Problem 66 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The foot of the perpendicular drawn from the origin, on the line, \(3 x+y=\lambda(\lambda \neq 0)\) is \(P\). If the linemeets \(x\)-axisat \(A\) and \(y\)-axis at \(B\), then the ratio \(B P: P A\) is [Online April \(15, \mathbf{2 0 1 8}]\) (a) \(9: 1\) (b) \(1: 3\) (c) \(1: 9\) (d) \(3: 1\)

Step-by-Step Solution

Answer

The ratio \(BP:PA\) is \(3:1\).

Step 1: Finding Intercepts of the Line

The given line equation is \(3x + y = \lambda\). To find where it meets the x-axis, set \(y = 0\), which gives the x-intercept \(A(\frac{\lambda}{3}, 0)\). Then, to find the y-intercept, set \(x = 0\), yielding \(y = \lambda\), so the point \(B(0, \lambda)\).

Step 2: Foot of the Perpendicular from the Origin

The given line's general equation is \(3x + y = \lambda\). A perpendicular from the origin will have a slope \(-\frac{1}{3}\) (negative reciprocal of line's slope). The equation of the perpendicular line from the origin is \(y = -\frac{1}{3}x\).

Step 3: Finding the Foot of Perpendicular

Solve the two simultaneous equations: \(3x + y = \lambda\) and \(y = -\frac{1}{3}x\). Substitute, giving \(3x - \frac{1}{3}x = \lambda\). Simplifying, \(\frac{10}{3}x = \lambda\), so \(x = \frac{3\lambda}{10}\). Substitute back into \(y = -\frac{1}{3}x\), obtaining \(y = -\frac{\lambda}{10}\). Thus, \(P\left( \frac{3\lambda}{10}, -\frac{\lambda}{10} \right)\).

Step 4: Calculating Distances PA and PB

From the points, \(P\left( \frac{3\lambda}{10}, -\frac{\lambda}{10} \right)\), \(A\left( \frac{\lambda}{3}, 0 \right)\), and \(B(0, \lambda)\), use the distance formula. For \(PA\), modify it into \(\sqrt{\left(\frac{\lambda}{3} - \frac{3\lambda}{10}\right)^2 + (0 + \frac{\lambda}{10})^2}\). This gives \(PA = \frac{\lambda}{6}\). For \(PB\), it's \(\sqrt{\left(0 - \frac{3\lambda}{10}\right)^2 + \left(\lambda + \frac{\lambda}{10}\right)^2}\), simplifying to \(PB = \frac{\lambda}{2}\).

Step 5: Finding the Ratio BP : PA

With \(PB = \frac{\lambda}{2}\) and \(PA = \frac{\lambda}{6}\), the ratio \(BP:PA = \frac{\lambda/2}{\lambda/6} = 3:1\).

Key Concepts

Perpendicular Foot

The concept of the perpendicular foot involves finding the point on a line where a perpendicular segment from a given point, often the origin (0,0), meets the line. It is a very common concept in coordinate geometry and helps in solving various geometric problems.
To determine the perpendicular foot, we first need to find the slope of the given line. For a line like \(3x + y = \lambda\), rewrite it in slope-intercept form, \(y = -3x + \lambda\), showing that the slope is \(-3\).
The perpendicular line from the origin will have a slope that is the negative reciprocal of the original line. In this case, the perpendicular slope is \(-\frac{1}{3}\), because multiplying \(-3\) by \(-\frac{1}{3}\) results in a slope of \(1\), the condition for perpendicular lines.
The equation of the line with this slope passing through the origin (point \((0,0)\)) is \(y = -\frac{1}{3}x\). Solving this together with the original line equation, \(3x + y = \lambda\), provides us the coordinates of the perpendicular foot \(P\). This intersection gives a systematic point \(P\), in our example \(\left( \frac{3\lambda}{10}, -\frac{\lambda}{10} \right)\).

Intercepts

Intercepts are the points where a line crosses the coordinate axes, namely the x-axis and y-axis. Understanding intercepts is crucial in coordinate geometry, as it helps describe the line's position and shape.
Finding the x-intercept involves setting \(y = 0\) in the line equation. From the line equation \(3x + y = \lambda\), by setting \(y\) to zero, we can solve for \(x\) and find the x-intercept, \(A\), at \(\left( \frac{\lambda}{3}, 0 \right)\).
Similarly, finding the y-intercept involves setting \(x = 0\). By substituting \(x = 0\) into the equation \(3x + y = \lambda\), we find \(y = \lambda\), giving the y-intercept, \(B\), at \((0, \lambda)\).
These intercepts provide significant points where the line crosses the axes, forming the basis for further geometric constructions, such as drawing a triangle and determining intercepts related ratios.

Distance Formula

The distance formula is a fundamental tool in coordinate geometry used to calculate the distance between two points in Cartesian coordinates. The formula is rooted in the Pythagorean theorem and is given by the expression:

• \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Here, \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points in question.
To find distances such as between points \(P\left( \frac{3\lambda}{10}, -\frac{\lambda}{10} \right)\) and \(A\left( \frac{\lambda}{3}, 0 \right)\), we substitute into the formula. Calculate the differences in x-values and y-values, square these differences, sum them, and take the square root. For these points specifically, this results in \(PA = \frac{\lambda}{6}\).
For another pair, such as \(P\left( \frac{3\lambda}{10}, -\frac{\lambda}{10} \right)\) and \(B(0, \lambda)\), follow the same method to find the distance \(PB = \frac{\lambda}{2}\). These calculations allow us to compare distances, crucial for problems involving ratios or optimizations.