A line through A(-5, -4) meets the line x + 3y + 2 = 0, 2x + y + 4 = 0 and x - y - 5 = 0 at the poin

Q: Why can't I just use y - y_1 = m(x - x_1)? It seems more intuitive.

It is intuitive, but it forces you to solve for intersection points in terms of m, leading to messy expressions like \frac{...}{1+3m}. The parametric form is elegant because it treats the distance r as the variable, which is exactly what the problem asks for. It bypasses the need to find the coordinates of the intersection points entirely!

Q: What if the line is vertical? Does the parametric form fail?

Great catch! If the line is vertical, \theta = 90^\circ. In that case, \cos \theta = 0 and \sin \theta = 1. The parametric equations still hold perfectly. The math is robust and handles all slopes, including the undefined ones, without breaking a sweat.

Q: How did you know to look for a perfect square at the end?

In JEE Advanced problems, when you see a quadratic equation in \sin \theta and \cos \theta equal to zero, always check if it's a perfect square. It's a common pattern to ensure a unique solution for the slope. If it wasn't a perfect square, you would have had two possible slopes, which is also a valid possibility in other problems!

Question

A line through $A(-5, -4)$ meets the line $x + 3y + 2 = 0$, $2x + y + 4 = 0$ and $x - y - 5 = 0$ at the points $B, C$ and $D$ respectively. If $(15/AB)^2 + (10/AC)^2 = (6/AD)^2$, find the equation of the line.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving a problem; we are uncovering a hidden symmetry. Imagine you are standing at point $A(-5, -4)$.

Around you, like silent sentinels, stand three lines: $L_1$, $L_2$, and $L_3$. A mysterious line cuts through your position, piercing these sentinels at points $B$, $C$, and $D$.

We are given the condition: 
$$\left( \frac{15}{AB} ight)^2 + \left( \frac{10}{AC} ight)^2 = \left( \frac{6}{AD} ight)^2$$

At first glance, this looks like a nightmare of distance formulas and radical signs. But pause. In JEE Advanced, the most intimidating equations often hide the most elegant simplifications. We are going to use the **Parametric Tool**.

### The Parametric Revelation

Why do we avoid the standard $y = mx + c$? Because it forces us to find the coordinates of $B$, $C$, and $D$, which is a trap.

Instead, let us define our line by its angle $	heta$ with the positive $x$-axis. Any point on this line at a distance $r$ from $A$ can be written as:
$$x = -5 + r \cos 	heta, \quad y = -4 + r \sin 	heta$$

For the first line, $x + 3y + 2 = 0$, we substitute our parametric coordinates:
$$(-5 + AB \cos 	heta) + 3(-4 + AB \sin 	heta) + 2 = 0$$

Look at the constants: $-5 - 12 + 2 = -15$. Moving it to the right, we get $AB(\cos 	heta + 3 \sin 	heta) = 15$. Suddenly, the distance $AB$ is isolated:
$$\frac{15}{AB} = \cos 	heta + 3 \sin 	heta$$

### The Algebraic Dance

We repeat this process for the other two lines. For the second line, $2x + y + 4 = 0$, the substitution yields:
$$2(-5 + AC \cos 	heta) + (-4 + AC \sin 	heta) + 4 = 0$$

The constants $-10 - 4 + 4$ simplify to $-10$. Thus, $AC(2 \cos 	heta + \sin 	heta) = 10$, or:
$$\frac{10}{AC} = 2 \cos 	heta + \sin 	heta$$

Finally, for the third line, $x - y - 5 = 0$, we get:
$$(-5 + AD \cos 	heta) - (-4 + AD \sin 	heta) - 5 = 0$$

The constants $-5 + 4 - 5$ result in $-6$. So, $AD(\cos 	heta - \sin 	heta) = 6$, which gives:
$$\frac{6}{AD} = \cos 	heta - \sin 	heta$$

### The Trigonometric Climax

Now, we substitute these expressions into our given condition:
$$(\cos 	heta + 3 \sin 	heta)^2 + (2 \cos 	heta + \sin 	heta)^2 = (\cos 	heta - \sin 	heta)^2$$

Expand the squares. The left side becomes:
$$(\cos^2 	heta + 9 \sin^2 	heta + 6 \sin 	heta \cos 	heta) + (4 \cos^2 	heta + \sin^2 	heta + 4 \sin 	heta \cos 	heta)$$

Combining these, we get $5 \cos^2 	heta + 10 \sin^2 	heta + 10 \sin 	heta \cos 	heta$. The right side is simply $\cos^2 	heta + \sin^2 	heta - 2 \sin 	heta \cos 	heta$.

Bringing everything to one side, we arrive at:
$$4 \cos^2 	heta + 9 \sin^2 	heta + 12 \sin 	heta \cos 	heta = 0$$

This is the moment of truth. Look at the coefficients: $4$ is $2^2$, $9$ is $3^2$, and $12$ is $2(2)(3)$. It is a perfect square:
$$(2 \cos 	heta + 3 \sin 	heta)^2 = 0$$

This implies $2 \cos 	heta + 3 \sin 	heta = 0$, or $	an 	heta = -\frac{2}{3}$.

### The Final Victory

We have found the slope $m = -\frac{2}{3}$. With the point $A(-5, -4)$ and the slope, we use the point-slope form:
$$y - (-4) = -\frac{2}{3}(x - (-5))$$

Multiplying by $3$, we get $3y + 12 = -2x - 10$. Rearranging, we arrive at the final, beautiful equation:
**$2x + 3y + 22 = 0$**

You have navigated the geometry, mastered the parametric form, and conquered the trigonometry. This is the essence of JEE Advanced—not just calculating, but seeing the structure beneath the chaos.

Animated Solution for Mathematics - Straight Lines: A line through $A (- 5, - 4)$ meets the line $x + 3 y + 2 = 0$ , $2 x + y + 4 = 0$ and $x - y - 5 = 0$ at the points $B, C$ and $D$ respectively. If $(15/ A B)^{2} + (10/ A C)^{2} = (6/ A D)^{2}$ , find the equation of the line.

Visualized Solution

Visualizing the Geometry

The Parametric Tool

Finding Distance $A B$

Finding Distance $A C$

Finding Distance $A D$

Applying the Condition

Expanding the Squares

Simplifying the Equation

Identifying the Perfect Square

Finding the Slope

The Final Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Parametric Revelation

The Algebraic Dance

The Trigonometric Climax

The Final Victory

Similar Questions

The equations of two sides $A B$ and $A C$ of a triangle $A B C$ are $4 x + y = 14$ and $3 x - 2 y = 5$ , respectively. The point $(2, - 4/3)$ divides the third side $B C$ internally in the ratio $2 : 1$ . The equation of the side $B C$ is

The straight lines $l_{1}$ and $l_{2}$ pass through the origin and trisect the line segment of the line $L : 9 x + 5 y = 45$ between the axes. If $m_{1}$ and $m_{2}$ are the slopes of the lines $l_{1}$ and $l_{2}$ , then the point of intersection of the line $y = (m_{1} + m_{2}) x$ with $L$ lies on

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

A straight line $L$ is perpendicular to the line $5 x - y = 1$ . The area of the triangle formed by the line $L$ and the coordinate axes is 5. Find the equation of the line $L$ .

A line cuts the x-axis at $A (7, 0)$ and the y-axis at $B (0, - 5)$ . A variable line $P Q$ is drawn perpendicular to $A B$ cutting the x-axis in $P$ and the y-axis in $Q$ . If $A Q$ and $B P$ intersect at $R$ , find the locus of $R$ .

If in a parallelogram $A B D C$ , the coordinates of $A$ , $B$ and $C$ are respectively $(1, 2)$ , $(3, 4)$ and $(2, 5)$ then the equation of the diagonal $A D$ is:

A straight line through the point $A (3, 4)$ is such that its intercept between the axes is bisected at $A$ . Its equation is

The vertices of a triangle are $A (- 1, - 7), B (5, 1)$ and $C (1, 4)$ . The equation of the bisector of the angle $∠ A B C$ is .........

Slope of a line passing through $P (2, 3)$ and intersecting the line, $x + y = 7$ at a distance of 4 units from $P$ , is

If a straight line passing through the point $P (- 3, 4)$ is such that its intercepted portion between the coordinate axes is bisected at $P$ , then its equation is :