Thread: problem with the method of finding the equation of locus

1. problem with the method of finding the equation of locus

Consider three points A(1,0), B(4,0) and C (6,0). L1 is a line with slope m passing through A. L2 is a line with slope n passing through C. L1 and L2 intersect at P.
(A) find the equations of L1 and L2
ans: y = m(x-1), y = n(x-6)
(B) if B is equidistant from L1 and L2, prove that 9m^2 + 5 m^2 n^2 - 4n^2 =0
(C) As m and n vary, find the equation of the locus of P.
ans: x^2 + y^2 - 20x + 64 = 0 --------------- (*)

i have no problem in (A) and(B) , but in (C), I don't understand why i should substitute the value m = y/(x-1) and n = y/(x-6) back in equation (*). m = y/(x-1) is the equation of L1 while n = y/(x-6) is the equation of L2. The interception of L1 and L2 of course describle the location of P. but is (*) true for all condition (I mean if the relationship between m and n in (*) is true under any condition)? If it is, why is it related to the locus of P?
from my textbook it says:
If a point which moves under certain conditions describes a path, and
(i) all points satisfying the conditions lie on the path,
(ii) every point on the path stisfies the conditions,
then the path is called the locus of the point.
General procedure for finding the equation of a locus:
1. let P(x,y) be a point on the locus
2. write down the given condition as a relation connecting x and y.
3. simplify the above relation to obtain the required equation of the locus.

but in this case how to use the so-called general procedure to find the equation of locus of P? i can't find a similar question or working example in my textbook. really confuse@@@@ would any one please tell me the basic concept of the way to find the equation of locus?

2. Hello, afeasfaerw23231233!

I got part (B ) . . .

Consider three points: .$\displaystyle A(1,0),\;B(4,0),\;C(6,0)$

$\displaystyle L_1$ is a line with slope $\displaystyle m$ passing through $\displaystyle A.$
$\displaystyle L_2$ is a line with slope $\displaystyle n$ passing through $\displaystyle C.$
$\displaystyle L_1$ and $\displaystyle L_2$ intersect at $\displaystyle P.$

(A) Find the equations of $\displaystyle L_1$ and $\displaystyle L_2$

Ans: .$\displaystyle y \:= \:m(x-1),\;\;y \:=\: n(x-6)$ . . . . Right!

(B) If $\displaystyle B$ is equidistant from $\displaystyle L_1$ and $\displaystyle L_2$,
prove that: .$\displaystyle 9m^2 + 5 m^2 n^2 - 4n^2 \:=\:0$

The distance from a point $\displaystyle (x_1.y_1)$ to a line $\displaystyle ax + by + c\:=\:0$

. . is given by: .$\displaystyle d \;=\;\frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$

We have the point $\displaystyle B(4,0)$

We want: .$\displaystyle d_1\:=\:BL_1$

$\displaystyle L_1$ has the equation: .$\displaystyle mx - y - m \:=\:0$

Hence: .$\displaystyle d_1\;=\;\frac{|4m - 0 - m|}{\sqrt{m^2+1}} \:=\;\frac{|3m|}{\sqrt{m^2+1}}$

We want: .$\displaystyle d_2\:=\:BL_2$

$\displaystyle L_2$ has the equation: .$\displaystyle nx - y - 6n \:=\:0$

Hence: .$\displaystyle d_2\;=\;\frac{|4n - 0 - 6n|}{\sqrt{n^2+1}} \;=\;\frac{|2n|}{\sqrt{n^2+1}}$

Since $\displaystyle d_1\:=\:d_2\!:\;\;\frac{|3m|}{\sqrt{m^2+1}} \:=\:\frac{|2n|}{\sqrt{n^2+1}} \quad\Rightarrow\quad |3m|\sqrt{n^2+1} \:=\:|2n|\sqrt{m^2+1}$

Square both sides: .$\displaystyle 9m^2(n^2+1) \:=\:4n^2(m^2+1) \quad\Rightarrow\quad 9m^2n^2 + 9m^2\:=\:4m^2n^2 + 4n^2$

Therefore: .$\displaystyle 9m^2 + 5m^2n^2 - 4n^2 \:=\:0$

3. yes i know how to do (B). but my question is i don't know why i should sub the value m = y/(x-1) and n = y/(x-6) back in equation 9m^2 + 5 m^2 n^2 - 4n^2 =0 in order to get the locus of P

4. Originally Posted by afeasfaerw23231233
yes i know how to do (B). but my question is i don't know why i should sub the value m = y/(x-1) and n = y/(x-6) back in equation 9m^2 + 5 m^2 n^2 - 4n^2 =0 in order to get the locus of P
Well the $\displaystyle 9m^2 + 5 m^2 n^2 - 4n^2 =0$ equation is the one that gives the locus in terms of m and n. You have two more equations relating m and n to x and y. Since you want an equation for the locus in terms of x and y you need to substitute these equations into your locus equation.

-Dan