RD Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Ex 22.3 Solutions
Question - 1 : - What does theequation (x – a) 2 + (y – b) 2 = r2 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 = r2
The given equation (x – a)2 +(y – b)2 = r2 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 = r2
(x – c)2 + y2 = r2
x2 + c2 – 2cx + y2 =r2
x2 + y2 -2cx = r2 –c2
Hence, the transformed equation is x2 +y2 -2cx = r2 – c2
Question - 2 : - What does theequation (a – b) (x2 + y2) – 2abx = 0 become if the origin isshifted to the point (ab / (a-b), 0) without rotation?
Answer - 2 : -
Given:
The equation (a – b) (x2 +y2) – 2abx = 0
The given equation (a – b) (x2 +y2) – 2abx = 0 can be transformedinto new equation by changing x by [X + ab / (a-b)] and y by Y
Question - 3 : - Find what thefollowing equations become when the origin is shifted to the point (1, 1)?
(i) x2 +xy – 3x – y + 2 = 0
(ii) x2 –y2 – 2x + 2y = 0
(iii) xy – x – y + 1 = 0
(iv) xy – y2 –x + y = 0
Answer - 3 : -
(i) x2 + 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
x2 + 1 + 2x + xy+ x + y + 1 – 3x – 3 – y – 1 + 2 = 0
Upon simplification we get,
x2 + xy = 0
∴ Thetransformed equation is x2 + xy= 0.
(ii) x2 –y2 – 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
x2 + 1 + 2x – y2 – 1 – 2y – 2x – 2 + 2y + 2 = 0
Upon simplification we get,
x2 – y2 = 0
∴ Thetransformed equation is x2 – y2 = 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 – y2 –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 – y2 –1 – 2y – x – 1 + y + 1 = 0
Upon simplification we get,
xy – y2 = 0
∴ Thetransformed equation is xy – y2 =0.
4. At what point the origin beshifted so that the equation x2 +xy – 3x + 2 = 0 does not contain any first-degree term and constant term?
Solution:
Given:
The equation x2 +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,
x2 + p2 + 2px + xy + py + qx + pq – 3x – 3p+ 2 = 0
x2 + xy + x(2p +q – 3) + y(q – 1) + p2 + 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 p2 +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 p2 +pq – 3p – q + 2 = 0.
Hence, the point to which origin must be shifted is (p, q) = (1,1).
Question - 4 : - Verify that the area of the triangle with vertices (2, 3), (5, 7) and (-3 -1) remains invariant under the translation of axes when the origin is shifted to the point (-1, 3).
Answer - 4 : -
Given:
The points (2, 3), (5, 7), and (-3, -1).
The area of triangle with vertices (x1, y1),(x2, y2), and (x3, y3) is
= ½ [x1(y2 – y3) + x2(y3 -y1)+ x3(y1 – y2)]
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.