Maths problem.. (1 Viewer)

[epicFAILx]

Can someone help me understand this problem:

Find the equation of the line which passes through the point P(1,-2) and the point of intersection of y= x and y= 2-x


Where do i begin?!!

HELP

 

[SpiralFlex]

Firstly! Find the point of intersection.









Point of intersection is .

Now this question is rather an odd one, usually we would find the gradient then use the 2 points formula. However in this case, we can inspect what the equation is. The reason we cannot use the gradient formula is because the denominator will be equal to zero. (which is undefined)

Plot and on your graph. Draw a straight line through both, you will see the equation of the straight line is

 

[epicFAILx]

Oh! Now i understand. Thankyou

 

[epicFAILx]

Oh! Now i understand..wait a sec

..actually.. how come the y was changed too x??

 

[SpiralFlex]

Oh! Now i understand. Thankyou

No worries.

 

[SpiralFlex]

..actually.. how come the y was changed too x??

You gave me two equations, and

See how they are both equal to ?

If they are both equal to , we can say that they are equal to each other.

 

[epicFAILx]

OH. Right... Haha.

THanks

 

[SpiralFlex]

OH. Right... Haha.

THanks

All the best with your questions.

 

[epicFAILx]

thanks

 

[Drongoski]

This happens to be a simple case. In general if L₁ = 0 and L₂ = 0 are equations of the 2 intersecting lines in General Form, then any line thru their intersection is of form: L₁ + kL₂ = 0 for some constant k.

Here the general form of line thru their intersection is: (x+y-2) + k(x-2) = 0
Since (1,-2) lies on this line: (1-2-2) + k(1+2) = 0 ==> k = 1

.: (x+y-2) + 1(x-y ) = 0

i.e. x = 1

In general this method is easier. Perversely, it seems harder for this question.

 

[cutemouse]

This happens to be a simple case. In general if L₁ = 0 and L₂ = 0 are equations of the 2 intersecting lines in General Form, then any line thru their intersection is of form: L₁ + kL₂ = 0 for some constant k.

Here the general form of line thru their intersection is: (x+y-2) + k(x-2) = 0
Since (1,-2) lies on this line: (1-2-2) + k(1+2) = 0 ==> k = 1

.: (x+y-2) + 1(x-y ) = 0

i.e. x = 1

In general this method is easier. Perversely, it seems harder for this question.

This guy speaks the truth.

 

[funnytomato]

This happens to be a simple case. In general if L₁ = 0 and L₂ = 0 are equations of the 2 intersecting lines in General Form, then any line thru their intersection is of form: L₁ + kL₂ = 0 for some constant k.

Here the general form of line thru their intersection is: (x+y-2) + k(x-2) = 0
Since (1,-2) lies on this line: (1-2-2) + k(1+2) = 0 ==> k = 1

.: (x+y-2) + 1(x-y ) = 0

i.e. x = 1

In general this method is easier. Perversely, it seems harder for this question.

nice, but the HSC doesn't test this as often , which they should since it's in the syllabus AND taught at school