Let L1 and L2 be perpendicular lines, and let F be a point at a distance of 18 from line L1 and a distance 25 from line L2. There are two distinct points, P and Q, that are each equidistant from F, from line L1, and from line L2. Find the area of the triangle FPQ.

The area of FPQ is 210.

Expert Answers

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Consider a coordinate system for which `L_1 ` is the x-axis and `L_2 ` is the y-axis. Then, F has coordinates `( 25 , 18 ) . ` A point equidistant from `L_1 ` and `L_2` has the coordinates with equal absolute values, `(x, +-x) . ` Consider points in the first or third quadrants, `( x , x ) .`

Then, `( d ( ( x , x ) , F ) )^2 = ( x - 25 )^2 + ( x - 18 )^2 = x^2 . ` It is a quadratic equation for `x , ` `x^2 - 2 * 43 x + 25^2 + 18^2 = 0 . ` Its roots are

`x_( 1 , 2 ) = 43 +- sqrt ( 43^2 - 25^2 - 18^2 ) = 43 +- sqrt ( 900 ) = 43 +- 30 ,`

so `x_1 = 13 ` and `x_2 = 73 .`

To determine the area of the triangle `( 25 , 18 ) , ( 13 , 13 ) , ( 73 , 73 ) , ` use cross product:

`A = 1 / 2 lt 73 - 13 , 73 - 13 gt xx lt 73 - 25 , 73 - 18 gt =`

`= 1 / 2 | [ 60 , 48 ] , [ 60 , 55 ] | = 1 /2 * 60 ( 55 - 48 ) = 30 * 7 = 210 .`

You can try to find points of the form `( x, -x ) .`

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