Question:
The bisector of the acute angle between the straight lines \(3x – 4y – 3 = 0\) and \(12x + 5y + 6 = 0\) passes through which one of the following points?
सरल रेखाओं \(3x – 4y – 3 = 0\) और \(12x + 5y + 6 = 0\) के बीच के न्यून कोण का समद्विभाजक निम्नलिखित में से किस बिंदु से होकर गुजरता है?
Options:
- (5, 3)
- (-3, 6)
- (2, 7)
- (-1, 4)
- (5, 3)
- (-3, 6)
- (2, 7)
- (-1, 4)
Correct Answer: c)
Explanation:
Make constants positive: \(-3x + 4y + 3 = 0\) and \(12x + 5y + 6 = 0\).
Check \(a_1 a_2 + b_1 b_2 = (-3)(12) + (4)(5) = -36 + 20 = -16 < 0\).
Since \(a_1 a_2 + b_1 b_2 < 0\), the acute angle bisector corresponds to the '+' sign:
\[\frac{-3x + 4y + 3}{\sqrt{(-3)^2 + 4^2}} = \frac{12x + 5y + 6}{\sqrt{12^2 + 5^2}}\]
\[\frac{-3x + 4y + 3}{5} = \frac{12x + 5y + 6}{13}\]
\[13(-3x + 4y + 3) = 5(12x + 5y + 6)\]
\[-39x + 52y + 39 = 60x + 25y + 30 \implies 99x – 27y – 9 = 0 \implies 11x – 3y – 1 = 0\]
Testing point (2, 7):
\[11(2) – 3(7) – 1 = 22 – 21 – 1 = 0\]
Thus, the bisector passes through (2, 7).
Hence, the correct option is (c).
अचर पदों को धनात्मक करने पर: \(-3x + 4y + 3 = 0\) तथा \(12x + 5y + 6 = 0\)।
\(a_1 a_2 + b_1 b_2 = -36 + 20 = -16 < 0\)।
अतः न्यूनकोण समद्विभाजक ‘+’ चिह्न लेने पर प्राप्त होता है:
\[\frac{-3x + 4y + 3}{5} = \frac{12x + 5y + 6}{13} \implies 11x – 3y – 1 = 0\]
बिंदु (2, 7) रखने पर:
\[11(2) – 3(7) – 1 = 22 – 22 = 0\]
अतः यह बिंदु (2, 7) से होकर गुजरता है।
अतः सही विकल्प (c) है।