Question

The distance of a point (2, 5) from the line $3x+y+4=0$ measured along the line $L_{1}$ and $L_{2}$ are same. If slope of line $L_{1}$ is $\frac{3}{4}$, then slope of the line $L_{2}$ is

• A. $-\frac{3}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $0$

Hint:

Logic Tip: The symmetry of the isosceles triangle condition immediately means that the bisector of the angle between $L_1$ and $L_2$ is perpendicular (or parallel) to the base line $L$. Setting up the angle equality is the standard and most robust way to solve this.

Answer 1

The Correct Option is D

Concept:
If the distances from a point $A(2,5)$ to a given line along two distinct lines $L_1$ and $L_2$ are equal, then the line $3x+y+4=0$ forms an isosceles triangle with $L_1$ and $L_2$. Because the triangle is isosceles, the angle $\theta$ made by $L_1$ with the given line is equal to the angle made by $L_2$ with the given line. We can use the angle between two lines formula: $\tan \theta = |\frac{m_1 - m_2}{1 + m_1m_2}|$.
Step 1: Identify the slopes of the known lines.
Let the given line be $L: 3x + y + 4 = 0$. Its slope is: $$m = -3$$ Let the slope of line $L_1$ be $m_1 = \frac{3}{4}$. Let the slope of line $L_2$ be $m_2$, which we need to find.
Step 2: Equate the angles made by $L_1$ and $L_2$ with L.
Since the distances are the same, the lines $L_1$ and $L_2$ are equally inclined to $L$. $$|\tan \theta_1| = |\tan \theta_2|$$ $$\left| \frac{m - m_1}{1 + mm_1} \right| = \left| \frac{m - m_2}{1 + mm_2} \right|$$
Step 3: Substitute the values and solve for $m_2$.
Substitute $m = -3$ and $m_1 = \frac{3}{4}$: $$\left| \frac{-3 - \frac{3}{4{1 + (-3)\left(\frac{3}{4}\right)} \right| = \left| \frac{-3 - m_2}{1 + (-3)m_2} \right|$$ $$\left| \frac{\frac{-15}{4{1 - \frac{9}{4 \right| = \left| \frac{-3 - m_2}{1 - 3m_2} \right|$$ $$\left| \frac{\frac{-15}{4{\frac{-5}{4 \right| = \left| \frac{3 + m_2}{1 - 3m_2} \right|$$ $$|3| = \left| \frac{3 + m_2}{1 - 3m_2} \right|$$
Step 4: Solve the absolute value equation.
This gives two cases: Case 1: $3 = \frac{3 + m_2}{1 - 3m_2}$ $$3(1 - 3m_2) = 3 + m_2$$ $$3 - 9m_2 = 3 + m_2$$ $$10m_2 = 0 \implies m_2 = 0$$ Case 2: $-3 = \frac{3 + m_2}{1 - 3m_2}$ $$-3(1 - 3m_2) = 3 + m_2$$ $$-3 + 9m_2 = 3 + m_2$$ $$8m_2 = 6 \implies m_2 = \frac{3}{4}$$ Since $m_2 = \frac{3}{4}$ is the slope of $L_1$, the slope of the distinct line $L_2$ must be $0$.