11th NCERT/CBSE Straight Lines Exercise 10.5 (supplementary) Questions 2
Hi

Question (1)

Find the new co-ordinates of the points in each of the following cases if the origin is shifted to the point (-3, -2) by translation of axes
Origin is shifted to (h, k) = (-3, -2) ⇒ h = - 3, k = -2

Solution

(i) (1, 1), (x, y) = (1, 1)
New co-ordinate of point (x', y') = (1, 1)
New co-ordinate of point (x', y') = (x-h, y-k)
= (1+3, 1+2) = 4, 3)

(ii) (0, 1), (x, y) = (0, 1)
New co-ordinate of point (x', y') = (x-h, y-k)
=(0 + 3, 1 + 2) = (3, 3)

(iii) (5, 0), (x, y) = (0, 1)
New co-ordinate of point (x', y') = (x - h, y - k)
=(5 + 3, 0 + 2) = (8, 2)

(iv) ( -1, -2), (x, y) = (-1, -2)
New co-ordinate of point (x', y') =(x - h, y - k)
=(-1+3, -2+2) = (2, 0)
(v) (3, -5), (x, y) = (3, -5)
New co-ordinate of point (x', y') = (x-h, y-k)
=(3+3, -5+2) = (6, -3)

Question (2)

Find what the following equation become when origin is shifted to the point (1,1)

Solution

(1) x2 + xy -3y2 - y + 2 = 0
The origin is shifted to (h, k) = (1, 1)
∴ x' = x - h, y' = y - k
x' = x - 1 , y' = y - 1
x = x' +1 , y = y' + 1
Replace x = x'+1 and y = y' + 1 in equation we get
x2 + xy - 3y2 - y + 2 = 0
(x'+1)2 + (x'+1)(y'+1) - 3(y'+1)2 - (y' +1) +2 = 0
x'2 - 3y'2 +3x'
+x'y' - 6y' = 0
we will rewrite as
x2 - 3y2 + xy +3x - 6y = 0

(ii) xy - y2 - x + y = 0
(x'+1)(y'+1) - (y'+1)2 - (x'+1) + (y'+1) = 0
x'y' + x' + y' +1 - y'2 - 2y' - 1 - x' - 1 + y' + 1 = 0
x'y' - y'2 = 0
xy - y2 = 0

(iii) xy - x - y +1=0
(x'+1) (y'+1) - (x'+1) - (y'+1) + 1 = 0
x'y' + x' + y' +1 - x' -1 -y' -1+1 = 0
x'y' = 0
xy = 0

Exercise 10.4 ⇐
⇒ Exercise 10.6