Given the pencil of lines a(2x + 5y +4) + b(3x-2y+25)=0.Find that line of pencil from which the point M(4,5) is at greatest distance

To find the line from the pencil of lines represented by the equation \( a(2x + 5y + 4) + b(3x - 2y + 25) = 0 \) that is at the greatest distance from the point \( M(4, 5) \), we can use the concept of distance from a point to a line. The distance \( d \) from a point \( (x_0, y_0) \) to a line given by the equation \( Ax + By + C = 0 \) is calculated using the formula:

\( d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \)

In our case, we need to express the pencil of lines in the standard form \( Ax + By + C = 0 \). Let's rewrite the given equation:

\( a(2x + 5y + 4) + b(3x - 2y + 25) = 0 \)

Expanding this, we have:

\( 2ax + 5ay + 4a + 3bx - 2by + 25b = 0 \)

Combining like terms gives us:

\( (2a + 3b)x + (5a - 2b)y + (4a + 25b) = 0 \)

Here, we can identify:

• A = \( 2a + 3b \)
• B = \( 5a - 2b \)
• C = \( 4a + 25b \)

Next, we substitute the coordinates of point \( M(4, 5) \) into the distance formula:

\( d = \frac{|(2a + 3b)(4) + (5a - 2b)(5) + (4a + 25b)|}{\sqrt{(2a + 3b)^2 + (5a - 2b)^2}} \)

Now, simplifying the numerator:

\( d = \frac{|8a + 12b + 25a - 10b + 4a + 25b|}{\sqrt{(2a + 3b)^2 + (5a - 2b)^2}} \)

This simplifies to:

\( d = \frac{|37a + 27b|}{\sqrt{(2a + 3b)^2 + (5a - 2b)^2}} \)

To maximize the distance \( d \), we need to maximize the numerator \( |37a + 27b| \) while minimizing the denominator \( \sqrt{(2a + 3b)^2 + (5a - 2b)^2} \). The maximum distance occurs when the line is perpendicular to the line connecting point \( M \) to the origin of the pencil of lines.

The direction vector of the line connecting \( M(4, 5) \) to the origin can be represented as \( (4, 5) \). The slope of this line is \( \frac{5}{4} \), and the slope of the line perpendicular to it is \( -\frac{4}{5} \). Thus, we can set up the condition for perpendicularity:

\( \frac{5a - 2b}{2a + 3b} = -\frac{4}{5} \)

Cross-multiplying gives:

\( 5(5a - 2b) = -4(2a + 3b) \)

Expanding and rearranging leads to:

\( 25a - 10b + 8a + 12b = 0 \)

Thus, we have:

\( 33a + 2b = 0 \)

From this, we can express \( b \) in terms of \( a \):

\( b = -\frac{33}{2}a \)

Substituting this back into the equation of the line gives us the specific line that is at the greatest distance from point \( M(4, 5) \). This line can be derived by substituting \( b \) into the original pencil equation, leading to the desired line equation.

In summary, the line from the pencil that is at the greatest distance from the point \( M(4, 5) \) can be found by maximizing the distance formula and ensuring the line is perpendicular to the vector from \( M \) to the origin. This approach combines geometric intuition with algebraic manipulation to arrive at the solution.