A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordina

Q: Wait, why do we assume the slope $m$ is negative? Could it be positive?

That is a fantastic observation! If the slope were positive, the line would pass through the first quadrant but would either not intersect the positive axes at all (if it passes through the origin) or would intersect the negative axes. The problem explicitly states it cuts the *positive* coordinate axes. For a line to pass through $(8, 2)$ and hit the positive $x$ and $y$ axes, it must lean to the left, which forces the slope $m$ to be strictly negative. Always let the geometry dictate your constraints!

Q: Can I solve this using derivatives instead of AM-GM?

Absolutely! You could define $S(m) = 10 - \frac{2}{m} - 8m$, differentiate with respect to $m$, set $S'(m) = 0$, and solve for $m$. You would get the same answer. However, in the JEE Advanced exam, time is your most precious currency. AM-GM allows you to bypass the calculus machinery entirely, turning a potentially messy derivative problem into a two-line algebraic solution. Master the shortcut, but keep the calculus as your backup plan.

Q: Why are we allowed to use AM-GM? Isn't it only for positive numbers?

You have hit the nail on the head! You are absolutely right—AM-GM is strictly for positive numbers. This is exactly why we checked the signs in step 7. Since $m < 0$, the term $-\frac{2}{m}$ is positive, and $-8m$ is also positive. Because both terms are positive, the inequality holds perfectly. Never apply AM-GM blindly; always verify the positivity of your terms first!

Question

A straight line $L$ with negative slope passes through the point $(8, 2)$ and cuts the positive coordinate axes at points $P$ and $Q$. Find the absolute minimum value of $OP + OQ$, as $L$ varies, where $O$ is the origin.

Sigma Percentile · Accepted Answer

### The Geometry of the Pivot

Imagine you are standing in a coordinate plane. You have a fixed anchor point, $A(8, 2)$, sitting comfortably in the first quadrant.

Now, imagine a straight line $L$ passing through this point. This line is not static; it is a pivot that rotates around $A$.

It has a constraint: it must cut the positive $x$-axis at point $P$ and the positive $y$-axis at point $Q$. As you rotate this line, the points $P$ and $Q$ slide along the axes. Our goal is to find the configuration where the sum of these distances, $OP + OQ$, is at its absolute minimum.

### Defining the Line

To capture this motion mathematically, we use the point-slope form of a line. We know the line passes through $(8, 2)$ and has a slope $m$.

Since the line must cut the positive axes, it must lean to the left, which tells us immediately that $m < 0$. The equation of our line is given by:

$$(y - 2) = m(x - 8)$$

To find the $x$-intercept $P$, we set $y = 0$. The equation becomes $-2 = m(x - 8)$, which simplifies to $x = 8 - \frac{2}{m}$. This is the distance $OP$.

Similarly, to find the $y$-intercept $Q$, we set $x = 0$, yielding $y = 2 - 8m$. This is the distance $OQ$. We now have our objective function:

$$S = OP + OQ = \left(8 - \frac{2}{m}ight) + (2 - 8m)$$

### The Algebraic Insight

Let us group the terms:

$$S = 10 + \left(-\frac{2}{m} - 8might)$$

We are looking for the minimum of this sum. Notice the structure of the variable part: $\left(-\frac{2}{m} - 8might)$. This is the classic signature of the **Arithmetic Mean-Geometric Mean (AM-GM) inequality**.

Before we proceed, we must respect the rules. AM-GM requires the terms to be positive. Since $m < 0$, let us look at our terms: $-\frac{2}{m}$ is positive because a negative divided by a negative is positive. Similarly, $-8m$ is positive because a negative times a negative is positive.

### The AM-GM Masterclass

We apply the inequality: $\frac{x + y}{2} \ge \sqrt{xy}$. Let $x = -\frac{2}{m}$ and $y = -8m$.

The arithmetic mean is $\frac{-\frac{2}{m} - 8m}{2}$. The geometric mean is:

$$\sqrt{\left(-\frac{2}{m}ight) 	imes (-8m)} = \sqrt{16} = 4$$

Look at the beauty of this cancellation! The $m$ in the numerator and the $m$ in the denominator vanish.

So, we have:

$$\frac{-\frac{2}{m} - 8m}{2} \ge 4$$

Multiplying by $2$, we find that the variable part $-\frac{2}{m} - 8m \ge 8$. The minimum value of this variable component is exactly $8$.

### The Final Result

We return to our total sum: $S = 10 + \left(-\frac{2}{m} - 8might)$. Since the minimum of the variable part is $8$, the minimum of the total sum is:

$$S_{min} = 10 + 8 = 18$$

There we have it. By visualizing the geometry and applying the elegance of AM-GM, we have tamed the variable slope. The absolute minimum value of $OP + OQ$ is **18**.

Animated Solution for Mathematics - Straight Lines: A straight line $L$ with negative slope passes through the point $(8, 2)$ and cuts the positive coordinate axes at points $P$ and $Q$ . Find the absolute minimum value of $O P + O Q$ , as $L$ varies, where $O$ is the origin.

Enter Numerical Value:

Visualized Solution

Visualizing the Fixed Point

Drawing the Line $L$

Equation of Line $L$

Finding the $x$ -intercept $P$

Finding the $y$ -intercept $Q$

Defining the Objective Function

Formulating the Sum $S$

Analyzing the Variable Terms

Applying AM-GM Inequality

Executing AM-GM

Minimum Value of the Variable Part

Final Minimum Value of $O P + O Q$

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of the Pivot

Defining the Line

The Algebraic Insight

The AM-GM Masterclass

The Final Result

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