MHTCET 26 Apr 2025 Shift 1 Paper

Concept:The internal angle bisector of $\angle A$ divides the opposite side $BC$ in the ratio of the adjacent sides $AB:AC$.Explanation:Compute the lengths of sides $AB$ and $AC$.$$AB=\sqrt{(3-1)^2+(5+1)^2+(3-0)^2}=\sqrt{4+36+9}=7$$$$AC=\sqrt{(-11-1)^2+(-5+1)^2+(6-0)^2}=\sqrt{144+16+36}=14$$Thus, $AB:AC=7:14=1:2$.Let the angle bisector meet $BC$ at point $D$.By the angle bisector theorem, $BD:DC=AB:AC=1:2$.Using the section formula for internal division,$$D=\frac{2B+1C}{3}=\left(\frac{6-11}{3},\frac{10-5}{3},\frac{6+6}{3}\right)=\left(-\frac{5}{3},\frac{5}{3},4\right)$$Direction ratios of $AD$ are:$$D-A=\left(-\frac{8}{3},\frac{8}{3},4\right)$$These are proportional to $(-2,2,3)$.So, the equation of the angle bisector through $A(1,-1,0)$ is:$$\frac{x-1}{-2}=\frac{y+1}{2}=\frac{z}{3}$$This is equivalent to:$$\frac{1-x}{2}=\frac{y+1}{2}=\frac{z}{3}$$Answer:Option A: $\frac{1-x}{2}=\frac{y-(-1)}{2}=\frac{z}{3}$