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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!

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    i can't see any equation of the curve – 2017-02-27
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    Don't you mean that $P$ lies on *the line segment* $AB$? – 2017-02-27

3 Answers 3

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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!

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    What are they?All of circle, hyperbola, and ellipse? – 2017-02-27
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    you can Control my solution there a two other equations – 2017-02-27
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    $$xy+x-2y=0$$ and $$x^2+y^2-2x-2y+2=0$$ are any restrictions given in your Problem? – 2017-02-27
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    Thank you, you provided a perfectly clear explanation. I wouldn't have thought express m in terms of x and y. – 2017-02-27
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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 $a=2x,b=2y$

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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.

enter image description here

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

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