## Co-ordinate Geometry : IITJEE Mains practice: some random problems again

Problem 1:

The line $Ax+By+C=0$ cuts the circle $x^{2}+y^{2}+ax+by+c=0$ at P and Q. The line $A^{'}x+B^{'}y+C^{'}=0$ cuts the circle $x^{2}+y^{2}+a^{'}x+b^{'}y+c^{'}=0$ at R and S. If P, Q, R and S are concyclic, prove that $\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0$.

Solution I;

An equation of a circle through P and Q is $x^{2}+y^{2}+ax+by+c +\lambda (Ax+By+C)=0$…call this equation I.

And, an equation of a circle through R and S is $x^{2}+y^{2}+a^{'}x+b^{'}y + c^{'}+\mu (A^{'}x+B^{'}y+C^{'})=0$…call this equation II.

If P, Q, R and S are concyclic, then I and II represent the same circle for same values of $\lambda$ and $\mu$. $\Longrightarrow a+ \lambda A=a^{'}+\mu A^{'}$ or $a-a^{'} + \lambda A - \mu A^{'}=0$

so also, $b + \lambda B = b^{'} + \mu B^{'}$ or $b-b^{'}+\lambda B - \mu B^{'}=0$ $c + \lambda C = c^{'} + \mu C^{'}$ or $c-c^{'} + \lambda C - \mu C^{'}=0$.

Eliminating $\lambda$ and $\mu$, we get the following: $\left | \begin{array}{ccc} a-a^{'} & A & -A^{'}\\ b-b^{'} & B & -B^{'}\\ c-c^{'} & C & -C^{'} \end{array} \right |=0$, that is, $\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0$

Problem II:

A straight line is such that the algebraic sum of the perpendiculars falling upon it from any number of fixed points is zero. Show that it always passes through a fixed point.

Solution II:

Let $(x_{i},y_{i})$ where $i=1,2,3,\ldots$ be n fixed points. Let $ax+by+c=0$ be the given line. Thus, as per given hypthesis, we have $\sum_{i=1}^{n}\frac{ax{i}+by_{i}+c}{\sqrt{(a^{2}+b^{2})}}=0 \Longrightarrow a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}y_{i}+nc=0 \Longrightarrow a\overline{x}+b\overline{y}+c=0$ where $\overline{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i}$ and $\overline{y}=\frac{}{}\sum_{i=1}^{n}y_{i}$

which shows that the given line passes through the fixed point $(\frac{1}{n}\sum_{i=1}^{n}x_{i}, \frac{1}{n}\sum_{i=1}^{n}y_{i})$.

Problem III:

The straight lines $L \equiv ax+by+c=0$ and $L_{1} \equiv a_{1}x+b_{1}y+c_{1}=0$ are intersecting. Find the straight line $L_{2}$ such that L is the bisector of the angle between $L_{1}$ and $L_{2}$.

Solution III:

Let the equation of the line $L_{2}$ be $L_{1}+ \lambda L=0 \Longrightarrow (a_{1}+\lambda a)x+(b_{1}+\lambda b)y+\lambda c=0$ where the slopes of $L_{2}, L, L_{1}$ are respectively $-\frac{a_{1}+\lambda a}{b_{1}+\lambda b}, \frac{-a}{b}, -\frac{a_{1}}{b_{1}}$.

Since L is the bisector of the angle between $L_{2}$ and $L_{1}$ we have $\frac{-(\frac{a_{1}+\lambda a}{b_{1}+\lambda b})+\frac{a}{b}}{1+\frac{a(a_{1}+\lambda a)}{b(b_{1}+\lambda b)}}=\frac{-\frac{a}{b}+\frac{a_{1}}{b_{1}}}{1+\frac{aa_{1}}{bb_{1}}}$ $\Longrightarrow \frac{-b(a_{1}+\lambda a)+a(a_{1}+\lambda b)}{b(b_{1}+\lambda b)+a(a_{1}+\lambda a)}=-\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}$ $\Longrightarrow \frac{ab_{1}-a_{1}b}{\lambda (a^{2}+b^{2})+aa_{1}+bb_{1}} = -\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}$ $\Longrightarrow \lambda = - \frac{2(aa_{1}+bb_{1})}{a^{2}+b^{2}}$

Hence, the equation of the required line $L_{1}$ is $(a^{2}+b^{2})(a_{1}x+b_{1}y+c_{1})=2(aa_{1}+bb_{1})(ax+by+c)$.

Problem IV:

If a, b are real numbers and $c>0$, find the locus represented by $|ay-bx|=c\sqrt{(x-a)^{2}+(y-b)^{2}}$.

PS: Please draw a right angled triangle PMA, with right angle at M, and P being $(x,y)$ and A being $(a,b)$.

Solution IV:

Let $x=a+r\cos {\theta}$ and $y=b+r\sin {\theta}$, then the given equation becomes $a\sin {\theta}-b\cos {\theta}=c$. $\Longrightarrow r\sin{(\theta-\alpha)}=c$ where $r=\sqrt{a^{2}+b^{2}}$ and $\tan {(\alpha)}=\frac{b}{a}$ which is the slope of $ay-bx$, which in turn implies $\frac{c}{r}=\sin (\theta -\alpha) \leq 1$ $\Longrightarrow c \leq r$, or $c \leq \sqrt{a^{2}+b^{2}}$. The given equation now becomes $\frac{|ay-bx|}{\sqrt{a^{2}+b^{2}}}=\frac{c}{\sqrt{a^{2}+b^{2}}}\sqrt{(x-a)^{2}+(y-b)^{2}}$….call this as relation I.

If M is the foot of the perpendicular from a point P(x,y) on the line $ay-bx=0$ and A is the point $(a,b)$ which clearly lies on this line, then from relation I, we have $\frac{PM}{PA}=\frac{c}{\sqrt{a^{2}+b^{2}}}=\sin {(\theta - \alpha)}$. Hence, the locus of P is a straight line through the point $(a,b)$ inclined at an angle $\arcsin {\frac{c}{\sqrt{a^{2}+b^{2}}}}$ with the line $ay-bx=0$.

Problem V:

Find the co-ordinates of the orthocentre of the triangle formed by the lines $y=0$ and $(1+t)x-ty+t(1+t)=0$ and $(1+u)x-uy+u(1+u)=0$, where $t \neq u$, and show that for all values of t and u, the orthocentre lies on the line $x+y=0$.

Solution V:

Let the equation of the side BC be $y=0$. Then, the coordinates of B and C are $(-t,0)$ and $(-u,0)$, respectively, where $(1+t)x-ty+t(1+t)=0$ and $(1+u)x-uy+u(1+u)=0$ are equations of AB and AC, respectively.

PS: Please draw the diagram on your own for a better understanding of the solution presented.

Now, equation of BE is $y={\frac{-u}{1+u}}(x+t)$…let us call this equaiton I.

And, equation of CF is $y=\frac{-t}{1+t}(x+u)$…let us call this equation II.

Solving I and II, we get the following: $x(\frac{u}{1+u}-\frac{t}{1+t})=\frac{tu}{1+t}-\frac{tu}{1+u}$, which in turn implies that $x=tu$ and $y=-tu$, so that the orthocentre is the point $(tu,-tu)$ which lies on the line $x+y=0$.

Cheers,

Nalin Pithwa

### One Comment

1. Rajender Sehgal