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List-I List-II
A) The number of integral values of ‘a’ for which point $(a, a^{2})$ lies completely inside the triangle formed by lines x = 0, y = 0,
2y + x = 3.
P) 0
B) The number of values of ‘a’ of the form $\frac{K}{3}$ where $K \in I$ so that point $(a, a^{2})$ lies between the lines x + y = 2 and 4x+4y–3=0. Q) 1
C) The reflection of point (t – 1, 2t + 2) in a line is (2t + 1, t) then the slope of line is R) 2
D) In a triangle ABC, the bisector of angles B and C lie along the lines y = x and y = 0. If A is (1, 2) then $\sqrt{10}$ d(A, BC) equals to(where d(A, BC) denotes the perpendicular distance of A from BC). S) 4

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A–P, B–R, C–Q, D–S

A–P, B–R, C–S, D–Q

A–Q, B–R, C–P, D–S

A–P, B–S, C–Q, D–R

answer is A.

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Detailed Solution

(a) $2 a^{2} + a - 3 = 0$
$(2 a + 3) (a - 1) = 0$
Question Image
$\Rightarrow a \in (0, 1)$
No. of integral values of a = 0
(b) $a^{2} + a - 2 = 0$
$\Rightarrow (a + 2) (a - 1) = 0 \Rightarrow a = - 2, 1$

$4 a^{2} + 4 a - 3 = 0 (2 a - 1) (2 a + 3) = 0 \Rightarrow a = \frac{1}{2}, - \frac{3}{2} \Rightarrow a \in (- 2, \frac{3}{- 2}) \cup (\frac{1}{2}, 1)$
Values of a of form $\frac{K}{3}$ are $\frac{- 5}{3}, \frac{2}{3}$
(c) Slope of line joining (t-1, 2t+2) and (2t+1,t) is $\frac{2 t + 2 - t}{t - 1 - 2 t - 1} = - 1$
Slope of perpendicular bisectors of points is 1.
(d) Images of A w.r.t. y = x and y = 0 lies on BC which are (2, 1), (1, -2)
Question Image
Equation of BC is y = 3x – 5
Perpendicular distance of A from
$B C = \frac{| 3 - 2 - 5 |}{\sqrt{10}} d (A, B C) = \frac{4}{\sqrt{10}} \Rightarrow \sqrt{10} d (A, B C) = 4$

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