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JEE Main 2023
Straight Lines
Straight Lines and Pair of Straight Lines
Hard

Question

Let a variable line of slope m>0m>0 passing through the point (4,9)(4,-9) intersect the coordinate axes at the points AA and BB. The minimum value of the sum of the distances of AA and BB from the origin is

Options

Solution

Key Concepts and Formulas

  • Equation of a Straight Line (Point-Slope Form): A line passing through the point (x1,y1)(x_1, y_1) with a slope mm can be represented as yy1=m(xx1)y - y_1 = m(x - x_1).
  • Intercepts with Coordinate Axes: The x-intercept is the point where the line crosses the x-axis (where y=0y=0), and the y-intercept is the point where the line crosses the y-axis (where x=0x=0).
  • AM-GM Inequality: For non-negative real numbers aa and bb, a+b2ab\frac{a+b}{2} \ge \sqrt{ab}. Equality holds when a=ba=b.

Step-by-Step Solution

Step 1: Formulate the Equation of the Line and Determine Intercepts

First, we need to find the equation of the line and then determine the coordinates of points AA and BB where it intersects the axes.

  • Equation of the Line: The line passes through the point (x1,y1)=(4,9)(x_1, y_1) = (4, -9) and has a slope mm. Using the point-slope form, the equation of the line is: y(9)=m(x4)y - (-9) = m(x - 4) y+9=m(x4)y + 9 = m(x - 4) Explanation: This step translates the given information (a point and a slope) into an algebraic equation representing the line.

  • Coordinates of Point A (x-intercept): Point AA lies on the x-axis, so its y-coordinate is 00. Substitute y=0y=0 into the line's equation: 0+9=m(xA4)0 + 9 = m(x_A - 4) 9=m(xA4)9 = m(x_A - 4) Since m>0m > 0, we can divide by mm: xA4=9mx_A - 4 = \frac{9}{m} xA=4+9mx_A = 4 + \frac{9}{m} So, point AA is (4+9m,0)\left(4 + \frac{9}{m}, 0\right). Explanation: By setting y=0y=0, we find the x-coordinate where the line crosses the x-axis. Since m>0m>0, 4+9m4 + \frac{9}{m} will always be positive, meaning AA is on the positive x-axis.

  • Coordinates of Point B (y-intercept): Point BB lies on the y-axis, so its x-coordinate is 00. Substitute x=0x=0 into the line's equation: yB+9=m(04)y_B + 9 = m(0 - 4) yB+9=4my_B + 9 = -4m yB=94my_B = -9 - 4m So, point BB is (0,94m)(0, -9 - 4m). Explanation: By setting x=0x=0, we find the y-coordinate where the line crosses the y-axis. Since m>0m>0, 4m-4m is negative, making 94m-9 - 4m always negative. This means BB is on the negative y-axis.

Step 2: Express the Sum of Distances from the Origin in Terms of mm

We need to find the sum of the distances of AA and BB from the origin O(0,0)O(0,0). Let this sum be S(m)S(m).

  • Distance OA: OA=4+9mOA = \left|4 + \frac{9}{m}\right| Since m>0m > 0, 4+9m4 + \frac{9}{m} is always positive. OA=4+9mOA = 4 + \frac{9}{m}

  • Distance OB: OB=94mOB = |-9 - 4m| Since m>0m > 0, 94m-9 - 4m is always negative. Therefore, its absolute value is its negation: OB=(94m)=9+4mOB = -(-9 - 4m) = 9 + 4m

  • Sum of Distances S(m): S(m)=OA+OB=(4+9m)+(9+4m)S(m) = OA + OB = \left(4 + \frac{9}{m}\right) + (9 + 4m) S(m)=13+4m+9mS(m) = 13 + 4m + \frac{9}{m} Explanation: Distances are always non-negative. We use the absolute value to ensure this. The expression for S(m)S(m) is now a function of the slope mm, which we need to minimize. The domain for mm is m>0m>0.

Step 3: Optimize the Function S(m)S(m) Using AM-GM Inequality

To find the minimum value of S(m)S(m), we will use the AM-GM inequality on the terms 4m4m and 9m\frac{9}{m} (since m>0m>0, both terms are positive).

4m+9m24m9m\frac{4m + \frac{9}{m}}{2} \ge \sqrt{4m \cdot \frac{9}{m}} 4m+9m2364m + \frac{9}{m} \ge 2\sqrt{36} 4m+9m2×64m + \frac{9}{m} \ge 2 \times 6 4m+9m124m + \frac{9}{m} \ge 12

The equality holds when 4m=9m4m = \frac{9}{m}, which gives 4m2=9m=324m^2 = 9 \Rightarrow m = \frac{3}{2}.

Thus, the minimum value of 4m+9m4m + \frac{9}{m} is 1212.

Explanation: AM-GM provides an elegant way to find the minimum of sums of positive terms, especially when their product is constant. It confirms the critical value of mm and the minimum value of the variable part of S(m)S(m).

Step 4: Calculate the Minimum Value of S(m)S(m)

Substitute the optimal value of 4m+9m=124m + \frac{9}{m} = 12 back into the expression for S(m)S(m).

S(m)=13+4m+9m=13+12=25S(m) = 13 + 4m + \frac{9}{m} = 13 + 12 = 25

Therefore, the minimum value of the sum of the distances of AA and BB from the origin is 2525.

Common Mistakes & Tips

  • Absolute Values for Distances: Always remember that distances are non-negative.
  • Domain of mm: The condition m>0m > 0 is crucial.
  • AM-GM Applicability: Recognize when AM-GM can be used for quick optimization.

Summary

This problem demonstrates a classic application of coordinate geometry and the AM-GM inequality. The key steps involve: setting up the equation of the line, finding the x and y intercepts in terms of the slope mm, formulating the total distance function S(m)S(m), and minimizing S(m)S(m) using the AM-GM inequality.

The minimum sum of distances is 25\boxed{25}, which corresponds to option (D). The final answer is \boxed{25}.

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