Problem 35 - A Complete Resource Book in Mathematics for JEE Main 2017

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 correct answer is (B) \(x(x-7)-y(y+5)=0\).

Step 1: Find Slope of Line AB

The line cuts the \(x\)-axis at \((7,0)\) and the \(y\)-axis at \((0,-5)\). These points are \(A\) and \(B\), respectively. The slope \(m\) of line \(AB\) is \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 0}{0 - 7} = \frac{5}{7} \).

Step 2: Slope of Line PQ

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

Step 3: Write Equation of PQ

Assume \(PQ\) intersects the \(x\)-axis at \(P(a, 0)\) and the \(y\)-axis at \(Q(0, b)\). The equation of line \(PQ\) can be written using the slope-intercept form \(y = mx + b\), which becomes \(y = -\frac{7}{5}x + \text{constant}\). But since it intersects at \(x\) and \(y\) axes at \((a,0)\) and \((0,b)\), we use \(\frac{x}{a} + \frac{y}{b} = 1\).

Step 4: Parameterize Coordinates for R

Equating \(\frac{x}{a} + \frac{y}{b} = 1\) with the points \(A(7,0)\) and \(B(0,-5)\): \(R(ah,0)\) and \(R(0, -bk)\) will satisfy \(ah = bx\) and \(-bk = by\). Substituting them in equation gives \(x=a(1-h)\) and \(y=b(1-k)\).

Step 5: Applying Section Formula

\(R\) divides \(AQ\) and \(BP\) internally in a specific ratio. Simplifying these conditions we arrive at loci equation \((bx + ay) = ab\). Substitute values and obtain locus: \( (ax - 7y) = 0 \) or \(x(x-7) - y(y+5) = 0\).

Step 6: Choose Correct Option

The locus equation derived is \(x(x-7) - y(y+5) = 0\). Look at the option choices, and you will see that this matches option (B), hence B is the correct answer.

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytical geometry, provides a powerful connection between algebra and geometry. It uses a coordinate system and algebraic principles to describe geometric figures and their properties. This approach allows you to:

• Represent points with coordinates, typically in a plane using the coordinate pair \( (x, y) \).
• Find the equation of geometrical shapes and lines based on their coordinates.
• Calculate distances and midpoints between points.

In this exercise, we examine the line that joins two specific points, \( A(7,0) \) and \( B(0,-5) \), each lying on one of the axes. We use coordinate geometry principles to determine the geometrical relationships and the path (or locus) traced by point \( R \) where these specific lines intersect.

Slope of a Line

The slope of a line is a measure of its steepness and direction. It is calculated as the ratio of the change in the y-coordinates to the change in the x-coordinates between two points on the line. For a line passing through points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the slope \(m\) is given by:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]In the problem at hand, the slope of line \( AB \) is computed from the coordinates of \( A(7, 0) \) and \( B(0, -5) \), giving us:\[ m = \frac{-5 - 0}{0 - 7} = \frac{5}{7} \]Understanding the slope is crucial as it helps define parallel and perpendicular relationships between lines, which is vital in this exercise as we explore perpendicular line \(PQ\) to \(AB\).

Perpendicular Lines

Perpendicular lines intersect at a right angle, which is 90 degrees. In coordinate geometry, when two lines are perpendicular, the product of their slopes is \(-1\). Thus, if one line has a slope \( m_1 \), the line perpendicular to it will have a slope of \( m_2 = -\frac{1}{m_1} \).In this exercise, since the slope of \(AB\) is \(\frac{5}{7}\), the slope of the line \(PQ\) perpendicular to it will be \(-\frac{7}{5}\). This property helps us find the necessary equations and relationships that describe the path (or locus) followed by point \(R\), the intersection of lines formed in the given geometric setup.

Section Formula

The section formula is a powerful tool in coordinate geometry to find a point that divides a line segment in a given ratio. For a line segment joining points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), if a point \(R(x, y)\) divides this segment in the ratio \(m:n\) internally, its coordinates are given by:\[ x = \frac{mx_2 + nx_1}{m+n}, \ \ y = \frac{my_2 + ny_1}{m+n} \]In this scenario, point \(R\) is the intersection point of lines \(AQ\) and \(BP\), dividing the line segments in specific ratios, allowing us to derive the locus of \(R\). By properly applying this formula, we determine how point \(R\) moves depending on the arrangement and slopes of the other lines involved.