A line is drawn through the point (1, 2) to meet the coordinate axes at P and Q such that it forms a

Q: Why do we reject the positive root $m=2$?

Great question! If you plug $m=2$ into our intercept formulas, you get $P(0, 0)$ and $Q(0, 0)$. A triangle with vertices at the origin and two other points that are also the origin is not a triangle at all—it's just a point! We need a non-degenerate triangle, so we must reject $m=2$.

Q: Is there a faster way to solve this without calculus?

Yes! You can use the AM-GM inequality. The area expression $A = \frac{1}{2} (4 - m - \frac{4}{m})$ can be rewritten. Since $m$ is negative, let $m = -k$ where $k > 0$. Then $A = \frac{1}{2} (4 + k + \frac{4}{k})$. By AM-GM, $k + \frac{4}{k} \ge 2\sqrt{k \cdot \frac{4}{k}} = 4$. This confirms the minimum area occurs when $k = \frac{4}{k}$, or $k^2 = 4$, so $k=2$, meaning $m=-2$.

Q: Why does the midpoint trick work?

It's a beautiful property of lines passing through a fixed point. For a line passing through $(x_0, y_0)$ to minimize the area of the triangle formed with the axes, the point $(x_0, y_0)$ must be the midpoint of the segment $PQ$. This is a standard result in coordinate geometry that saves precious time in competitive exams.

Question

A line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $OPQ$, where $O$ is the origin. If the area of the triangle $OPQ$ is least, then the slope of the line $PQ$ is

Options:

(A) -1/4

(B) -4

(C) -2

(D) -1/2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We start by anchoring our line. A line passing through $A(1, 2)$ with an unknown slope $m$ is defined by the point-slope form:
$$y - 2 = m(x - 1)$$

To form a triangle with the axes, this line must intersect the x-axis at point $P$ and the y-axis at point $Q$. To find $P$, we set $y = 0$:
$$0 - 2 = m(x - 1) \ \Rightarrow x = 1 - \frac{2}{m}$$
Thus, the x-intercept is $P = (1 - \frac{2}{m}, 0)$.

Similarly, for $Q$, we set $x = 0$:
$$y - 2 = m(0 - 1) \ \Rightarrow y = 2 - m$$
Thus, the y-intercept is $Q = (0, 2 - m)$. We have successfully translated the geometric constraint into algebraic coordinates.

### The Area Function

The area $A$ of the right-angled triangle $	riangle OPQ$ is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Substituting our intercepts, we get:
$$A = \frac{1}{2} \left( 1 - \frac{2}{m} ight) (2 - m)$$

Let's expand this expression to analyze the function:
$$A = \frac{1}{2} \left[ 2 - m - \frac{4}{m} + 2 ight] = \frac{1}{2} \left[ 4 - m - \frac{4}{m} ight]$$

This is the function we need to minimize. Note that for the triangle to exist in the first quadrant, the intercepts must be positive, which implies we must choose a **negative slope** $m$.

### The Calculus Approach

To find the minimum, we differentiate $A$ with respect to $m$:
$$\frac{dA}{dm} = \frac{1}{2} \left[ 0 - 1 - 4(-1)m^{-2} ight] = \frac{1}{2} \left[ -1 + \frac{4}{m^2} ight]$$

Setting $\frac{dA}{dm} = 0$ gives us:
$$-1 + \frac{4}{m^2} = 0 \ \Rightarrow m^2 = 4 \ \Rightarrow m = \pm 2$$

Since $m=2$ would not form a triangle in the first quadrant, we must choose **$m = -2$**. The negative slope confirms that the line must tilt downwards to enclose a region in the first quadrant.

### The Elegant Shortcut: The Midpoint Trick

There is a powerful theorem in coordinate geometry: for a line passing through a fixed point $(x_0, y_0)$ to minimize the area of the triangle formed with the axes, the fixed point must be the **midpoint** of the segment $PQ$.

If $(1, 2)$ is the midpoint of $PQ$, then the intercepts must be $P(2, 0)$ and $Q(0, 4)$. The slope is then calculated as:
$$m = \frac{4 - 0}{0 - 2} = -2$$

This confirms our previous result using a much faster method. Understanding these underlying structures allows you to solve complex problems with **maximum efficiency**.

Animated Solution for Mathematics - Straight Lines: A line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $O P Q$ , where $O$ is the origin. If the area of the triangle $O P Q$ is least, then the slope of the line $P Q$ is

Select Answer:

Visualized Solution

Coordinate System & Fixed Point

Forming $△ O P Q$

Equation of Line $P Q$

Coordinates of Point $P$

Coordinates of Point $Q$

Area of $△ O P Q$

Simplifying the Area

Minimizing the Area

Calculating $\frac{d A}{d m}$

Equating Derivative to Zero

Choosing the Correct $m$

Pro-Tip: The Midpoint Trick

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Area Function

The Calculus Approach

The Elegant Shortcut

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