Locus of a point on a straight line with variable gradients.

Question

A straight line of gradient m passes through the point (1,1) and cuts the x- and y- axes at A and B respectively. The point P lies on AB and is such that AP:PB = 1:2 .

Show that, as m varies, P moves on the curve whose equation is 3xy - x - 2y = 0.

So far, my working is as follows:

If the straight line passes through the point (1,1) , then the straight line should have the following equation:

$y = m(x-1) + 1$

which means that:

$A = (\frac{(m-1)}{m} , 0)$

$ B = (0, (-m +1))$

Therefore, the points of P should be:

X = $\frac{2(0) + 1(\frac{(m-1)}{m})}{3}$

Y = $\frac{2(m-1) + 1(0)}{3}$

Unfortunately, I'm not sure how to continue from here.

Thanks in advance for your assistance!

Answer 1

we have the coordinate of $$A(1-1/m,0)$$ and $$B(0,1-m)$$ from $$2AP=BP$$ we get after squaring $$4((x-1+1/m)^2+y^2)=x^2+(y-1+m)^2$$ plugging $$m=\frac{y-1}{x-1}$$ we get $$4\, \left( x-1+{\frac {x-1}{y-1}} \right) ^{2}+4\,{y}^{2}-{x}^{2}- \left( y-1+{\frac {y-1}{x-1}} \right) ^{2} =0$$ simplifying and facrotzing we get $$\left( 3\,yx-x-2\,y \right) \left( yx+x-2\,y \right) \left( {x}^{2} +{y}^{2}-2\,x-2\,y+2 \right) =0$$ and one factor is your equation!

Answer 2

That's not correct way to start with. Let P(say x,y) be locus is point on line AB. Given condition is AP:PB =1:2 and let's say A(a,0) and B(0,b) so if you take distance AP and PB.

$\dfrac{AP}{PB}=\dfrac 12$

$2AP=PB$ so squaring on both sides gives $4(AP)^2 = (PB)^2$

Now if you know distance between two points (say $M(x_1,y_1) and N(x_2,y_2)$ ) is $MN=\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2}$

So by solving $4(AP)^2 = (PB)^2$ you will get equation in terms of $a,b,x,y$

Also $P$ divides $AB$ in the ratio 1:2 then $(x,y)=\left(\frac a2 ,\frac b2\right)$ so replace $a=2x,b=2y$

Answer 3

An elementary solution

The equation of a straight line going through the point $(1,1)$ is

$$y=mx-m+1.$$

If $y=0$ then $0=mx-m+1$ from which equation

$$x=\frac{m-1}{m}.$$

If $x=0$ then

$$y=1-m.$$

So, we have two points: one on the $y$ axis and one on the $x$ axis:

$$\left(\frac{m-1}m,0\right) \text{ and }\ (0,1-m).$$

The coordinates of the point dividing the segment above the required way

$$x=\frac23\frac{m-1}{m}\ \text{ and }\ y=\frac13(1-m).$$

Hence

$$m=\frac1{1-\frac32x} \ \text{ and } \ m=1-3y.$$

That is, for the dividing point, we have

$$\frac1{1-\frac32x}=1-3y.$$

Finally,

$$-2y+3xy-x=0.$$

I did a little experiment with my dynamic geometry program and found the trace of the dividing point as shown below.

This is in line with the equation of the thick black curve:

$$y=\frac x{3x-2}.$$