Problem 33 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A line through \(A(-5,-4)\) meets the lines \(x+3 y+2=0\), \(2 x+y+4=0\) and \(x-y-5=0\) at the point \(B, C\) and \(D\), respectively. If \(\left(\frac{15}{A B}\right)^{2}+\left(\frac{10}{A C}\right)^{2}=\left(\frac{6}{A D}\right)^{2}\), the equa- tion of the line is (A) \(2 x+3 y+22=0\) (B) \(2 x-3 y+22=0\) (C) \(3 x+2 y+22=0\) (D) \(3 x-2 y+22=0\)

Step-by-Step Solution

Answer

The equation of the line is \(2x + 3y + 22 = 0\).

Step 1: Find the general equation of a line

Let's assume the equation of the line through point \(A(-5,-4)\) is \(y + 4 = m(x + 5)\), where \(m\) is the slope. Simplify this equation to express it in standard form:

Step 2: Express intersection points with given lines

Determine the intersection points (\(B\), \(C\), and \(D\)) of the line with the given lines:1. For \(x + 3y + 2 = 0\): Substitute \(y = mx + b\) in the equation to find \(x_B\) and \(y_B\).2. Repeat for \(2x + y + 4 = 0\) to find \(x_C\) and \(y_C\).3. Do the same for \(x - y - 5 = 0\) to find \(x_D\) and \(y_D\).Substitute these values back to find \(B\), \(C\), and \(D\).

Step 3: Calculate distances AB, AC, and AD

Calculate the distance of each intersection from point \(A(-5,-4)\) using the distance formula:\[ AB = \sqrt{(x_B + 5)^2 + (y_B + 4)^2} \]\[ AC = \sqrt{(x_C + 5)^2 + (y_C + 4)^2} \]\[ AD = \sqrt{(x_D + 5)^2 + (y_D + 4)^2} \]

Step 4: Solve the given equation

Substitute the distances from Step 3 into the equation \[ \left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right)^2 \]Solve the equation for the slope \(m\) by expressing \(AB, AC, AD\) in terms of \(m\) and solving the resulting quadratic equation.

Step 5: Derive equation of the line

Once \(m\) is identified, substitute it back into the line equation from Step 1 to find the standard form of the equation. Compare this with the choices provided:- Multiply and simplify the equation to a recognizable form- Match with option \((A)\), \((B)\), \((C)\), or \((D)\)

Key Concepts

Equation of a Line

An equation of a line represents all the points on the line in a mathematical format. For a line passing through a point like \(A(-5, -4)\), it can be derived using the slope-intercept form. This is given as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the known point.
The slope \(m\) explains how slanted the line is, and in this exercise, we start by assuming \(y + 4 = m(x + 5)\). Simplifying this equation can express it in the standard format, \(Ax + By + C = 0\).

• Ensures you know the coordinates of a known point.
• Simplify to standard form, essential for comparison.

Understanding this equation is crucial, as it links the various components needed for further calculations.

Distance Formula

The distance formula helps to determine the length between two points, say \((x_1, y_1)\) and \((x_2, y_2)\). The formula is:\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
The given exercise used this formula to find distances \(AB, AC,\) and \(AD\) from point \(A(-5, -4)\) to points \(B, C,\) and \(D\). Calculating these distances is crucial for solving the relation \[\left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right)^2\]

• Substitute the coordinates to get the lengths.
• Make sure the calculations are precise for correctness.

This provides the groundwork needed to solve for the slope \(m\) and move toward finding the line equation.

Intersection Points

Intersection points are where two lines cross each other. To find these points, substitute the line equation into the given line equations like \(x + 3y + 2 = 0\) to solve for coordinates \(B, C,\) and \(D\).

• This involves solving simultaneous equations.
• The calculated intersection points serve as coordinates for distance calculations.

By determining these intersections accurately, you can proceed to compute various distances from point \(A\) to these intersections, forming an equation involving these measures.

Slope

The slope, represented by \(m\), is a key component as it affects the angle at which a line tilts. It's essentially the 'rise over run,' or change in \(y\) relative to change in \(x\).
The slope helps in formulating the line equation and in this context, is determined from simultaneous equations. Using the distance relationship:\[\left(\frac{15}{AB}\right)^2 + \left(\frac{10}{AC}\right)^2 = \left(\frac{6}{AD}\right)^2\]
This formula is vital to rearrange and solve for \(m\).

• A positive slope means the line rises with increasing \(x\).
• Finding the right \(m\) leads directly to the correct line equation.

Determining \(m\) accurately is the primary step in composing the final equation.

Quadratic Equation

Quadratic equations are part of the challenge when solving for the slope \(m\). These equations take the form \(ax^2 + bx + c = 0\) and require factoring, completing the square, or using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In this exercise, solving the derived equation might involve rearranging terms into a quadratic. This is due to expressing distances and other parameters into functions of \(m\), leading to solutions that help identify the correct equation of the line.

• Use the quadratic formula when direct methods are cumbersome.
• Simplify and verify the selected solutions.

By establishing \(m\) as roots of such an equation, the task simplifies to forming the complete correct line equation.