Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 a

Q: Wait, why is it $y = -x$ and not $y = x$?

Great question! In the fourth quadrant, the x-coordinate is positive ($x > 0$) and the y-coordinate is negative ($y < 0$). The distance from the y-axis is $|x| = x$, and the distance from the x-axis is $|y| = -y$. For these distances to be equal, we must have $x = -y$, which rearranges to $y = -x$.

Q: Can I just use Cramer's rule to solve for $x$ and $y$?

You absolutely could, but it is a massive trap! Cramer's rule is computationally heavy and prone to algebraic errors. In JEE Advanced, time is your most valuable resource. Using the geometric property $y = -x$ allows you to solve the problem in seconds, bypassing the need for determinants entirely.

Q: What if $a$ or $b$ were zero?

The problem statement explicitly defines $a, b, c,$ and $d$ as non-zero numbers. This ensures that the lines are well-defined and that we don't encounter division-by-zero errors when isolating $x$.

Question

Let $a, b, c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4ax + 2ay + c = 0$ and $5bx + 2by + d = 0$ lies in the fourth quadrant and is equidistant from the two axes then

Options:

(A) $3bc - 2ad = 0$

(B) $3bc + 2ad = 0$

(C) $2bc - 3ad = 0$

(D) $2bc + 3ad = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on the Cartesian plane, looking at the intersection of two lines. The problem specifies that the intersection point $P(x, y)$ lies in the fourth quadrant. In the fourth quadrant, the $x$-coordinate is positive ($x > 0$) and the $y$-coordinate is negative ($y < 0$). The distance to the $y$-axis is $|x|$, and the distance to the $x$-axis is $|y|$. Since $x > 0$ and $y < 0$, we have $|x| = x$ and $|y| = -y$. Equating these distances, we get $x = -y$, or simply: $$y = -x$$ This simple relationship is our golden ticket to solving the problem efficiently. ### The Trap of Complexity Many students, upon seeing two linear equations, immediately reach for heavy artillery like Cramer's rule or cross-multiplication. While these methods are mathematically sound, they are often a trap in competitive exams. They lead to messy determinants and a high probability of calculation errors. Instead, let's embrace the elegance of substitution by using the relationship $y = -x$ immediately. ### The Algebraic Dance Let's take our first line: $4ax + 2ay + c = 0$. By substituting $y = -x$, the equation transforms into: $$4ax + 2a(-x) + c = 0$$ Simplifying this, we get $4ax - 2ax + c = 0$, which reduces to $2ax + c = 0$. Solving for $x$, we find: $$x = \frac{-c}{2a}$$ Now, let's apply the same logic to the second line: $5bx + 2by + d = 0$. Substituting $y = -x$ gives us: $$5bx + 2b(-x) + d = 0$$ This simplifies to $5bx - 2bx + d = 0$, or $3bx + d = 0$. Solving for $x$ again, we get: $$x = \frac{-d}{3b}$$ ### The Final Synthesis We now have two expressions for the same $x$-coordinate of the intersection point $P$. Since they represent the same point, they must be equal: $$\frac{-c}{2a} = \frac{-d}{3b}$$ The negative signs on both sides cancel out, leaving us with $\frac{c}{2a} = \frac{d}{3b}$. Now, a simple cross-multiplication brings us to the finish line: $$3bc = 2ad$$ Rearranging this to match standard forms, we get **$3bc - 2ad = 0$**. By focusing on the geometric reality of the point rather than blindly applying formulas, we have navigated the problem with precision and speed.

Animated Solution for Mathematics - Straight Lines: Let $a, b, c$ and $d$ be non-zero numbers. If the point of intersection of the lines $4 a x + 2 a y + c = 0$ and $5 b x + 2 b y + d = 0$ lies in the fourth quadrant and is equidistant from the two axes then

Select Answer:

Visualized Solution

Visualizing the Intersection

Decoding the Geometry

The JEE Shortcut

Solving for $x$ from Line 1

Substituting into Line 2

Solving for $x$ from Line 2

Equating the $x$ Values

Cross-Multiplication

Final Rearrangement

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Trap of Complexity

The Algebraic Dance

The Final Synthesis

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