RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates Ex-22.1 |
RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates Ex-22.2 |

**What does theequation (x – a) ^{ 2} + (y – b)^{ 2} = r^{2} becomewhen the axes are transferred to parallel axes through the point (a-c, b)?**

**Answer
1** :

Given:

The equation, (x – a)^{ 2} +(y – b)^{ 2} = r^{2}

The given equation (x – a)^{2} +(y – b)^{2} = r^{2} can be transformed into the newequation by changing x by x – a + c and y by y – b, i.e. substitution of x by x+ a and y by y + b.

((x + a – c) – a)^{2} +((y – b ) – b)^{2} = r^{2}

(x – c)^{2} + y^{2} = r^{2}

x^{2} + c^{2} – 2cx + y^{2} =r^{2}

x^{2} + y^{2} -2cx = r^{2} –c^{2}

Hence, the transformed equation is x^{2} +y^{2} -2cx = r^{2} – c^{2}

**What does theequation (a – b) (x ^{2} + y^{2}) – 2abx = 0 become if the origin isshifted to the point (ab / (a-b), 0) without rotation?**

**Answer
2** :

Given:

The equation (a – b) (x^{2} +y^{2}) – 2abx = 0

The given equation (a – b) (x^{2} +y^{2}) – 2abx = 0 can be transformedinto new equation by changing x by [X + ab / (a-b)] and y by Y

**Find what thefollowing equations become when the origin is shifted to the point (1, 1)?****(i) x ^{2} +xy – 3x – y + 2 = 0**

**Answer
3** :

**(i) x ^{2} + xy – 3x – y + 2 = 0**

Firstly let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1)^{2} + (x +1) (y + 1) – 3(x + 1) – (y + 1) + 2 = 0

x^{2} + 1 + 2x + xy+ x + y + 1 – 3x – 3 – y – 1 + 2 = 0

Upon simplification we get,

x^{2} + xy = 0

∴ Thetransformed equation is x^{2} + xy= 0.

**(ii) x ^{2} –y^{2} – 2x + 2y = 0**

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1)^{2} – (y + 1)^{2} – 2(x + 1) + 2(y + 1) = 0

x^{2} + 1 + 2x – y^{2} – 1 – 2y – 2x – 2 + 2y + 2 = 0

Upon simplification we get,

x^{2} – y^{2} = 0

∴ Thetransformed equation is x^{2} – y^{2} = 0.

**(iii) xy – x – y + 1 = 0**

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1) (y + 1) – (x + 1) – (y + 1) + 1 = 0

xy + x + y + 1 – x – 1 – y – 1 + 1 = 0

Upon simplification we get,

xy = 0

∴ Thetransformed equation is xy = 0.

**(iv) xy – y ^{2} –x + y = 0**

Let us substitute the value of x by x + 1 and y by y + 1

Then,

(x + 1) (y + 1) – (y + 1)^{2} –(x + 1) + (y + 1) = 0

xy + x + y + 1 – y^{2} –1 – 2y – x – 1 + y + 1 = 0

Upon simplification we get,

xy – y^{2} = 0

∴ Thetransformed equation is xy – y^{2} =0.

**4. At what point the origin beshifted so that the equation x ^{2} +xy – 3x + 2 = 0 does not contain any first-degree term and constant term?**

**Solution:**

Given:

The equation x^{2} +xy – 3x + 2 = 0

We know that the origin has been shifted from (0, 0) to (p, q)

So any arbitrary point (x, y) will also be converted as (x + p,y + q).

The new equation is:

(x + p)^{2} + (x +p)(y + q) – 3(x + p) + 2 = 0

Upon simplification,

x^{2} + p^{2} + 2px + xy + py + qx + pq – 3x – 3p+ 2 = 0

x^{2} + xy + x(2p +q – 3) + y(q – 1) + p^{2} + pq – 3p– q + 2 = 0

For no first degree term, we have 2p + q – 3 = 0 and p – 1 = 0,and

For no constant term we have p^{2} +pq – 3p – q + 2 = 0.

By solving these simultaneous equations we have p = 1 and q = 1from first equation.

The values p = 1 and q = 1 satisfies p^{2} +pq – 3p – q + 2 = 0.

Hence, the point to which origin must be shifted is (p, q) = (1,1).

**Answer
4** :

Given:

The points (2, 3), (5, 7), and (-3, -1).

The area of triangle with vertices (x_{1}, y_{1}),(x_{2}, y_{2}), and (x_{3}, y_{3}) is

= ½ [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} -y_{1})+ x_{3}(y_{1} – y_{2})]

The area of given triangle = ½ [2(7+1) + 5(-1-3) – 3(3-7)]

= ½ [16 – 20 + 12]

= ½ [8]

= 4

Origin shifted to point (-1, 3), the new coordinates of thetriangle are (3, 0), (6, 4), and (-2, -4) obtained from subtracting a point(-1, 3).

The new area of triangle = ½ [3(4-(-4)) + 6(-4-0) – 2(0-4)]

= ½ [24-24+8]

= ½ [8]

= 4

Since the area of the triangle before and after thetranslation after shifting of origin remains same, i.e. 4.

∴ We cansay that the area of a triangle is invariant to shifting of origin.

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