Question

When the coordinate axes are rotated through an angle of 135°,a point P changes to (4, −3). Find the
original coordinates of P

Answer 1

Answer:

Therefore the original coordinate of P is $(-\frac{1}{\sqrt2},\frac{7}{\sqrt2})$.

Step-by-step explanation:

If the coordinate axes rotates at an angle θ. Then original coordinate P(x,y)  change into P'(x',y')

$x= x'cos\theta-y'sin\theta$

$y=x'sin\theta +y'cos \theta$

Given that a point changes to (4,-3).

Here x' = 4 and y'= -3 and θ = 135°

Let (x,y) be the original coordinate of P

Then

x= 4 cos 135° - (-3) sin 135°

 $=4.(-\frac{1}{\sqrt2})+3.(\frac1 {\sqrt2})$

 $=-\frac{1}{\sqrt2}$

and y = 4 sin 135°+(-3) cos 135°

          $=4.(\frac{1}{\sqrt2})-3.(-\frac1 {\sqrt2})$

           $=\frac{7}{\sqrt 2}$

Therefore the original coordinate of P is $(-\frac{1}{\sqrt2},\frac{7}{\sqrt2})$