
A variable line L is drawn through \[O\left( 0,0 \right)\] to meet lines \[{{L}_{1}}:2x+3y=5\] and \[{{L}_{2}}:2x+3y=10\] at point P, Q and \[2.OP.OQ=OR.OP+OR.OQ\] . Locus of R is
A. $9x+6y=20$
B. $6x-9y=20$
C. $6x+9y=20$
D. $9x-6y=20$
Answer
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Hint: For solving this problem we need to have a clear understanding of coordinate geometry, the equation of lines, the concept of locus. Employing the formulas for the equation of lines in polar form and equating it according to the given equation, we can easily find the required locus.
The polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by the length of its position vector r and the angle $\theta $ between it and the positive direction of x-axis.
Now the equation of a straight line in polar coordinate form can be written as
$L:\dfrac{x}{\cos \theta }=\dfrac{y}{\sin \theta }=r$
Complete step by step solution:
A locus is a set of points, in geometry, which satisfies a given condition or situation for a shape or a figure. Basically, a locus is a set of all points, whose location satisfies or is determined by one or more specified conditions. In other words, the set of the points that satisfy some property is often called the locus of a point satisfying this property.
As given in the problem, the line L passes through the origin \[O\left( 0,0 \right)\] and it meets the lines \[{{L}_{1}}:2x+3y=5\] and \[{{L}_{2}}:2x+3y=10\] at point P, Q. So, using the formula of polar coordinates we can say that,
\[OP=\dfrac{5}{2\cos \theta +3\sin \theta }\] and $OQ=\dfrac{10}{2\cos \theta +3\sin \theta }$ and $OR=r$
Using the condition of \[2.OP.OQ=OR.OP+OR.OQ\] , we can write that
$\Rightarrow 2\times \dfrac{5}{2\cos \theta +3\sin \theta }\times \dfrac{10}{2\cos \theta +3\sin \theta }=r\times \dfrac{5}{2\cos \theta +3\sin \theta }+r\times \dfrac{10}{2\cos \theta +3\sin \theta }$
Simplifying this we can get that
$\Rightarrow 20=6r\cos \theta +9r\sin \theta $
which can be also written as $6x+9y=20$ , due to the fact that $x=r\cos \theta $ and $y=r\sin \theta $ .
So, the correct answer is “Option C”.
Note: These problems might seem simple but the calculation of finding the locus of a point can be lengthy in some cases. One must solve these problems very carefully and must not use incorrect formulas of the polar coordinate system. Cartesian coordinate systems can also be used but the calculation can tend to be very lengthy so it is always suggested to use polar coordinate form in these types of problems.
The polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by the length of its position vector r and the angle $\theta $ between it and the positive direction of x-axis.
Now the equation of a straight line in polar coordinate form can be written as
$L:\dfrac{x}{\cos \theta }=\dfrac{y}{\sin \theta }=r$
Complete step by step solution:
A locus is a set of points, in geometry, which satisfies a given condition or situation for a shape or a figure. Basically, a locus is a set of all points, whose location satisfies or is determined by one or more specified conditions. In other words, the set of the points that satisfy some property is often called the locus of a point satisfying this property.
As given in the problem, the line L passes through the origin \[O\left( 0,0 \right)\] and it meets the lines \[{{L}_{1}}:2x+3y=5\] and \[{{L}_{2}}:2x+3y=10\] at point P, Q. So, using the formula of polar coordinates we can say that,
\[OP=\dfrac{5}{2\cos \theta +3\sin \theta }\] and $OQ=\dfrac{10}{2\cos \theta +3\sin \theta }$ and $OR=r$
Using the condition of \[2.OP.OQ=OR.OP+OR.OQ\] , we can write that
$\Rightarrow 2\times \dfrac{5}{2\cos \theta +3\sin \theta }\times \dfrac{10}{2\cos \theta +3\sin \theta }=r\times \dfrac{5}{2\cos \theta +3\sin \theta }+r\times \dfrac{10}{2\cos \theta +3\sin \theta }$
Simplifying this we can get that
$\Rightarrow 20=6r\cos \theta +9r\sin \theta $
which can be also written as $6x+9y=20$ , due to the fact that $x=r\cos \theta $ and $y=r\sin \theta $ .
So, the correct answer is “Option C”.
Note: These problems might seem simple but the calculation of finding the locus of a point can be lengthy in some cases. One must solve these problems very carefully and must not use incorrect formulas of the polar coordinate system. Cartesian coordinate systems can also be used but the calculation can tend to be very lengthy so it is always suggested to use polar coordinate form in these types of problems.
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