Question

From $\left( {1,4} \right)$ you travel $5\sqrt 2$ units by making ${135^0}$ angles with positive x-axis (anti-clock wise direction) and then 4 units by making ${120^0}$ angle with positive x-axis (clockwise) to reach Q. Find the co-ordinates of point Q.
A. $\left( { + 6,9 - 2\sqrt 3 } \right)$
B. $\left( { - 6,9 - 2\sqrt 3 } \right)$
C. $\left( { - 6,9 + 2\sqrt 3 } \right)$
D. $\left( { + 6,9 + 2\sqrt 3 } \right)$

Answer 1

Hint : In this question, consider the unknown points as variables and rotate the point by an angle of ${135^0}$ by travelling $5\sqrt 2$ units. Then rotate the obtained point by angle of ${120^0}$ by travelling 4 units by using the concept of translation and rotation of axes. So, use this concept to reach the solution of the given problem.

Complete step by step solution :
Let the initial position be $P\left( {{x_0},{y_0}} \right) = P\left( {1,4} \right)$
Let the unknown points be $R\left( {{x_1},{y_1}} \right)$ and $Q\left( {{x_2},{y_2}} \right)$.
Point $R\left( {{x_1},{y_1}} \right)$ can be obtained by travelling $5\sqrt 2$ units by making ${135^0}$ angles with positive x-axis (anti-clock wise direction) from point $P\left( {{x_0},{y_0}} \right) = P\left( {1,4} \right)$.
We know that if a point $\left( {{x_a},{y_a}} \right)$ has travelled $r$ units by making an angle of $\theta$ with the positive x-axis (anti-clock wise direction) then that point will be $\left( {{x_a} + r\cos \theta ,{y_a} + r\sin \theta } \right)$.
So, we have

$$\Rightarrow {x_1} = {x_0} + r\cos \left( {{{135}^0}} \right) \\\ \Rightarrow {x_1} = 1 + 5\sqrt 2 \left( { - \dfrac{1}{{\sqrt 2 }}} \right) \\\ \therefore {x_1} = 1 - 5 = - 4 \\\$$

And

$$\Rightarrow {y_1} = {y_0} + r\sin \left( {{{135}^0}} \right) \\\ \Rightarrow {y_1} = 4 + 5\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}} \right) \\\ \therefore {y_1} = 4 + 5 = 9 \\\$$

Hence, $R\left( {{x_1},{y_1}} \right) = \left( { - 4,9} \right)$
Point $Q\left( {{x_2},{y_2}} \right)$ can be obtained by travelling 4 units by making an angle of with positive x-axis (clock wise direction) then from point $R\left( {{x_1},{y_1}} \right) = \left( { - 4,9} \right)$.
We know that if a point $\left( {{x_a},{y_a}} \right)$ has travelled $r$ units by making an angle of $\theta$ with the positive x-axis (clock wise direction) then that point will be$\left( {{x_a} + r\cos \left( { - \theta } \right),{y_a} + r\sin \left( { - \theta } \right)} \right)$.
So, we have

$$\Rightarrow {x_2} = {x_1} + r\cos \left( { - \theta } \right) \\\ \Rightarrow {x_2} = - 4 + 4\cos \left( { - {{120}^0}} \right) \\\ \Rightarrow {x_2} = - 4 + 4\left( { - \dfrac{1}{2}} \right) \\\ \therefore {x_2} = - 4 - 2 = - 6 \\\$$

And

$$\Rightarrow {y_2} = {y_1} + r\sin \left( { - \theta } \right) \\\ \Rightarrow {y_2} = 9 + 4\sin \left( { - {{120}^0}} \right) \\\ \Rightarrow {y_2} = 9 - 4\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\\ \therefore {y_2} = 9 - 2\sqrt 3 \\\$$

Hence, $Q\left( {{x_2},{y_2}} \right) = \left( { - 6,9 - 2\sqrt 3 } \right)$
Thus, the correct option is B. $\left( { - 6,9 - 2\sqrt 3 } \right)$

Note : If a point $\left( {{x_a},{y_a}} \right)$ has travelled $r$ units by making an angle of $\theta$ with the positive x-axis (anti-clock wise direction) then that point will be $\left( {{x_a} + r\cos \theta ,{y_a} + r\sin \theta } \right)$.If a point $\left( {{x_a},{y_a}} \right)$ has travelled $r$ units by making an angle of $\theta$ with the positive x-axis (clockwise direction) then that point will be$\left( {{x_a} + r\cos \left( { - \theta } \right),{y_a} + r\sin \left( { - \theta } \right)} \right)$.