Let A(1, 0), B(2, -1) and C(7/3, 4/3) be three points. If the equation of the bisector of the angle

Q: Why do we need unit vectors? Can't we just add $\vec{BA}$ and $\vec{BC}$ directly?

If you add $\vec{BA}$ and $\vec{BC}$ directly, the resulting vector will be biased towards the longer side. By normalizing them into unit vectors, you create a rhombus where all sides are equal, ensuring the diagonal perfectly bisects the angle.

Q: Is there another way to solve this without vectors?

Yes, you could use the angle bisector theorem or the distance formula, but those methods often involve cumbersome square roots and sign considerations. The vector method is the most elegant and robust approach for JEE.

Q: Why did we get $\alpha=3$ and $\beta=1$?

The problem states the equation is $\alpha x + \beta y = 5$. After deriving the equation $3x + y = 5$, we simply equate the coefficients. It is a direct comparison.

Question

Let $A(1, 0), B(2, -1)$ and $C(\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $ABC$ is $\alpha x + \beta y = 5$, then the value of $\alpha^2 + \beta^2$ is

Options:

(A) 13

(B) 10

(C) 5

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the equation of the angle bisector of $\angle ABC$ with vertices $B(2, -1)$, $A(1, 0)$, and $C(\frac{7}{3}, \frac{4}{3})$, we utilize the geometric property that the diagonal of a rhombus bisects the vertex angle.

First, we define the vectors originating from point $B$:

$$\vec{BA} = A - B = (1 - 2, 0 - (-1)) = (-1, 1)$$

$$\vec{BC} = C - B = \left(\frac{7}{3} - 2, \frac{4}{3} - (-1)\right) = \left(\frac{1}{3}, \frac{7}{3}\right)$$

### Normalizing the Vectors

To ensure the bisector is perfectly centered, we must normalize these vectors into unit vectors $\hat{u}_1$ and $\hat{u}_2$. This prevents the longer vector from biasing the direction of the resultant sum.

The magnitude of $\vec{BA}$ is $|\vec{BA}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$. Thus, the unit vector is:

$$\hat{u}_1 = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$$

The magnitude of $\vec{BC}$ is $|\vec{BC}| = \sqrt{(\frac{1}{3})^2 + (\frac{7}{3})^2} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$. Thus, the unit vector is:

$$\hat{u}_2 = \left(\frac{1/3}{5\sqrt{2}/3}, \frac{7/3}{5\sqrt{2}/3}\right) = \left(\frac{1}{5\sqrt{2}}, \frac{7}{5\sqrt{2}}\right)$$

### The Master Equation

The direction of the angle bisector is given by the sum of these unit vectors, $\vec{V} = \hat{u}_1 + \hat{u}_2$:

$$\vec{V} = \left(-\frac{5}{5\sqrt{2}} + \frac{1}{5\sqrt{2}}, \frac{5}{5\sqrt{2}} + \frac{7}{5\sqrt{2}}\right) = \left(-\frac{4}{5\sqrt{2}}, \frac{12}{5\sqrt{2}}\right)$$

We can simplify the direction vector by scaling it by $-\frac{5\sqrt{2}}{4}$, which yields the direction ratio $(1, -3)$. Consequently, the slope of the bisector is $m = \frac{-3}{1} = -3$.

### Final Calculation

Using the point-slope form with point $B(2, -1)$, the equation of the line is:

$$y - (-1) = -3(x - 2)$$

$$y + 1 = -3x + 6$$

$$3x + y = 5$$

Comparing this to the form $\alpha x + \beta y = 5$, we identify $\alpha = 3$ and $\beta = 1$. The final result is:

$$\alpha^2 + \beta^2 = 3^2 + 1^2 = **10**$$

Animated Solution for Mathematics - Straight Lines: Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

Select Answer:

Visualized Solution

Visualizing the Points

The Angle Bisector Concept

Defining Vector $B A$

Unit Vector $\overset{u}{^}_{1}$

Defining Vector $B C$

Unit Vector $\overset{u}{^}_{2}$

Direction of the Bisector

Simplifying the Direction

Equation of the Bisector

Standard Form

Final Calculation

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Normalizing the Vectors

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Points

The Angle Bisector Concept

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BA

Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u^1​

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u^2​

Direction of the Bisector

Simplifying the Direction

Equation of the Bisector

Standard Form

Final Calculation

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Normalizing the Vectors

The Master Equation

Final Calculation

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