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 axesOrigin 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) xThe 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

x

(x'+1)

x'

we will rewrite as

x

(ii) xy - y

(x'+1)(y'+1) - (y'+1)

x'y' + x' + y' +1 - y'

x'y' - y'

xy - y

(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