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
591.3k+ views
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

