Problem:

The line 5x 3y=30 meets the x axis at A and y axis at B. The perpendicular bisector of AB meet the line x-y=0 at P. The area of triangle PAB is? Explain in details.

Solution:

Step 1: Finding the coordinates of A and B
We know that the line 5x 3y=30 meets the x-axis at A and y-axis at B.

When y=0, 5x=30, x=6. Hence A is (6,0).

When x=0, 3y=30, y=10. Hence B is (0,10).

Step 2: Finding the midpoint of AB
Using the coordinates of A and B, we can find the midpoint of AB.

Midpoint of AB = ( (6+0)/2, (0+10)/2 ) = (3,5)

Step 3: Finding the equation of the perpendicular bisector of AB
The perpendicular bisector of AB passes through the midpoint of AB and is perpendicular to AB. Hence, the slope of the perpendicular bisector is the negative reciprocal of the slope of AB.

The slope of AB = (10-0)/(0-6) = -5/3
The slope of the perpendicular bisector = 3/5 (negative reciprocal of -5/3)

Using the coordinates of the midpoint of AB, we can find the equation of the perpendicular bisector as follows:

y - 5 = (3/5)(x - 3)
=> 5y - 25 = 3x - 9
=> 3x - 5y + 16 = 0

Step 4: Finding the coordinates of P
The perpendicular bisector of AB meets the line x-y=0 at P. Hence, we can find the coordinates of P by solving the two equations simultaneously.

3x - 5y + 16 = 0
x - y = 0

Solving these equations, we get x = 4 and y = 4. Hence, P is (4,4).

Step 5: Finding the area of triangle PAB
We can find the lengths of PA, PB, and AB using the distance formula.

PA = sqrt( (4-3)^2 + (4-5)^2 ) = sqrt(2)
PB = sqrt( (6-0)^2 + (0-10)^2 ) = 2*sqrt(37)
AB = sqrt( (6-0)^2 + (0-10)^2 ) = 2*sqrt(34)

Using the formula for the area of a triangle with sides a,b,c, we can find the area of triangle PAB as follows:

s = (sqrt(2) + 2*sqrt(37) + 2*sqrt(34))/2
Area of PAB = sqrt( s(s-sqrt(2))(s-2*sqrt(37))(s-2*sqrt(34)) )

Hence, the area of triangle PAB is sqrt( 1120 - 38*sqrt(37) ) square units.