The line segment joining P1 (0,-5) and P2 (4,1) is extended beyond P1 to Pr so that Pr is 4 times as far from P2 as from P1. Find the coordinates of Pr

Guest Nov 23, 2014

#1**+10 **

Let the coordinates of Pr = (x, y)

So we have

Four times the distance from the x coordinate of Pr to the x coordinate of P1 = the distance from the x coordinate of Pr to the x coordinate of P2.......expressed mathematically, we have

4(x - 0) = x - 4

4x = x - 4

3x = -4

x = -4/3

And four times the distance from the y coordinate of Pr to the y coordinate of P1 = the distance from the y coordinate of Pr to the y coordinate of P2.......expressed mathematically.......

4(y-(-5)) = y - 1

4(y+5) = y - 1

4y + 20 = y - 1

3y = -21

y = -7

So Pr = (-4/3, -7)

Proof

Distance from Pr to P1

√[(-4/3 - 0)^2 + (-7- (-5))^2] = √[(-4/3)^2 + (-2)^2] = about 2.4037

Distance from Pr To P2

√[(-4/3 - 4)^2 + (-7- 1)^2] = √[(-16/3 )^2 + (-8)^2] = about 9.6148

And 9.6148034012373047793 / 4 = 2.4037

CPhill
Nov 23, 2014

#1**+10 **

Best Answer

Let the coordinates of Pr = (x, y)

So we have

Four times the distance from the x coordinate of Pr to the x coordinate of P1 = the distance from the x coordinate of Pr to the x coordinate of P2.......expressed mathematically, we have

4(x - 0) = x - 4

4x = x - 4

3x = -4

x = -4/3

And four times the distance from the y coordinate of Pr to the y coordinate of P1 = the distance from the y coordinate of Pr to the y coordinate of P2.......expressed mathematically.......

4(y-(-5)) = y - 1

4(y+5) = y - 1

4y + 20 = y - 1

3y = -21

y = -7

So Pr = (-4/3, -7)

Proof

Distance from Pr to P1

√[(-4/3 - 0)^2 + (-7- (-5))^2] = √[(-4/3)^2 + (-2)^2] = about 2.4037

Distance from Pr To P2

√[(-4/3 - 4)^2 + (-7- 1)^2] = √[(-16/3 )^2 + (-8)^2] = about 9.6148

And 9.6148034012373047793 / 4 = 2.4037

CPhill
Nov 23, 2014