Can you apply the distance formula?

Geometry Level 2

Line segment $M$ has endpoints $Q(3,1)$ and $R(2,4)$ . Line segment $M'$ has endpoints $Q'(4,2)$ and $R'(7,3)$ and is created by rotating line segment $M$ around point $P$ . What are the coordinates of $P$ ?

(4,1)

(8,-3)

(4,-4)

(0,0)

1 solution

Aaron Tsai
May 5, 2016

Relevant wiki: Distance Formula

Let $(x,y)$ be the coordinates of $P$ . The distance between $P$ and the endpoints of $M$ and $M'$ are the same, regardless of the rotation. We can set up the system of equations

$\sqrt{(x-3)^{2}+(y-1)^{2}}=\sqrt{(x-4)^{2}+(y-2)^{2}}$

$\sqrt{(x-2)^{2}+(y-4)^{2}}=\sqrt{(x-7)^{2}+(y-3)^{2}}$

Simplifing the first equation, we have

$x^{2}-6x+9+y^{2}-2y+1=x^{2}-8x+16+y^{2}-4y+4$

$2x+2y=10$

$x+y=5$

Simplifying the second equation, we have

$x^{2}-4x+4+y^{2}-8y+16=x^{2}-14x+49+y^{2}-6y-9$

$10x-2y=38$

$5x-y=19$

Adding these two equations together, we get

$6x=24$

$x=4$

$\implies$ $y=1$

Therefore, point $P$ is located at $(4,1)$ .

I tried it in this way; The line M and M’ intersect at point P thus their solution gives us P M : 3x+y-10=0 M’: x-3y+2=0 The goal here is to solve x-3y+2=0 and 3x+y-10=0 for the variables x and y. The solutions to your equations are: x= 14/5 and y=8/5 But they aren't in the options, so where am I wrong?

Ashwin Kumar - 5 years, 1 month ago

Line (segment) $M$ and $M'$ do not intersect at point $P$ . $P$ is just the center of rotation.

Aaron Tsai - 5 years, 1 month ago

Oh sorry, i mistook the statement "created by rotating line segment M AROUND point P ." as "created by rotating line segment M ABOUT point P ".My bad. I would suggest that a diagram would eliminate this mistake from happening to others (I'm just saying it , not that everyone is as stupid as me to make that silly mistake ;D )

Ashwin Kumar - 5 years, 1 month ago

Can you apply the distance formula?

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