Locus of intersection of lines.

Question

We have a line which touches $x$ axis at $A=(7,0)$ and $B=(0,-5)$. A variable line PQ which touches $x$ axis at P and $y$ axis at Q is perpendicular to the line $AB$. Now we have to find the locus of R which is the point of intersection of $AQ$ and $BP$.

Now first of all equation of AB can be written as $$AB:5x-7y=35$$ Taking $P=(p,0) $ and $Q=(0,q)$

Equation of $PQ$ can be written as $$PQ: qx+py=qp$$

Since PQ is perpendicular to AB using $$slope(PQ) \cdot slope (AB)=-1$$ $$\Rightarrow 5q=7p \tag1$$

Now writing the equation of $AQ$ using $(7,0)$ and $(0,q)$ $$AQ: qx+7y-7q=0$$

Similarly writing the equation of $BP$ using $(0,-5)$ and $(p,0)$ $$BP: 5x-py-5p=0$$

Now since $R = (h,k)$ is the intersection of $AQ$ and $BP$ solving the two equations we get

$$h=\dfrac{35p+7pq}{35+pq} \tag2$$ $$k=\dfrac{35q-5pq}{35+pq} \tag3$$

I just don't know how can I go ahead and use equation $(1),(2)$ and $(3)$ in order to relate the terms to get the locus.

Please help me further.

Answer 1

If you divide the equations, you get $$\frac hk=\frac{35p+7pq}{35q-5pq}$$

Now substitute $$q=\frac{7p}{5}$$ into this and simplify so $$\frac hk=\frac{25+7p}{35-5p}$$

Now rearrange this to get $$p=\frac{35h-25k}{7k+5h}$$ and substitute this into the equation of the line $BP$ which is $5h-pk-5p=0$, and after some simplification you end up with $$h^2+k^2-7h+5k=0$$

Which basically gives you the locus of a circle. $$(h-\frac{7}{2})^2+(k+\frac{5}{2})^2=\frac{74}{4}$$

Note that this circle has a centre which is the midpoint of $AB$, and passes through $A$, $B$ and the origin $O$.

Answer 2

Into the equations for h and k substitute p for q: $q=\frac75p$ and $$h=\frac{35p+\frac{49}5p^2}{35+\frac75p^2}$$ $$k=\frac{49p-7p^2}{35+\frac75p^2}$$ This is a parametric equation in p. If we convert this to an implicit function and remove p (I did this in Wolfram Alpha) we get: $$\frac{-777323750h+111046250 h^2+555231250k+111046250 k^2}{625}=0$$ $$h^2-7h+k^2+5k=0$$ $$(h-\tfrac72)^2+(k+\tfrac52)^2=\tfrac{37}2$$ which, of course, is a circle centred on $(\frac72,-\frac52)$ with radius $\sqrt{\frac{37}2}$.

Answer 3

We note that $P$ is an orthocenter of the triangle $ABQ$. So $\angle BRQ = 90^{\circ}$ and thus $R$ describes a circle with diameter $AB=\sqrt{74}$. So the locus of point $R(x,y)$ is $$(x-{7\over 2})^2+(y+{5\over 2})^2 = {74\over 4}$$