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Question 980425: Can someone please solve these 3 coordinate geometry in the x, y plane questions for me? It would really help. E out. Preferably if you show your working out too so I can see how it's done.
2) the line PQ is a diameter of the circle center (-4,-2). The line l(small L) passes through P and is perpendicular to PQ. Given that Q is (10,4), find the equation of l.
Thank you!!
Answer by Cromlix(4381)   (Show Source):
You can put this solution on YOUR website! Hi there,
First find the coordinates of P
As Centre of circle is (-4,-2)
and point Q is (10,4)
Using 'C' for centre of circle
Distance QC/Distance CP = 1/1
c - q /p - c = 1/1
Cross multiply
p - c = c - q
Collect like terms
p = 2c - q
p = 2(-4,-2) - (10, 4)
p = (-18,-8)
This is the coordinates of P (-18,-8)
Gradient of CQ (which will equal
gradient of PQ)
Gradient = y2 - y1/ x2 - x1
Using C (-4,-2) and Q(10,4)
Gradient = 4 - (-2)/10 - (-4)
Gradient = 6/14 or 3/7
If line l is perpendicular to PQ
then their gradients multiply
together to give -1
m1 x m2 = -1
So, line l will have a gradient
of -7/3
Using coords of P (-18,-8)
y - b = m(x - a)
y - (-8) = -7/3(x - (-18)
y + 8 = -7/3(x + 18)
y + 8 = -7/3x - 126/3
y = -7/3x - 126/3 - 24/3(8)
y = -7/3x - 50
Hope this helps:-)
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