
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Hint: A locus is a set of points, in geometry, which satisfies a given condition or situation for a shape or a figure. Here we are to find the set of points which $P$could be, and the equation it follows. So, here a variable line is passing through A, B and point $(1,3)$. Now, $OAPB$ is a rectangle that gets completed, where O is the origin, and P is any variable point. Since it is a rectangle, so it’s opposite two sides have to be equal in length. Now, we are to find a suitable equation for P from the given conditions, assuming two variables, x and y.
Complete answer:
Given, a variable line is drawn through the point $(1,3)$.
It meets the x-axis at A and y-axis at B. Therefore, let the coordinate of A be $(h,0)$ and B be $(0,k)$.
Now, since $OAPB$ is a rectangle, it’s opposite sides must be equal.
So, we have, $OA = PB$ and $OB = PA$.
Here, $OA$ is the distance between the origin $O$ and the point $A$.
And, $OB$ is the distance between the origin $O$ and the point $B$.
Now, $OA = h$ and $OB = k$ since coordinates of points A and B are $(h,0)$ and $(0,k)$.
So, we get,
$PB = h$ and $PA = k$.
Therefore, we can say that, distance of P from the y-axis is $h$ units and from the x-axis is $k$ units.
So, the coordinate of P is$(h,k)$.
Now, from the intercept formula, we can write the equation of the variable line as,
$\dfrac{x}{a} + \dfrac{y}{b} = 1$
Now, the intercepts of the line are $h$ and $k$. So, we get,
$ \Rightarrow \dfrac{x}{h} + \dfrac{y}{k} = 1 - - - - (1)$
Now, $(1,3)$ lies on the line represented by equation $\left( 1 \right)$.
So, substituting $x = 1$ and $y = 3$ in $(1)$, we get,
$ \Rightarrow \dfrac{1}{h} + \dfrac{3}{k} = 1$
Now, replacing $(h,k)$ by $(x,y)$ to get the locus of P in $(x,y)$ and to match the options, we get,
$ \Rightarrow \dfrac{1}{x} + \dfrac{3}{y} = 1$
So, the locus of $P$is $\dfrac{1}{x} + \dfrac{3}{y} = 1$.
Hence, option C is the correct answer.
Note:
We must know the intercept form of an equation in two variables to solve the problem. We should also note the above method as the standard way of solving such locus problems. We should take care of the calculations to be sure of the final locus of point P. The locus of a given point tells us about the path it is revolving. We also must know the properties of rectangles in order to attempt the question.
Complete answer:
Given, a variable line is drawn through the point $(1,3)$.
It meets the x-axis at A and y-axis at B. Therefore, let the coordinate of A be $(h,0)$ and B be $(0,k)$.
Now, since $OAPB$ is a rectangle, it’s opposite sides must be equal.
So, we have, $OA = PB$ and $OB = PA$.
Here, $OA$ is the distance between the origin $O$ and the point $A$.
And, $OB$ is the distance between the origin $O$ and the point $B$.
Now, $OA = h$ and $OB = k$ since coordinates of points A and B are $(h,0)$ and $(0,k)$.
So, we get,
$PB = h$ and $PA = k$.
Therefore, we can say that, distance of P from the y-axis is $h$ units and from the x-axis is $k$ units.
So, the coordinate of P is$(h,k)$.
Now, from the intercept formula, we can write the equation of the variable line as,
$\dfrac{x}{a} + \dfrac{y}{b} = 1$
Now, the intercepts of the line are $h$ and $k$. So, we get,
$ \Rightarrow \dfrac{x}{h} + \dfrac{y}{k} = 1 - - - - (1)$
Now, $(1,3)$ lies on the line represented by equation $\left( 1 \right)$.
So, substituting $x = 1$ and $y = 3$ in $(1)$, we get,
$ \Rightarrow \dfrac{1}{h} + \dfrac{3}{k} = 1$
Now, replacing $(h,k)$ by $(x,y)$ to get the locus of P in $(x,y)$ and to match the options, we get,
$ \Rightarrow \dfrac{1}{x} + \dfrac{3}{y} = 1$
So, the locus of $P$is $\dfrac{1}{x} + \dfrac{3}{y} = 1$.
Hence, option C is the correct answer.
Note:
We must know the intercept form of an equation in two variables to solve the problem. We should also note the above method as the standard way of solving such locus problems. We should take care of the calculations to be sure of the final locus of point P. The locus of a given point tells us about the path it is revolving. We also must know the properties of rectangles in order to attempt the question.
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