This will be the last lesson in the Coordinate Geometry Basics series. I’ll be closing with a few solved examples relating to translation and rotation of axes.

**Example 1 **Find the new coordinates of the point (3, 4)

(i) when the origin is shifted to the point (1, 3).

(ii) when the axes are rotated by an angle θ anticlockwise, where tanθ = 4/3

(iii) when the origin is shifted to (1, –2), and the axes are rotated by 90° in the clockwise direction.

**Solution **(i) We’ll directly use the formula derived in the previous lesson: x = X + h, y = Y + k

We have, 3 = X + 1 and 4 = Y + 3, or X = 2 and Y = 1. Therefore, the coordinates with respect to the shifted origin are **(2, 1)**.

As simple as that! The next two are equally simple.

(ii) In this case, we need to calculate the values of sinθ and cosθ first. They’ll come out to be 4/5 and 3/5 respectively. Now let’s use our formulas x = Xcosθ – Ysinθ and y = Xsinθ + Ycosθ.

That is, 3 = (3X – 4Y)/5 and 4 = (4X + 3Y)/5. Solving for X and Y, we get X = 5 and Y = 0. Therefore, the new coordinates will be **(5, 0)**.

(iii) We didn’t talk about simultaneous rotation as well as translation. But turns out it is quite easy. We can find the new coordinates by first shifting the origin, followed by rotation, or the other way around.

We can also combine the two formulas straight away, i.e. x = Xcosθ – Ysinθ + h and y = Xsinθ + Ycosθ + k, and solve for X and Y to obtain the new coordinates. (You may try doing it separately and compare the answers)

Note that the axes are rotated clockwise in this case, but our formulas consider anticlockwise direction. So we’ve to take θ = –90°

Let’s calculate now: 3 = Xcos(–90°) – Ysin(–90°) + 1 and 4 = Xsin(–90°) + Ycos(–90°) – 2.

This gives us X = – 6 and Y = 2. Therefore, the final coordinates are **(****–6****, 2)**

The next few problems will talk about equations of curves with respect to the new coordinate systems.

**Example 2 **Find the new equation of the following curves after the coordinates are transformed as indicated:

(i) x + 3y = 6, when the origin is shifted to the point (–4, 1).

(ii) Find the equation of the curve x^{2} + y^{2} = 4, when the axes are rotated by an angle of 60° in the anticlockwise direction.

(iii) Find the equation of the curve x^{2} – y^{2} = 10, when the axes are rotated by an angle of 45° in the clockwise direction.

**Solution **(i) In this case we do not need to find the new coordinates. We only need to find the relation between them (that’s what an equation is). So we’ll simply replace the old coordinates with the new ones in the given equation.

Using the expressions for shifting of origin, we have x = X – 4 and y = Y + 2. Substituting the values in the given equation, we get X – 4 + 3(Y + 1) = 6. Or, **X + 3Y = 7**.

And that’s all. We’ve obtained the relation between the new coordinates, which is nothing but the equation of that curve with respect to the new origin.

(ii) This one is quite similar to the previous one, except that we’re rotating the axes instead of translating them. Using the formulas, we have x = Xcos60° – Ysin60° and y = Xsin60° + Ycos60°. Let’s substitute in the given equation. We get, (Xcos60° – Ysin60°)^{2} + (Xsin60° + Ycos60°)^{2} = 4. This leads to X^{2}cos^{2}60° + Y^{2}sin^{2}60° – 2XYcos60°sin60° + X^{2}sin^{2}60° + Y^{2}cos^{2}60° + 2XYsin60°cos60° = 4 or **X ^{2} + Y^{2} = 4**.

Hmm. Nothing happened. Strange.

(iii) This one is quite similar to the previous one, except that we’re rotating the axes instead of translating them. Using the formulas, we have x = Xcos45° + Ysin45° and y = –Xsin45° + Ycos45°. On substituting in the given equation, we get (Xcos45° + Ysin45°)^{2} – (–Xsin45° + Ycos45°)^{2} = 10. This on simplification gives us **XY = 5**.

**Example 3 **To what point should the origin be shifted so that the equation x^{2} + y^{2} – 4x + 6y – 4 = 0 becomes free of the first degree terms? (i.e. -4x and 6y)

**Solution **Let the origin be shifted to the point (h, k). Let’s transform the equation, and see what happens. Replacing x by X + h and y by Y + k, we get X^{2} + Y^{2} + X(-4 + 2h) + Y(6 + 2k) + h^{2} + k^{2} – 4h + 6k – 4 = 0.

Now the first degree terms in this equation have the coefficients (–4 + 2h) and (6 + 2k). Since we want to get rid of them, we’ll equate both to them to zero. By doing that, we get **h = 2 **and **k = -3.**

The new equation will look like X^{2} + Y^{2} – 17 = 0. (No first degree terms, yay!)

**Example 4 **By what angle should the axes be rotated so that the equation 3x^{2} + 2xy + y^{2} = 1 becomes free of the xy term?

**Solution **This one is similar to the previous one, except that now we’ve got to rotate the axes. Let’s do the hard work. The transformed equation will become 3(Xcosθ – Ysinθ)^{2} + 2(Xcosθ – Ysinθ)(Xsinθ + Ycosθ) + (Xsinθ + Ycosθ)^{2} = 1. (Whoa!)

X^{2}(3cos^{2}θ + 2sinθcosθ + sin^{2}θ) + XY(-4sinθcosθ +2cos^{2}θ – 2sin^{2}θ) + Y^{2}(3sin^{2}θ + cos^{2}θ – 2sinθcosθ) = 1

We don’t want the XY term, so we’ll equate its coefficient to zero. We get 2cos2θ = 2sin2θ or tan2θ = 1, or θ = **45°**.

That’ll be all for coordinate geometry basics. We now have all the tools we need to sail through the next set of lessons. Hope you enjoyed!

The next stop will be Straight Lines. I’ll meet you there. Goodbye!

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