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Question

Mathematics Question on Coordinate Geometry

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 APB\triangle APB is 10. If the locus of P is ax+by=15ax + by = 15, then 5a+2b5a + 2b is:

A

125\frac{12}{5}

B

65-\frac{6}{5}

C

4

D

6

Answer

125\frac{12}{5}

Explanation

Solution

The points are:

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

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

Area=12hk1 111 231.\text{Area} = \frac{1}{2} \begin{vmatrix} h & k & 1 \\\ -1 & 1 & 1 \\\ 2 & 3 & 1 \end{vmatrix}.

Equating this to 10:

12hk1 111 231=10.\frac{1}{2} \begin{vmatrix} h & k & 1 \\\ -1 & 1 & 1 \\\ 2 & 3 & 1 \end{vmatrix} = 10.

Expand the determinant:

12[h(13)k(12)+1(32)]=10,\frac{1}{2} \left[h(1 - 3) - k(-1 - 2) + 1(-3 - 2)\right] = 10, 12[2h+3k5]=10.\frac{1}{2} \left[-2h + 3k - 5\right] = 10.

Simplify:

2h+3k5=20,-2h + 3k - 5 = 20, 2h+3k=25.-2h + 3k = 25.

The locus of PP is obtained by replacing hh with xx and kk with yy:

2x+3y=25.-2x + 3y = 25.

Rewriting in the form ax+by=15ax + by = 15:

2x5+3y5=15,\frac{-2x}{5} + \frac{3y}{5} = 15, 25x+35y=15.-\frac{2}{5}x + \frac{3}{5}y = 15.

Here:

a=65,b=95.a = -\frac{6}{5}, \, b = \frac{9}{5}.

Finally, calculate 5a+2b5a + 2b:

5a+2b=5(65)+2(95),5a + 2b = 5\left(-\frac{6}{5}\right) + 2\left(\frac{9}{5}\right), 5a+2b=6+185=30+185=125.5a + 2b = -6 + \frac{18}{5} = \frac{-30 + 18}{5} = \frac{-12}{5}.

Thus, the value of 5a+2b5a + 2b is:

125.\frac{-12}{5}.