|
Question 885998: P(x,2x)
The ordinate of a point P is twice the abscissa. This point is equidistant from B(-3,1) and F(8,-2). Find the coordinates of P.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the point P is equal to (a,b) where b is equal to 2 * a.
the point P can therefore be represented as (a,2a).
the point P is equidistant from the points B(-3,1) and F(8,-2)
this means that the length of the line between P and B is equal to the length of the line between P and F.
the equation for the length of a line is:
L = sqrt((x2-x1)^2 + (y2-y1)^2)
Assign one the points for each line to (x1,y1) and assign the other point for each line to (x2,y2)
for the first line, you get:
(x1,y1) = (-3,1)
(x2,y2) = (a,2a)
for the second line, you get:
(x1,y1) = (8,-2)
(x2,y2) = (a,2a)
the point (x2,y2) is the point that's common to both lines.
for each line you want to find the length of the line.
for the first line, the equation for L becomes:
L1 = sqrt((a+3)^2 + (2a-1)^2)
for the second line, the equation for L becomes:
L2 = sqrt((a-8)^2 + (2a+2)^2)
simplify both these equations to get:
L1 = sqrt(2a^2 + 6a + 9 + 4a^2 - 4a + 1)
L2 = sqrt(2a^2 - 16a + 64 + 4a^2 + 8a + 4)
simplify these equation further to get:
L1 = sqrt(6a^2 + 2a + 10)
L2 = sqrt(6a^2 - 8a + 68)
set these equations equal to each other to get:
sqrt(6a^2 + 2a + 10) = sqrt(6a^2 - 8a + 68
square both sides of this equation to get:
6a^2 + 2a + 10 = 6a^2 - 8a + 68
solve for a to get:
a = 5.8
2a = 11.6
those are your solutions for a and 2a and those are the x and y coordinates for point P.
your 2 lines are:
L1 = the line between the points (-3,1) and (5.8,11.6)
L2 = the line between the points (8,-2) and (5.8,11.6)
to confirm the lengths of these lines are equal, then solve for the lengths of both lines.
you will find that:
length of L1 is equal to sqrt(189.8)
length of L2 is equal to sqrt(189.8)
|
|
|
| |
