Let ABC be a triangle with A(-3, 1) and ∠ACB = θ, 0

Q: Wait, why does the midpoint of $AC$ lie on the median through $B$?

The median is defined as the line segment connecting a vertex to the midpoint of the opposite side. Since the median passes through $B$ and the midpoint of $AC$, the midpoint must satisfy the equation of the median line. It is a fundamental property of triangles!

Q: Can we assume the angle bisector is the line $BC$?

No! The angle bisector is a line that divides the angle at $C$ into two equal parts. It does not necessarily coincide with the side $BC$. The side $BC$ is just one of the lines forming the angle $\theta$. The bisector is a distinct line passing through $C$.

Q: Why do we calculate $\tan(\theta/2)$ first?

Because the angle bisector splits the angle $\theta$ into two equal parts, the angle between the side $AC$ and the bisector is exactly $\theta/2$. This allows us to use the slopes of $AC$ and the bisector to find $\tan(\theta/2)$ directly, which is much easier than trying to find the slope of $BC$ first.

Question

Let $ABC$ be a triangle with $A(-3, 1)$ and $\angle ACB = 	heta, 0 < 	heta < \frac{\pi}{2}$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of angle bisector of $C$ is $7x - 4y - 1 = 0$, then $	an 	heta$ is equal to:

Options:

(A) $\frac{1}{2}$

(B) $\frac{3}{4}$

(C) $\frac{4}{3}$

(D) $2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Cartesian plane, looking at a triangle $ABC$. You know where vertex $A$ is, but $B$ and $C$ are shrouded in mystery.

You are given two clues: the equation of the median through $B$ and the equation of the angle bisector of $C$. This is not just a math problem; it is a detective story. We need to find the coordinates of $C$ and then use the power of trigonometry to find $	an 	heta$.

### Hunting for Vertex $C$

Let us assume the coordinates of $C$ are $(a, b)$. We know $C$ lies on the angle bisector $7x - 4y - 1 = 0$. This gives us our first constraint:
$$7a - 4b = 1$$

Now, consider the median through $B$. A median connects a vertex to the midpoint of the opposite side. So, the median from $B$ must pass through the midpoint $M$ of $AC$.

With $A(-3, 1)$ and $C(a, b)$, the midpoint $M$ is $(\frac{a-3}{2}, \frac{b+1}{2})$. Since $M$ lies on the median $2x + y - 3 = 0$, we substitute these coordinates into the equation:
$$2\left(\frac{a-3}{2}ight) + \frac{b+1}{2} - 3 = 0$$

Simplifying this, we get $(a-3) + \frac{b+1}{2} - 3 = 0$, which leads to $2a + b = 11$. Now we have a system of two linear equations:
$$7a - 4b = 1$$
$$2a + b = 11$$

Solving this system, we find $a = 3$ and $b = 5$. Therefore, vertex $C$ is at $(3, 5)$.

### The Geometry of the Bisector

Now that we have $C(3, 5)$, we can find the slope of $AC$. Using the slope formula, $m_1 = \frac{5-1}{3-(-3)} = \frac{4}{6} = \frac{2}{3}$.

The angle bisector equation is $7x - 4y - 1 = 0$, which can be rewritten as $y = \frac{7}{4}x - \frac{1}{4}$. Thus, the slope of the bisector is $m_2 = \frac{7}{4}$.

The angle between the side $AC$ and the bisector is exactly $\frac{	heta}{2}$. We use the formula for the tangent of the angle between two lines:
$$	an\left(\frac{	heta}{2}ight) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$$

Substituting our slopes, we get:
$$	an\left(\frac{	heta}{2}ight) = \left| \frac{\frac{7}{4} - \frac{2}{3}}{1 + (\frac{7}{4})(\frac{2}{3})} ight| = \left| \frac{\frac{21-8}{12}}{\frac{12+14}{12}} ight| = \frac{13}{26} = \frac{1}{2}$$

### The Trigonometric Bridge

We have found $	an(\frac{	heta}{2}) = \frac{1}{2}$, but the question asks for $	an 	heta$. We use the double-angle identity:
$$	an 	heta = \frac{2 	an(\frac{	heta}{2})}{1 - 	an^2(\frac{	heta}{2})}$$

Plugging in our value, we get:
$$	an 	heta = \frac{2(\frac{1}{2})}{1 - (\frac{1}{2})^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$

The journey is complete! By carefully translating geometric properties into algebraic constraints, we have unraveled the mystery of triangle $ABC$. The final result is **$	an 	heta = \frac{4}{3}$**.

Animated Solution for Mathematics - Straight Lines: Let $A B C$ be a triangle with $A (- 3, 1)$ and $∠ A C B = θ, 0 < θ < \frac{π}{2}$ . If the equation of the median through $B$ is $2 x + y - 3 = 0$ and the equation of angle bisector of $C$ is $7 x - 4 y - 1 = 0$ , then $tan θ$ is equal to:

Select Answer:

Visualized Solution

Analyze the Given Information

Locate Vertex $C (a, b)$

Define Midpoint $M$ of $A C$

Substitute $M$ into Median Equation

Simplify to Equation (ii)

Solve for $a$ and $b$

Find Slope of $A C$

Find Slope of Angle Bisector

Relate $θ$ and $θ /2$

Calculate $tan (θ /2)$

Apply Double Angle Formula

Final Computation

Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Hunting for Vertex $C$

The Geometry of the Bisector

The Trigonometric Bridge

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Select Answer:

Visualized Solution

Analyze the Given Information

Locate Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C(a,b)

Define Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M into Median Equation

Simplify to Equation (ii)

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Find Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC