The perpendicular bisectors of the sides of △JKL intersect at point P. If JP = 2x + 5 meters and KP = 4x − 13 meters, what is the length of LP to the nearest meter?

Answer 1

The perpendicular bisectors of the sides of a triangle meet at a point called the circumcenter. The circumcenter is the center of a circle that contains the three vertices of the triangle. In this case point P is equidistant from each of the vertices, J, K, and L. That means JP = KP = LP.

Since JP = KP, that means 2x+5 = 4x-13 => x = 9.

If x = 9, then JP = 23, and so does KP. But LP is also a radius of the circumcircle so has the same length, i.e. LP = 23.

Here is a Desmos graph where you can drag the vertices of a triangle, show its perpendicular bisectors, and display the radii.

desmos.com/calculator/nyzcfhnbo1

Answer 2

if P is near 180 degrees then

JK= 4x-13+ 2x+5= 6x-8

and LP = near 0 meters

if P is near 0 degrees JK= |4x-13 -2x-5|= |2x-18|

and LP= about 2x-18, if x>9

or LP = about 18-x if 9>x

in an equilateral triangle L would be the center of the triangle

let JK =JP = JP

= 2x+5 = 4x-13

2x=18

x=9

height of the triangle = 4.5tan60 = 4.5(sqr3)

and LP =2.25sqr3 = about 3.897 feet

=LJ =LK = about 3.9 feet

let x = 6, then KP =4(6)-13 =24-13 = 11

JP then = 2(6)+5 = 12+5=17

leaving 6< JK < 28

Let JK = 17

then sides are 11, 17 and 17

with P between 17 and 17

where 11^2 = 2(17^2) -2(17)(17)CosP

law of cosines: c^2 +=a^2 +b^2 -2abCosC where c=11, a=b=17, P=C

CosP= (121 -578)/-578= 457/578

P = Cos^-1(457/578) = about 37.753 degrees

other 2 angles J and K = (180-37.753)/2 = 71.1235 degrees

LP is less than 17 and less than the height of the triangle using 11 as the base

LP<(11/2)tan71/1235

LP<about 16

to calculate point L, tr