Question

Let A(-1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of $\triangle APB$ is 10. If the locus of P is $ax + by = 15$, then $5a + 2b$ is:

• A. $\frac{12}{5}$
• B. $-\frac{6}{5}$
• C. 4
• D. 6

Answer 1

The Correct Option is A

The points are:

\[ A(-1, 1), \, B(2, 3), \, P(h, k). \]

The area of \(\triangle PAB\) is given as 10. Using the determinant formula for the area of a triangle:

\[ \text{Area} = \frac{1}{2} \begin{vmatrix} h & k & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix}. \]

Equating this to 10:

\[ \frac{1}{2} \begin{vmatrix} h & k & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix} = 10. \]

Expand the determinant:

\[ \frac{1}{2} \left[h(1 - 3) - k(-1 - 2) + 1(-3 - 2)\right] = 10, \] \[ \frac{1}{2} \left[-2h + 3k - 5\right] = 10. \]

Simplify:

\[ -2h + 3k - 5 = 20, \] \[ -2h + 3k = 25. \]

The locus of \(P\) is obtained by replacing \(h\) with \(x\) and \(k\) with \(y\):

\[ -2x + 3y = 25. \]

Rewriting in the form \(ax + by = 15\):

\[ \frac{-2x}{5} + \frac{3y}{5} = 15, \] \[ -\frac{2}{5}x + \frac{3}{5}y = 15. \]

Here:

\[ a = -\frac{6}{5}, \, b = \frac{9}{5}. \]

Finally, calculate \(5a + 2b\):

\[ 5a + 2b = 5\left(-\frac{6}{5}\right) + 2\left(\frac{9}{5}\right), \] \[ 5a + 2b = -6 + \frac{18}{5} = \frac{-30 + 18}{5} = \frac{-12}{5}. \]

Thus, the value of \(5a + 2b\) is:

\[ \frac{-12}{5}. \]