Класс 11. Решения RD Sharma. Глава 22. Краткий обзор декартовой системы прямоугольных координат. Упражнение 22.3.
Вопрос 1. Во что превратится уравнение (x – a) 2 + (y – b) 2 = r 2 при переносе осей на параллельные оси через точку (a–c, b)?
Решение:
We are given,
(x – a)2 + (y – b)2 = r2
Putting x = X + a – c and y = Y + b, we get,
=> ((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
Therefore the required equation is X2 + Y2 – 2cX = r2 – c2.
Вопрос 2. Во что превратится уравнение (a – b) (x 2 + y 2 ) – 2abx = 0, если сместить начало координат в точку (ab/(a–b), 0) без вращения?
Решение:
We are given,
(a – b) (x2 + y2) – 2abx = 0
Putting x = X + [ab/(a–b)] and y = Y, we get,
=>
=>
=>
=>
=> X2 (a–b)2 + (ab)2 + 2abX (a–b)+Y2 (a–b)2 = 2abX (a–b)+2(ab)2
=> (a – b)2 (X2 + Y2) = a2b2
Therefore the required equation is (a – b)2 (X2 + Y2) = a2b2.
Вопрос 3. Найдите, во что превратятся следующие уравнения при смещении начала координат в точку (1, 1)?
(i) х 2 + ху - 3х - у + 2 = 0
Решение:
We are given,
x2 + xy – 3x – y + 2 = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (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
Therefore the required equation is X2 + XY = 0.
(ii) х 2 - у 2 - 2х + 2у = 0
Решение:
We are given,
x2 – y2 – 2x + 2y = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (X + 1)2 – (Y + 1)2 – 2(X + 1) + 2(Y + 1) = 0
=> X2 + 1 + 2X – Y2 – 1 – 2Y – 2X – 2 + 2Y + 2 = 0
=> X2 – Y2 = 0
Therefore the required equation is X2 – Y2 = 0.
(iii) ху – х – у + 1 = 0
Решение:
We are given,
xy – x – y + 1 = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (X + 1) (Y + 1) – (X + 1) – (Y + 1) + 1 = 0
=> XY + X + Y + 1 – X – 1 – Y – 1 + 1 = 0
=> XY = 0
Therefore the required equation is XY = 0.
(iv) ху – у 2 – х + у = 0
Решение:
We are given,
xy – y2 – x + y = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (X + 1) (Y + 1) – (Y + 1)2 – (X + 1) + (Y + 1) = 0
=> XY + X + Y + 1 – Y2 – 1 – 2Y – X – 1 + Y + 1 = 0
=> XY – Y2 = 0
Therefore, the required equation is XY – Y2 = 0.
Вопрос 4. В какую точку сместить начало координат так, чтобы уравнение x 2 + xy – 3x – y + 2 = 0 не содержало ни одного члена первой степени и постоянного члена?
Решение:
We are given,
x2 + xy – 3x – y + 2 = 0
Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,
=> (X + a)2 + (X + a)(Y + b) – 3(X + a) – (Y + b) + 2 = 0
=> X2 + a2 + 2aX + XY + aY + bX + ab – 3X – 3a – Y – b + 2 = 0
=> X2 + XY + X(2a + b – 3) + Y(a – 1) + a2 + ab – 3a – b + 2 = 0
As our transformed equation has no first-degree term, we have,
2a + b – 3 = 0 and a – 1 = 0
By solving these equations we have a = 1 and b = 1.
Therefore, the origin has been shifted to (1,1) from (0,0).
Вопрос 5. Проверить, что площадь треугольника с вершинами (2, 3), (5, 7) и (–3, –1) остается неизменной относительно переноса осей при смещении начала координат в точку (–1 , 3).
Решение:
Here, L.H.S. = A1 = Area of the triangle with vertices (2, 3), (5, 7) and (–3, –1)
=
=
=
=
= 4 sq. units
As the origin shifted to point (–1, 3), the new coordinates of the triangle are:
(X1, Y1) = (2–1, 3+3) = (1, 6)
(X2, Y2) = (5–1, 7+3) = (4, 10)
(X3, Y3) = (–3–1, –1+3) = (–4, 2)
Now, R.H.S. = A2 = Area of the triangle with vertices (1, 6), (4, 10) and (–4, 2)
=
=
=
=
= 4 sq. units
Therefore, A1 = A2.
Hence, proved.
Вопрос 6. Найдите, во что превратятся следующие уравнения при смещении начала координат в точку (1, 1)?
(i) х 2 + ху - 3у 2 - у + 2 = 0
Решение:
We are given,
x2 + xy – 3y2 – y + 2 = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (X + 1)2 + (X + 1) (Y + 1) – 3(Y + 1)2 – (Y + 1) + 2 = 0
=> X2 + 1 + 2X + XY + X + Y + 1 – 3Y2 – 3 – 6Y – Y – 1 + 2 = 0
=> X2 – 3Y2 + XY + 3X – 6Y = 0
Therefore, the required equation is X2 – 3Y2 + XY + 3X – 6Y = 0.
(ii) ху – у 2 – х + у = 0
Решение:
We are given,
xy – y2 – x + y = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (X + 1) (Y + 1) – (Y + 1)2 – (X + 1)+ Y + 1 = 0
=> XY + X + Y + 1 – Y2 – 1 – 2Y – X – 1 + Y + 1 = 0
=> XY – Y2 = 0
Therefore, the required equation is XY – Y2 = 0.
(iii) ху – х – у + 1 = 0
Решение:
We are given,
xy – x – y + 1 = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (X + 1) (Y + 1) – (Y + 1) – (X + 1) + 1 = 0
=> XY + X + Y + 1 – Y – 1 – X – 1 + 1 = 0
=> XY = 0
Therefore, the required equation is XY = 0.
(iv) х 2 - у 2 - 2х + 2у = 0
Решение:
We are given,
x2 – y2 – 2x + 2y = 0
Putting x = X + 1 and y = Y + 1, we get,
=> (X + 1)2 – (Y + 1)2 – 2(X + 1) + 2(Y + 1) = 0
=> X2 + 1 + 2X – Y2 – 1 – 2Y – 2X – 2 + 2Y + 2 = 0
=> X2 – Y2 = 0
Therefore, the required equation is X2 – Y2 = 0.
Вопрос 7. Найдите точку, в которую следует сместить начало координат после переноса осей, чтобы следующие уравнения не имели членов первой степени.
(i) х 2 + у 2 – 4х – 8у + 3 = 0
Решение:
We are given,
x2 + y2 – 4x – 8y + 3 = 0
Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,
=> (X + a)2 + (Y + b)2 – 4(X + a) – 8(Y + b) + 3 = 0
=> X2 + a2 + 2aX + Y2 + b2 + 2bY – 4X – 4a – 8Y – 8b + 3 = 0
=> X2 + Y2 + (2a – 4)X + (2b – 8)Y + (a2 + b2 – 4a – 8b +3) = 0
As our transformed equation has no first-degree term, we have,
2a – 4 = 0 and 2b – 8 = 0
By solving these equations we have a = 2 and b = 4.
Therefore, the origin has been shifted to (2,4) from (0,0).
(ii) х 2 + у 2 – 5х + 2у – 5 = 0
Решение:
We are given,
x2 + y2 – 5x + 2y – 5 = 0
Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,
=> (X + a)2 + (Y + b)2 – 5(X + a) + 2(Y + b) – 5 = 0
=> X2 + a2 + 2aX + Y2 + b2 + 2bY – 5X – 5a + 2Y + 2b – 5 = 0
=> X2 + Y2 + (2a – 5)X + (2b + 2)Y + (a2 + b2 – 5a + 2b – 5) = 0
As our transformed equation has no first-degree term, we have,
2a – 5 = 0 and 2b + 2 = 0
By solving these equations we have a = 5/2 and b = –1.
Therefore, the origin has been shifted to (5/2, –1) from (0,0).
(iii) х 2 – 12 х + 4 = 0
Решение:
We are given,
x2 – 12x + 4 = 0
Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,
=> (X + a)2 – 12(X + a) + 4 = 0
=> X2 + a2 + 2aX – 12X – 12a + 4 = 0
=> X2 + (2a – 12)X + (a2 – 12a + 4) = 0
As our transformed equation has no first-degree term, we have,
=> 2a – 12 = 0
=> a = 6
Therefore, the origin has been shifted to (6,b) from (0,0) where b is any arbitrary value.
Вопрос 8. Проверить, что площадь треугольника с вершинами (4, 6), (7, 10) и (1, –2) остается неизменной относительно переноса осей при смещении начала координат в точку (–2, 1).
Решение:
Here, L.H.S. = A1 = Area of the triangle with vertices (4, 6), (7, 10) and (1, –2)
=
=
=
=
= 6 sq. units
As the origin shifted to point (–2, 1), the new coordinates of the triangle are:
(X1, Y1) = (4–2, 6+1) = (2, 7)
(X2, Y2) = (7–2, 10+1) = (5, 11)
(X3, Y3) = (1–2, –2+1) = (–1, –1)
Now, R.H.S. = A2 = Area of the triangle with vertices (2, 7), (5, 11), (–1, –1)
=
=
=
=
= 6 sq. units
Therefore, A1 = A2.
Hence, proved.
