Match the items of column I with that of column II
Column-I | Column-II |
(i) If the line joining P (1, 3) and Q (5, 7) subtend right angle at R(x, y) such that area of ΔPQR = 2, the number of such points R is | [A] 2 |
(ii) Let cos2m (n!) πx equals to α. If x ∈ Q, then area of triangle with vertices (α, β), (–2, 1) and (2, 1) is (where β lies on the line at 1 unit distant from line joining these vertices) | [B] –3 |
(iii) A man starts from P(–3, 4) and reaches Q (0, 1) after touching x- axis at R (α, 0) such that PR + RQ is minimum then 5α is | [C] 3 |
(iv) The integral value of 'a' for which the images of the point (a, a–1) w.r.t. the line mirror 3x + y = 6a is a point (a2 + 1, a), is | [D] 4 |
Text Solution
Verified by Experts(i) [D]; (ii) [A]; (iii) [B]; (iv) [A]
Ans.
(i) [D]
(ii) [A]
(iii) [B]
(iv) [A]
Sol. (i)
PQ ≡ x – y + 2 = 0
d (PQ) = 4 
Δ = 
p. 4 
= 2
p = 
p = 
= 
x – y + 2 = 1, x – y + 2 = –1
x – y = –1, x – y = –3
(x –5) (x – 1) + (y – 7) (y – 3) = 0
using x + 1 = y, x 2 – 6x + 5 + (x –6) (x –2) = 0
2x 2 – 14x + 17 = 0 ⇒ 2 points
similarly for x – y = – 3 ⇒ 2 pts.
Hence total 4 points.
(ii) α = 1 (1, β ) , (–2, 1), (2, 1) β can be above or below
so Δ = 
4 (1) = 2
(iii)
For PR + RQ to be minimum Q ′ is mirror image of θ
with x-axis
So Q ′ ≡ (0, –1)
So PQ ′ ≡ 
x + (y + 1) = 0
for R ⇒ α = – 3/5
5 α = – 3
(iv) Mid-point of object and image will be on the line i.e. 
(a 2 + a + 1) + 
= 6a
⇒ 3a 2 + 5a + 2 = 12a
⇒ (3a –1) (a – 2) = 0 ⇒ a = 2, 1/3 a = 2
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