Straight line question ( K method) (1 Viewer)

[Twickel]

Hi
From a question I get this 20+23k=0 which should I move to the RHS the 23k or 20 the answer was 5x+6y-27=0 I got -5x-6y+27=0 so in general is there a rule in which to move to the RHS?

Also
y-6=4x-4

I got y-4x-2=0 the book has 4x-y+2=0

What am I doing wrong?

 

[bos1234]

The general rule is
ax+by+c = 0
the co-efficient of x which is a SHOULD be positive

So just take everything to the other side

 

[Twickel]

So make the co of x positive. What do I do in a situation like this
Find the equation of the straight line passing through the point ( -2,1) and through the point of intersection of 4x-y-1=0 and 2x-y+5=0.

For the first bit -8-1-1= -10 the second bit = 0 so how to solve it?

 

[Aerath]

It doesn't matter. Both your answers were right. The only thing is the number in front of x should be a positive integer. So throw everything on the other side.

 

[Twickel]

Find the equation of the straight line passing through the point ( -2,1) and through the point of intersection of 4x-y-1=0 and 2x-y+5=0.

For the first bit -8-1-1= -10 the second bit = 0 so how to solve it?

 

[Aerath]

Switch the equations around:
4x-y-1 = 0
2x-y+5 = 0
P(-2,1)

Therefore 2x-y+5+k(4x-y-1) = 0
Sub (-2,1)
-4-5+5+k(-8-1-1) = 0
0 +k(-10) = 0
k = 0

Therefore 2x-y+5+0(4x-y-1) = 0
2x-y+5 = 0

That's the answer. Of course, you could revert back to Yr 9 coordinate geometry, which is what I would've done, if I had gotten constant = different constant (unless the question specified: "Using the k-method....").

 

[conics2008]

dont bother with the K method, its useless.. Think of it logicaly.

first solve sim the two equations and then find gradient between the points you found and the given points and then use the point gradient formula.

 

[Twickel]

Ok thanks why did you switch the equations around?

 

[Aerath]

So I wouldn't get -10 = 0 (which is obviously false). By switching the equations around, I got k = 0, which could possibly be a solution (which it was).

 

[Twickel]

Thankyou very much.

 

[me121]

It doesn't matter. Both your answers were right. The only thing is the number in front of x should be a positive integer. So throw everything on the other side.

yes, both are right, and it doesn't matter. but i disagree with the number in front of x should be a positive integer. this is trivial, and a matter of opinion. unless they ask in the question that they want this, then you can write,
5x+6y-27=0
or
-5x-6y+27=0
or
-25x-30y+810=0

all these answers will (or at least should) get the same marks.

oh and what's this K method? I've never heard of it?

 

[Aerath]

I don't think so...the rules regarding general form (I think), are Ax+By+C = 0, where A, B and C are integers, and A is positive.

I really don't know how to explain the k-method.

 

[Twickel]

Good, thankyou.

ok last question. If ABC are collinear points AB=BC. If A is (4,5) and B is the point 91,-1) find the co ordinates of C.

 

[undalay]

Good, thankyou.

ok last question. If ABC are collinear points AB=BC. If A is (4,5) and B is the point 91,-1) find the co ordinates of C.

The difference of the x,y values of point C is equal to the difference between A and B.

e.g.

If A was (1,1) and B was (3,3) then C would be (5,5) Assuming ABC were collinear and B is in the middle

 

[Twickel]

The answer is -2 -7 so could you explain what you just said?

 

[undalay]

A (4,5)
B (1,-1)
C (-2, -7)

Looking at the X value of A.
It goes down by 3 at B
It goes down again by the sam amount to C

This is the same for Y value (only goes down by 6)

 

[Twickel]

Did not see that last post sorry if the B point was higher do I add the difference?

 

[lolokay]

yes, both are right, and it doesn't matter. but i disagree with the number in front of x should be a positive integer. this is trivial, and a matter of opinion. unless they ask in the question that they want this, then you can write,
5x+6y-27=0
or
-5x-6y+27=0
or
-25x-30y+810=0

all these answers will (or at least should) get the same marks.

oh and what's this K method? I've never heard of it?

The K method is used when you want to find the equation of the line that goes through the point of intersection of two lines, and another point.

If the two lines are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂, and the point is (m,n), you write it as

a₁x + b₁y + c₁ + k(a₂x + b₂y + c₂) = 0
then substitute in m and n for x and y to solve for k, then expand and simplify and you get the equation of the line you wanted.

 

[undalay]

Did not see that last post sorry if the B point was higher do I add the difference?

yes.

 

[munchiecrunchie]

Good, thankyou.

ok last question. If ABC are collinear points AB=BC. If A is (4,5) and B is the point 91,-1) find the co ordinates of C.

Since AB = AC, therefore point B is the midpoint of AC

let the co-ordinates of C be (x, y)

therefore, 1 = (4 + x ) / 2
2 = 4 + x
x = -2

- 1= (5 + y) / 2
- 2= 5 + y
y= - 7
therefore, the co-ordinates of point C are (-2, -7)