Letting L1=ax+by+c=0
and L2=px+qy+r=0.
The equation of anyline passing through the intersection of L1 and L2 is
L1+kL2=0 where k is a variable depending on just how the actual line through the point is.
Can anyone prove this because all i get is 0=0.
Find the point of intersection of the lines in terms of a,b,c,p,q and r.
If I did my math right, the coordinates are
x' = (rb-cq)/(aq-bp)
and
y' = (cp-ar)/(aq-bp)
Any line passing though the intersection can be written
y-y' = m (x - x')
where m is an arbitrary constant.
L1 + kL2 = 0 because of the way L1 and L2 are defined. It is also a general equation for a line
(a + kp)x + (b + kq)y +c + kr = 0
I think the statement proves itself, just by substituting for L1 and L2. By making both substitutions, you require that the equations of both lines are satisfied, so that the equation that results passes through the common point.
but how does
(a + kp)x + (b + kq)y +c + kr = 0
relate to
y-y' = m (x - x')?
It should be possible to rewrite
(a + kp)x + (b + kq)y +c + kr =0 in the other form, with the values of x' and y' that I derived. m will be a function of k and other constants.
y = [(a+kp)/(b+kq)]x + (-c-kr)/(b + kq)
In my first version, the y intercept would be at
y = y' + mx'= (cp-ar)/(aq-bp) + m (rb-cq)/(aq-bp)
I agree they do not look consistent. I may have made an error in the derivation.
To prove that the equation of any line passing through the intersection of two lines L1 and L2 can be written as L1+kL2=0, where k is a variable, we need to use the concepts of linear algebra and the properties of lines in a coordinate plane.
Let's start by considering the point of intersection of L1 and L2. This point satisfies both equations L1 and L2 simultaneously. Let's denote this point as (x0, y0).
Since (x0, y0) lies on L1, substituting these values into the equation L1=ax+by+c=0, we get:
a(x0) + b(y0) + c = 0
Similarly, since (x0, y0) lies on L2, substituting these values into the equation L2=px+qy+r=0, we get:
p(x0) + q(y0) + r = 0
Now, let's introduce a variable k into the equation to represent any line passing through this point (x0, y0).
We can write the equation of this line as L = L1 + kL2, where L represents the equation of the line passing through the intersection of L1 and L2.
Substituting the equations of L1 and L2 into the equation for L, we have:
L = (a(x0) + b(y0) + c) + k(p(x0) + q(y0) + r)
Now, let's simplify this equation:
L = a(x0) + b(y0) + c + kp(x0) + kq(y0) + kr
L = (a(x0)+kp(x0)) + (b(y0)+kq(y0)) + c + kr
L = (a+kp)(x0) + (b+kq)(y0) + c + kr
Since (a+kp), (b+kq), and (c+kr) are constants, we can redefine them as new constants A, B, and C, respectively. Therefore, we get:
L = A(x0) + B(y0) + C = 0
So, we have proven that the equation of any line passing through the intersection of L1 and L2 can be written as L1+kL2=0, where k is a variable. The equation L1+kL2=0 represents a family of lines passing through the intersection point (x0, y0).