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Find the distance between the following pair of points:
(i) (- 6, 7) and (-1, -5)
(ii) (a + b, b + c) and (a -b, c - b)
(iii) (a sin α, - b cos α) and (- a cos α, b sin α)
(iv) (a, 0) and (0, b)
(i) We have P (- 6, 7) and Q (- 1, - 5)
Here,
x1 = - 6, y1 = 7 and
x2 = -1, Y2 = - 5
(ii) We have P (a + b, b + c) and Q (a - b, c - b) here,
x1 = a + b, y1 = b + c and x2 = a - b, Y2 = c - b
(iii) We have P(a sinα, – b cos α) and Q(-a cos α, b sin α) here
x1 = a sin α, y1 = - b cos α and
x2 – a cos α, y2 = b sin α
(iv) We have P(a, 0) and Q (0, b)
x1 = a,y1 = 0, x2 = 0, y2 = b,
Find the value of a when the distance between the points (3, a) and (4, 1) is √10.
We have P (3, a) and Q(4, 1)
Squaring both sides
⇒ 10 = 2 + a2 – 2a
⇒ a2 – 2a + 2 – 10 = 0
⇒ a2 – 2a – 8 = 0
Splitting the middle team
⇒ a2 – 4a + 2a - 8 = 0
⇒ a(a - 4) + 2(a - 4) = 0
⇒ (a - 4) (a + 2) = 0
⇒ a = 4, a = - 2
If the points (2, 1) and (1, -2) are equidistant from the point (x, y) from (-3, 0) as well as from (3, 0) are 4.
We have P(2, 1) and Q(1,- 2) and R(X, Y)
Also, PR = QR
∴ PR = QR
⇒ x2 + 5 - 4x + y2 –2y = x2 + 5 - 2x + y2 + 4y
⇒ x2 + 5 - 4x + y2 – 2y = x2 + 5 - 2x + y2 + 4y
⇒ - 4x + 2x - 2y - 4y = 0
⇒ - 2x – 6y = 0
⇒ - 2(x + 3y) = 0
⇒ -2(x + 3y) = 0
⇒ x + 3y = 0/-2
⇒ x + 3y = 0
Hence Proved.
Find the value of x, y if the distances of the point (x, y) from (- 3, 0) as well as from (3, 0) are 4.
We have P(x, y), Q( -3, 0) and R(3, 0)
⇒ 16 = x2 + 9 + 6x + y2
⇒ x2 + y2 = 7 - 6x …… (1)
⇒16 = x2 + 9 – 6x + y2
⇒ x2 + y2 = 16 – 9 + 6x
⇒ x2+ y2 = 7 + 6x …. (2)
Equating (1) and (2)
7 - 6x = 7 + 6x
⇒ 7 – 7 = 6x + 6x
⇒ 0 = 12x
⇒ x = 12
Substituting the value of x = 0 in (2)
x2 + y2 = 7+ 6x
0 + y2 = 7 + 6 × 0
Y2 = 7
Y = + 7
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Chapter 14: Coordinate Geometry Exercise –...