Question

How do you find the point of intersection of the simultaneous equations $r = 2 - 3\cos \theta$ and $r = \cos \theta$ ?

Answer 1

Hint : In the given question, we need to solve two simultaneous equations in two variables. There are various methods to solve two given equations in two variables like substitution method, cross multiplication method, elimination method, matrix method and many more. The equations given in the question can be solved using any one of the above mentioned methods easily. But we will solve the equations using the substitution method.

Complete step by step solution:
In the question, we are given a couple of simultaneous linear equations in two variables.
$r = 2 - 3\cos \theta$
$r = \cos \theta$
In the substitution method, we substitute the value of one variable from an equation into another equation so as to get an equation in only one variable.
Now putting the value of $\cos \theta$ obtained from one equation into another. So, we equate the values of r from both the equations.
So, we get, $2 - 3\cos \theta = \cos \theta$
Shifting all the terms consisting $\cos \theta$ to the right side of the equation and all the constants to the right side of the equation, we get,
$\Rightarrow 2 = \cos \theta + 3\cos \theta$
$\Rightarrow 4\cos \theta = 2$
$\Rightarrow \cos \theta = \left( {\dfrac{1}{2}} \right)$
So, the value of $\cos \theta$ is $\left( {\dfrac{1}{2}} \right)$.
Putting the value of $\cos \theta$ in any of the two equations, we get,
$r = \cos \theta$
$\Rightarrow r = \left( {\dfrac{1}{2}} \right)$
Therefore, solution of the simultaneous linear equations $r = 2 - 3\cos \theta$ and $r = \cos \theta$ is $r = \left( {\dfrac{1}{2}} \right)$ and $\cos \theta = \left( {\dfrac{1}{2}} \right)$. So, the point of intersection of the equations $r = 2 - 3\cos \theta$ and $r = \cos \theta$ is the point $r = \left( {\dfrac{1}{2}} \right)$ and $\cos \theta = \left( {\dfrac{1}{2}} \right)$.
Now, we can find the cartesian coordinates of the points using the values of r and $\cos \theta$. So, the coordinates of the points in the Cartesian system are $\left( {r\cos \theta ,r\sin \theta } \right)$. We can find the value of $\sin \theta$ using ${\sin ^2}\theta + {\cos ^2}\theta = 1$. So, we get, $\sin \theta = \pm \dfrac{{\sqrt 3 }}{2}$ .
Hence, we get, $\left( {r\cos \theta ,r\sin \theta } \right) = \left( {\left( {\dfrac{1}{2} \times \dfrac{1}{2}} \right),\left( {\dfrac{1}{2} \times \left( { \pm \dfrac{{\sqrt 3 }}{2}} \right)} \right)} \right) = \left( {\dfrac{1}{4}, \pm \dfrac{{\sqrt 3 }}{4}} \right)$
Therefore, the point of intersection of the equations $r = 2 - 3\cos \theta$ and $r = \cos \theta$ are $\left( {\dfrac{1}{4},\dfrac{{\sqrt 3 }}{4}} \right)$ and $\left( {\dfrac{1}{4}, - \dfrac{{\sqrt 3 }}{4}} \right)$ .
So, the correct answer is “ $\left( {\dfrac{1}{4},\dfrac{{\sqrt 3 }}{4}} \right)$ and $\left( {\dfrac{1}{4}, - \dfrac{{\sqrt 3 }}{4}} \right)$ ”.

Note : Linear Equation in two variables: A equation consisting of 2 variables having degree one is known as Linear Equation in two variables. Standard form of Linear Equation in two variables is $ax + by + c = 0$ where a, b and c are the real numbers and a, b which are coefficients of x and y respectively are not equal to 0. Here we have variables as r and $\theta$ which are part of the polar coordinate system.