Vector Algebra
Distance from a point to an angle bisector line
nta_pyq_2025_apr
Grade 12
Question:
Let A, B, C be three points in the $xy$-plane, whose position vectors are given by $\sqrt{3}\hat{i}+\hat{j}$, $\hat{i}+\sqrt{3}\hat{j}$ and $a\hat{i}+(1-a)\hat{j}$ respectively with respect to the origin $O$. If the distance of the point $C$ from the line bisecting the angle between the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is $\dfrac{9}{\sqrt{2}}$, then the sum of all the possible values of $a$ is:
$2$
$\dfrac{9}{2}$
$1$
$0$
Step-by-Step Solution
Key Concept: Both $\overrightarrow{OA}$ and $\overrightarrow{OB}$ have equal magnitudes; the angle bisector is $y=x$, i.e. $x-y=0$. Distance of $C=(a,1-a)$ from this line is $|2a-1|/\sqrt{2}$; set it equal to $9/\sqrt{2}$ and use Vieta's theorem on the resulting quadratic.
$|\overrightarrow{OA}|=|\overrightarrow{OB}|=2$, so the angle bisector of $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is the line $y=x$.
Distance from $C=(a,1-a,0)$ to line $x-y=0$: $D=\dfrac{|a-(1-a)|}{\sqrt{2}}=\dfrac{|2a-1|}{\sqrt{2}}=\dfrac{9}{\sqrt{2}}$.
$(2a-1)^2=81$. Expanding: $2(4a^2-4a+1)=81(a^2+(1-a)^2)=81(2a^2-2a+1)$.
$8a^2-8a+2=162a^2-162a+81 \Rightarrow 154a^2-154a+79=0$.
Sum of values of $a=-\dfrac{-154}{154}=1$.
Correct Answer: 3