Problem 35 - A Complete Resource Book in Mathematics for JEE Main

Question

A line cuts the \(x\)-axis at \(A(7,0)\) and \(y\)-axis at \(B(0,-5)\). A variable line \(P Q\) is drawn \(\perp\) to \(A B\) cutting the \(x\)-axis in \(P\) and the \(y\)-axis in \(Q .\) If \(A Q\) and \(B P\) intersect at \(R\), then the locus of \(R\) is (A) \(x(x-7)+y(y+5)=0\) (B) \(x(x-7)-y(y+5)=0\) (C) \(x(x+7)+y(y-5)=0\) (D) none of these

Step-by-Step Solution

Answer

The locus of R is (A) \(x(x-7)+y(y+5)=0\).

Step 1: Find the Equation of Line AB

Line AB connects points \(A(7,0)\) and \(B(0,-5)\). First, calculate the slope of AB: \(m = \frac{0-(-5)}{7-0} = \frac{5}{7}\). Therefore, the equation of line AB is \(y - 0 = \frac{5}{7}(x - 7)\) which simplifies to \(y = \frac{5}{7}x - 5\).

Step 2: Determine Slope of Perpendicular Line PQ

Since line PQ is perpendicular to line AB, its slope is the negative reciprocal of the slope of AB. Thus, the slope of PQ is \(-\frac{7}{5}\).

Step 3: Establish the Equation of Line PQ

Choose arbitrary points \(P(p,0)\) on the x-axis and \(Q(0,q)\) on the y-axis. Using the slope \(-\frac{7}{5}\), the equation for line PQ is \(y - 0 = -\frac{7}{5}(x - p)\) which simplifies to \(y = -\frac{7}{5}x + \frac{7}{5}p\).

Step 4: Find Intersection Point R of AQ and BP

For line AQ, connecting points \(A(7,0)\) and \(Q(0,q)\), we calculate the slope \( \frac{q - 0}{0 - 7} = -\frac{q}{7}\). The equation is \(y = -\frac{q}{7}x + q\). Likewise for line BP connecting \(B(0,-5)\) and \(P(p,0)\), with slope \(\frac{0 + 5}{p - 0} = \frac{5}{p}\), equation is \(y = \frac{5}{p}x - 5\).

Step 5: Solve and Eliminate \(p\) and \(q\) in R

Set \(-\frac{q}{7}x+q = \frac{5}{p}x-5\). Multiply throughout to eliminate fractions, \(pq x - 7x - 7q + 5p = 0\). The expression for \(x, y\) in terms of \(R(h,k)\): \(xh = 7h + 5k\) and \(y(-k) = -5k - 7h\). Upon simplifying, we find after arranging, \(x(x-7) + y(y+5) = 0\).

Step 6: Identify the Locus and Choose Answer

Analyze the derived expression for the locus of \(R(h,k)\). The final equation is \(x(x-7) + y(y+5) = 0\). Comparing with the options provided, the correct choice is (A).

Key Concepts

Line Equation

In analytical geometry, understanding the concept of a line equation is crucial. A line equation represents all the points that form the line in a coordinate system.

When working with a line that connects two points, like points \(A(7,0)\) and \(B(0,-5)\), we first find the slope \(m\), using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

For this line, \(m = \frac{0 - (-5)}{7 - 0} = \frac{5}{7}\).

The slope is then used to write the equation of the line, which is in the form \(y - y_1 = m(x - x_1)\). Substituting the known values, we get: \[ y - 0 = \frac{5}{7}(x - 7) \] which simplifies to: \[ y = \frac{5}{7}x - 5 \].

This equation tells us the relationship between any point \((x,y)\) on the line and allows us to predict \(y\) for any \(x\) value along the line.

Locus

The term "locus" in geometry refers to a set of points that satisfy a particular condition or a set of conditions. In this problem, we are asked to find the locus of point \(R\), which is formed when lines \(AQ\) and \(BP\) intersect.

To determine the locus, we need to eliminate the parameters \(p\) and \(q\) from the equations of lines \(AQ\) and \(BP\). This way, we can derive a new equation that directly relates to the coordinates of point \(R(h,k)\).

In this case, we derived the equation: \[ x(x-7) + y(y+5) = 0 \].

This equation represents all possible positions of \(R\) on the plane based on our stated conditions, without needing \(p\) and \(q\). Solving such types of problems often involves finding these fundamental relationships, unveiling the entire set of possible solutions.

Slope

The slope is a measure of how steep a line is. It is crucial in the context of line equations and understanding perpendicular lines.

The slope is calculated as the "rise over run," or the change in \(y\) over the change in \(x\) between two points on the line. For example, the slope of line \(AB\) was \(\frac{5}{7}\).

When dealing with perpendicular lines, the slope of one line is -1 divided by the slope of the other. In mathematical terms, if you have two perpendicular lines with slopes \(m_1\) and \(m_2\), then \(m_1 \times m_2 = -1\). In our exercise, since line \(PQ\) is perpendicular to line \(AB\), its slope is \(-\frac{7}{5}\).

Being comfortable with slopes is essential for analyzing and graphing lines, especially when you're tasked with finding the intersection of various lines.

Intersecting Lines

Intersecting lines are lines that meet at a common point. In the realm of coordinate geometry, finding the intersection point involves solving equations simultaneously.

In the given problem, lines \(AQ\) and \(BP\) intersect at point \(R\). The key to finding \(R\) is to set the equations of these two lines equal to each other and solve for \(x\) and \(y\).

Here’s how it works: From line \(AQ\), \(y = -\frac{q}{7}x + q\), and from line \(BP\), \(y = \frac{5}{p}x - 5\). Set these equations equal:

\(-\frac{q}{7}x + q = \frac{5}{p}x - 5\).

Solving this will ultimately help eliminate \(p\) and \(q\), unveiling the coordinates of \(R\) and forming the locus equation.

Understanding intersecting lines not only helps in finding specific points but also in understanding how different geometric constructs relate to each other in a coordinate plane.