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

cheers

**0**

# 2004 Paper 1, Q1

Started by will_789, Apr 06 2006 09:47 AM

1 reply to this topic

### #1

Posted 06 April 2006 - 09:47 AM

### #2

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

y=-2

when y=-2, x= -3y-1 (Rearanged first equation to get x on its own)

-3(-2)-1=x

=6-1

=5

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

Now just use the gradient equation: m=(y

(-2-4) / (5-7)

= -6 /-2

=3

[ B ]

when perpendicular m

3 x m

m

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

y=(-1/3)x-1 m=-1/3 (Rearanged equations to get them in the form y=mx+C, which is a

y=(-2/5)x m= -2/5

Since, when perpendicular m

3 x -1/3 = -1 perpendicular

3 x -2/5 -1 not perpendicular

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

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

y=-2

when y=-2, x= -3y-1 (Rearanged first equation to get x on its own)

-3(-2)-1=x

=6-1

=5

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

Now just use the gradient equation: m=(y

_{2}-y_{1}) / (x_{2}-x_{1})(-2-4) / (5-7)

= -6 /-2

=3

[ B ]

when perpendicular m

_{1}m_{2}=-13 x m

_{2}=-1 (Gradient of AB X m_{2})m

_{2}= -1/3Use both equations from question to find the gradients of the lines:

y=(-1/3)x-1 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 m= -2/5

Since, when perpendicular m

_{1}m_{2}=-13 x -1/3 = -1 perpendicular

3 x -2/5 -1 not perpendicular

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

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