# Co ordinate Geometry Help urgently needed - exam in two days!!

• Nov 26th 2007, 05:58 PM
ToriBori
Co ordinate Geometry Help urgently needed - exam in two days!!
ok so i was working through a past paper earlier today and i came across this question:

An orienteering course has been marked by four points on a graph O(0,0), A(-200,500), B(300,600) and C(800,0)

a) a drink station is to be positioned at the intersection on the diagonals of the quadrilateral OABC
Find the coordinates of the drink station

ok so i looked through my maths resourse book and worked out that the diagonal lines are perpendicular to each other. i know this because i worked out the gradients of each diagonal and the equation, they are:

OB: (0,0) and (300,600)

600 - 0
300 - 0 = m = 2

midpoint is (150,300) and m= 2

y - 300 = 2 (x - 150)
= 2x - 300
y = 2x

AC: (-200,500) and (800,0)

0 - 500
800 - -200 = m = -0.5

midpoint is (300,250) and m= -0.5

y - 250 = -0.5 (x - 300)
= -0.5x + 150
y = -0.5x + 400

if we times the gradients we get -1 so the two diagonal lines are perdendicular. This is where i am stick: i dont know how to work out where perpendicular lines meet because they dont meet at their midpoints.
can someone help me please??? its really urgent as my maths exam is this thursday and i really need help!!!

ToriBori
• Nov 26th 2007, 11:57 PM
earboth
Quote:

Originally Posted by ToriBori
ok so i was working through a past paper earlier today and i came across this question:

An orienteering course has been marked by four points on a graph O(0,0), A(-200,500), B(300,600) and C(800,0)

a) a drink station is to be positioned at the intersection on the diagonals of the quadrilateral OABC
Find the coordinates of the drink station

ok so i looked through my maths resourse book and worked out that the diagonal lines are perpendicular to each other. i know this because i worked out the gradients of each diagonal and the equation, they are:

OB: (0,0) and (300,600)

600 - 0
300 - 0 = m = 2

midpoint is (150,300) and m= 2

y - 300 = 2 (x - 150)
= 2x - 300
y = 2x

AC: (-200,500) and (800,0)

0 - 500
800 - -200 = m = -0.5

midpoint is (300,250) and m= -0.5

y - 250 = -0.5 (x - 300)
= -0.5x + 150
y = -0.5x + 400

if we times the gradients we get -1 so the two diagonal lines are perdendicular. This is where i am stick: i dont know how to work out where perpendicular lines meet because they dont meet at their midpoints.
can someone help me please??? its really urgent as my maths exam is this thursday and i really need help!!!

ToriBori

Hello,

all your calculations are OK. Calculate now the coordinates of the intersection of the two lines:

$y = 2x~\wedge ~ y=-\frac12 \cdot x +400$

$2x=-\frac12 \cdot x +400~\implies~x=160$ Plug in this value into the first equation. The intersection is at P(160, 320).
• Nov 27th 2007, 05:06 AM
earboth
Hi,