Question -
Answer -
(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).