Match the following columns :
Column I | Column II |
(i) The image of a point A(3,8) in the line x + 3y – 7 = 0, is | [A] 11x–3y+9= 0 |
(ii) A point equidistant from the lines 4x+ 3y + 10 = 0, 5x – 12y + 26 = 0 & 7x + 24y – 50 = 0 | [B] (0, 0) |
(iii) The equation of the acute angle bisector in between the lines 3x – 4y + 7 = 0 & 12x + 5y – 2 = 0 | [C] 2x+ y – 4 = 0 |
(iv) A straight line through P(1, 2) is such that its intercept between the axes is bisected at P, then its equation is | [D] (–1, – 4) |
Text Solution
Verified by Experts(i) [D]; (ii) [B]; (iii) [A]; (iv) [C]
Ans.
(i) [D]
(ii) [B]
(iii) [A]
(iv) [C]
Sol. (i) x + 3y – 7 = 0......(i) {a = 1, b = 3
⇒ 
= 
= 
(ii) Check by option
(iii) 3x – 4y + 7 = 0 .....(i)
12x + 5y – 2 = 0 .......(ii)
– 12x – 5y + 2 = 0 ......(ii)
By (i) ⇒ a 1 = 3; b 1 = – 4
By (ii) ⇒ a 2 = –12 ; b 2 = – 5
⇒ (a 1 a 2 + b 1 b 2 ) = – 36 + 20 < 0
∴ acute angle bisector is
⇒ 
=+ 
(iv) Equation of line AB is
= (m + n)
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