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4.2 Q-2

Question Statement

Find the coordinates of point P in the original xy-coordinate system, given that the axes have been translated through point Oβ€²O'. The coordinates of point PP in the translated system are provided.


Background and Explanation

This problem involves translating the coordinates of a point from a translated coordinate system (with origin at Oβ€²O') back to the original xy-coordinate system. To do this, we need to reverse the translation by adding the coordinates of the new origin Oβ€²O' to the coordinates of point PP. The general formula for translating back to the original system is:

x=xβ€²+h,y=yβ€²+kx = x' + h, \quad y = y' + k

Where:

  • (xβ€²,yβ€²)(x', y') are the coordinates of point P in the translated system,
  • (h,k)(h, k) are the coordinates of the new origin Oβ€²O',
  • (x,y)(x, y) are the coordinates of point P in the original system.

Solution

i. For P(8, 10) and O(3, 4)

Here, we have:

  • P(8,10)P(8, 10), and
  • O(3,4)O(3, 4).

Now, apply the translation formula:

  • x=xβ€²+h=8+3=11x = x' + h = 8 + 3 = 11,
  • y=yβ€²+k=10+4=14y = y' + k = 10 + 4 = 14.

Thus, the coordinates of point P in the original system are:

P(11,14).P(11, 14).

ii. For P(-5, -3) and O(3, 4)

Here, we have:

  • P(βˆ’5,βˆ’3)P(-5, -3), and
  • O(3,4)O(3, 4).

Now, apply the translation formula:

  • x=xβ€²+h=βˆ’5+3=βˆ’2x = x' + h = -5 + 3 = -2,
  • y=yβ€²+k=βˆ’3+4=1y = y' + k = -3 + 4 = 1.

Thus, the coordinates of point P in the original system are:

P(βˆ’2,1).P(-2, 1).

iii. For P(-3/4, -7/6) and O’(1/4, -1/6)

Here, we have:

  • P(βˆ’34,βˆ’76)P\left(\frac{-3}{4}, \frac{-7}{6}\right), and
  • Oβ€²(14,βˆ’16)O'\left(\frac{1}{4}, \frac{-1}{6}\right).

Now, apply the translation formula:

  • x=xβ€²+h=βˆ’34+14=βˆ’24=βˆ’12x = x' + h = \frac{-3}{4} + \frac{1}{4} = \frac{-2}{4} = \frac{-1}{2},
  • y=yβ€²+k=βˆ’76+βˆ’16=βˆ’86=βˆ’43y = y' + k = \frac{-7}{6} + \frac{-1}{6} = \frac{-8}{6} = \frac{-4}{3}.

Thus, the coordinates of point P in the original system are:

P(βˆ’12,βˆ’43).P\left(\frac{-1}{2}, \frac{-4}{3}\right).

iv. For P(4, -3) and O’(-2, 3)

Here, we have:

  • P(4,βˆ’3)P(4, -3), and
  • Oβ€²(βˆ’2,3)O'(-2, 3).

Now, apply the translation formula:

  • x=xβ€²+h=4+(βˆ’2)=2x = x' + h = 4 + (-2) = 2,
  • y=yβ€²+k=βˆ’3+3=0y = y' + k = -3 + 3 = 0.

Thus, the coordinates of point P in the original system are:

P(2,0).P(2, 0).

Key Formulas or Methods Used

  • Translation of Coordinates:
x=xβ€²+h,y=yβ€²+kx = x' + h, \quad y = y' + k

Where (xβ€²,yβ€²)(x', y') are the coordinates of point P in the translated system, and (h,k)(h, k) are the coordinates of the translated origin Oβ€²O'.


Summary of Steps

  1. Identify the coordinates of point P and the origin Oβ€²O' in the translated system.
  2. Use the formula x=xβ€²+hx = x' + h and y=yβ€²+ky = y' + k to find the coordinates of P in the original system.
  3. Repeat for all given points and their respective origins.