Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
The bisector of angle between the straight lines $y - b = \frac{2m}{1-m^2}(x - a)$ and $y - b = \frac{2m'}{1-m'^2}(x - a)$ are:
$(y - b)(m + m') + (x - a)(1 - mm') = 0$
$(y - b)(m + m') - (x - a)(1 - mm') = 0$
$(y - b)(m - m') + (x - a)(1 + mm') = 0$
$(y - b)(1 - mm') - (x - a)(m + m') = 0$
Step-by-Step Solution
Key Concept: Angle bisectors between two lines have slopes derived from the formula combining the original slopes.
Given two lines with inclinations $2\theta_1$ and $2\theta_2$, their slopes are $m = \tan(2\theta_1)$ and $m' = \tan(2\theta_2)$. The angle bisectors have inclinations $\theta_1 + \theta_2$ or $\theta_1 + \theta_2 + \frac{\pi}{2}$. The slopes of the bisectors are $\frac{m + m'}{1 - mm'}$ and $\frac{mm' - 1}{m + m'}$, with equations $y - b = \frac{(m+m')}{(1-mm')}(x-a)$ and $y - b = \frac{(mm'-1)}{(m+m')}(x-a)$.
Correct Answer: 1,4