Question

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

• A. 25
• B. 30
• C. 15
• D. 10

Answer 1

The Correct Option is A

The equation of the line is:

\[ y + 9 = m(x - 4). \]

Find \( A \) and \( B \) (intersection points with the axes):

At \( y = 0 \), \( x = \frac{9}{m} + 4 \implies A\left(\frac{9}{m} + 4, 0\right) \).

At \( x = 0 \), \( y = -9 - 4m \implies B(0, -9 - 4m) \).

The sum of distances:

\[ OA + OB = \sqrt{\left(\frac{9}{m} + 4\right)^2} + \sqrt{(-9 - 4m)^2}. \]

Using AM-GM inequality, the minimum value occurs when:

\[ m = \frac{3}{2}. \]

Substitute \( m \) to get:

\[ OA + OB = 25. \]

Answer 2

To solve this problem, first note that we need to find the points \( A \) and \( B \) where the line of slope \( m > 0 \), passing through the point \( (4, -9) \), intersects the coordinate axes.

The equation of a line in point-slope form is given by:

\(y - y_1 = m(x - x_1)\)

Given point is \( (4, -9) \), and slope \( m \). So, the equation of the line becomes:

\(y + 9 = m(x - 4)\)

Rearranging this gives:

\(y = mx - 4m - 9\)

To find the x-intercept (point A), set \( y = 0 \):

\(0 = mx - 4m - 9\)

Solving for \( x \),

\(mx = 4m + 9\) \(x = \frac{4m + 9}{m}\)

So, point \( A \) is:

\(A \left(\frac{4m+9}{m}, 0\right)\)

To find the y-intercept (point B), set \( x = 0 \):

\(y = m(0) - 4m - 9\) \(y = -4m - 9\)

So, point \( B \) is:

\(B (0, -4m-9)\)

The distances from the origin to the points \( A \) and \( B \) are:

Distance \( OA \) is:

\(OA = \sqrt{\left(\frac{4m+9}{m}\right)^2 + 0^2} = \frac{4m+9}{m}\)

Distance \( OB \) is:

\(OB = \sqrt{0^2 + (-4m-9)^2} = |4m + 9|\)

Since \( m > 0 \), \( |4m + 9| = 4m + 9 \).

The sum of these distances is:

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

To minimize \( S \), differentiate it with respect to \( m \) and set the derivative to zero:

\(\frac{dS}{dm} = \frac{d}{dm}\left(\frac{4}{m} + \frac{9}{m} + 4m + 9\right)\) \(= -\frac{4}{m^2} - \frac{9}{m^2} + 4\) \(= -\frac{13}{m^2} + 4\)

Set the derivative to zero:

\(-\frac{13}{m^2} + 4 = 0\) \(4 = \frac{13}{m^2}\) \(m^2 = \frac{13}{4}\) \(m = \frac{\sqrt{13}}{2}\)

Compute the minimum value of \( S \) by substituting \( m = \frac{\sqrt{13}}{2} \):

\(S = \frac{4m+9}{m} + 4m + 9\) \(= \left(\frac{4\cdot\frac{\sqrt{13}}{2} + 9}{\frac{\sqrt{13}}{2}}\right) + 4\cdot\frac{\sqrt{13}}{2} + 9\) \(= \frac{2\sqrt{13} + 9}{\frac{\sqrt{13}}{2}} + 2\sqrt{13} + 9\) \(= 4 + 13 + 9\) \(= 25\)

Thus, the minimum value of the sum of the distances is 25.

Therefore, the correct answer is 25.