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# maths Q (not sure what topic this is) watch

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1. 1. Show that the following equation represents a single straight line

x²-4xy+2x+4y²-4y+1=0

2. same as 1 but represents a PAIR of straight lines (find their point of intersection)

2x²-xy+5x-y²+y+2=0

i need to know the method as well as the answers please, thanks
2. 1.
x²- 4xy + 2x + 4y² - 4y + 1 = x²- 4xy + 4y² + 2x - 4y + 1
= (x-2y)² + 2(x-2y) + 1 = 0

Let x-2y=z, then:
z² + 2z + 1 = 0
(z+1)² = 0
z = -1
x-2y = -1
=> 2y = x + 1, which is an equation of a straight line.
3. (Original post by manps)
1. Show that the following equation represents a single straight line

x²-4xy+2x+4y²-4y+1=0

2. same as 1 but represents a PAIR of straight lines (find their point of intersection)

2x²-xy+5x-y²+y+2=0

i need to know the method as well as the answers please, thanks
Any ideas on the 2nd part anyone?
4. Or simply differentiate (implicitly) wrt x:
2x-4x(dy/dx)-4y+2+8y(dy/dx)-4(dy/dx)=0
dy/dx(-4x+8y-4)+2x-4y+2=0
dy/dx=½
5. 2. 2x²-xy+5x-y²+y+2=0
(2x+y+1)(x-y+2)=0
=>2x+y+1=0 or x-y+2=0
Now, solve simultaneously to give (-1,1) for intersection point.
6. 2x+y+1=0 or x-y+2=0
y=-2x-1
y=x+2

x+2=-2x-1
3x+3=0
x=-1
y=1
7. (Original post by C4>O7)
2x²-xy+5x-y²+y+2=0
(2x+y+1)(x-y+2)=0
=>2x+y+1=0 or x-y+2=0
Nice!
8. Hehe
..I should get some sleep before gen/stud. exam tomorrow

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