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2004 Paper 1, Q1 - HSN forum

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2004 Paper 1, Q1


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#1 will_789

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Posted 06 April 2006 - 09:47 AM

Could someone show me how to do these questions, with all the working, if possible
cheers

#2 SncZ

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Posted 06 April 2006 - 08:21 PM

[ A ]

at point of intersection, use similtaneous equations:

x+3y+1=0
2x+5y=0

-2x-6y-2=0 (times whole equation by -2, so that the x terms cancels out )
2x+5y=0

Add the two equations, term for term (-2x+2x=0, -6y+5y=-y,-2+0=-2 )
-y-2=0
therefore.gif y=-2

when y=-2, x= -3y-1 (Rearanged first equation to get x on its own)
therefore.gif -3(-2)-1=x
=6-1
=5

therefore.gif co-ods (5,-2) , (7,4) <-- given to you

Now just use the gradient equation: m=(y2-y1) / (x2-x1)

therefore.gif (-2-4) / (5-7)
= -6 /-2
=3

[ B ]

when perpendicular m1m2=-1

therefore.gif 3 x m2=-1 (Gradient of AB X m2)
therefore.gif m2 = -1/3

Use both equations from question to find the gradients of the lines:

y=(-1/3)x-1 therefore.gif m=-1/3 (Rearanged equations to get them in the form y=mx+C, which is a must when finding the gradient).

y=(-2/5)x therefore.gif m= -2/5

Since, when perpendicular m1m2=-1

3 x -1/3 = -1 therefore.gif perpendicular
3 x -2/5 notequal.gif -1 therefore.gif not perpendicular

therefore, AB is only perpendicular to one line ( x+3y+1=0 )





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