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\[\left( {{x_a} + r\cos \left( { - \theta } \right),{y_a} + r\sin \left( { - \theta } \right)} \right)\].