## An immortal genius Professor Stephen Hawking is no more in this world

Some tributes that I can only humbly share …

## 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

## Some random problems/solutions in Coordinate Geometry II: IITJEE mathematics training

Question I:

Find the equation of the tangent to the circle $x^{2}-y^{2}-4x-8y+16=0$ at the point $(2+\sqrt{3},3)$. If the circle rolls up along this tangent by 2 units, find its equation in the new position.

Solution I:

The centre $C_{1}$ of the given circle is $(2,4)$ and its radius is 2. Equation of the tangent at $A(2+\sqrt{3},3)$ to the circle is

$x(2+\sqrt{3})+3y-2(x+2+\sqrt{3})-4(y+3)+16=0$

or $\sqrt{3}x-y-2\sqrt{3}=0$.

The slope of this line is $\sqrt{3}$ showing that it makes an angle of 60 degrees with the x-axis. After the circle rolls up along the tangent at A through a distance 2 units, its centre moves from $C_{1}$ to $C_{2}$. We now find the co-ordinates of $C_{2}$. Since $C_{1}C_{2}$ is parallel to the tangent at A and it passes through $C_{1}$$(2,4)$, its equation is $\frac{x-2}{\cos {\theta}} = \frac{y-4}{\sin {\theta}}$, where $\theta=60 \deg$; $C_{2}$ being at a distance 2 units on this line from $C_{1}$; its co-ordinates are

$(2\cos {\theta}+2, 2\sin{\theta}+4)$, that is, $(3, 4+\sqrt{3})$.

Hence, the equation of the circle in the new position is

$(x-3)^{2}+(y-(4+\sqrt{3})^{2})=2^{2}$, which in turn implies that

$x^{2}+y^{2}-6x-2(4+\sqrt{3})y+8(3+\sqrt{3})=0$.

Question 2:

A triangle has two of its sides along the axes, its third side touches the circle $x^{2}+y^{2}-2ax-2ay+a^{2}=0$. Prove that the locus of the circumcentre of the triangle is

$a^{2}-2a(x+y)+2xy=0$.

Solution 2:

The given circle has its centre at $C(a,a)$ and its radius is a so that it touches both the axes along which lie the two sides of the triangle. Let the third side be $\frac{x}{p} + \frac{y}{p}=1$.

So that A is $(p,a)$ and B is $(a,q)$ and the line AB touches the given circle. Since $\angle AOB$ is a right angle, AB is diameter of the circumcentre of the triangle AOB. So, the circumcentre $P(h,k)$ of the triangle AOB is the mid-point of AB,

that is, $2h=p$, $2k=q$.

Now, the equation of AB is $\frac{x}{p} + \frac{y}{q}=1$, which touches the given circle,

$\frac{a(p+q)-pq}{\sqrt{p^{2}+q^{2}}}=a$

$\Longrightarrow a^{2}(p+q)^{2}+p^{2}-2apq(p+q)=a^{2}(p^{2}+q^{2})$

$\Longrightarrow 2a^{2}-2a(p+q)+pq=0$

$2a^{2}-2a(2h+2k)+2h-2k=0$.

Hence, the locus of $P(h,k)$ is $a^{2}-2a(x+y)+2xy=0$.

Question 3:

A circle of radius 2 units rolls on the outerside of the circle $x^{2}+y^{2}+4x=0$, touching it externally. Find the locus of the centre of this outside circle. Also, find the equations of the common tangents of these two circles when the line joining the centres of the two circles make an angle of 60 degrees with x-axis.

Solution 3:

The centre C of the given circle is $(-2,0)$ and its radius is 2. Let $P(h,k)$ be the centre of the outer circle touching the given circle externally then $CP=2+2=4$, which in turn implies,

$(h+2)^{2}+k^{2}=4^{2}$

So, the locus of P is $(x+2)^{2}+y^{2}=16$, or $x^{2}+y^{2}+4x-12=0$.

Since the two circles touch each other externally,, there are 3 common tangents to these circles.

One will be perpendicular to the line joining the centres and the other two will be parallel to the line joining the centres as the radii of the two circles are equal, co-ordinates of P are given by

$\frac{h+2}{\cos{60 \deg}} = \frac{k-0}{\sin{60 \deg}}=4 \Longrightarrow h=0, k=2\sqrt{3}$,

co-ordinates of M, the mid-point of CP is $(-1,\sqrt{3})$.

Hence, the equation of the common tangent perpendicular to CP is

$y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x+1) \Longrightarrow x+\sqrt{3}y-2=0$.

Let the equation of the common tangent parallel to CP be $\sqrt{3}x-y+\lambda=0$.

Since it touches the given circle $\frac{}{}= \pm 2 \Longrightarrow \lambda = 2\sqrt{3} \pm 4$.

Hence, the other common tangents are $\sqrt{3}x -y \pm 4 + 2\sqrt{3}=0$.

Question 4:

If $S=0$ and $S^{'}=0$ are the equations of two circles with radii r and $r^{'}$ respectively, then show that the circles $\frac{S}{r} \pm \frac{S^{'}}{r}=0$ cut orthogonally.

Solution 4:

Let the line of centres of the given circle be taken as the x-axis and its mid-point as the origin…Note this is the key simplifying assumption.

If the distance between the centres is 2a, the co-ordinates of the centre are $(a,0)$ and $(-a,0)$. Hence, we get the following:

$S \equiv (x-a)^{2}+y^{2}-r^{2}=0$, that is,

$S^{'} \equiv x^{2}+y^{2}-2ax +a^{2}-r^{2}=0$

and $S^{'} \equiv x^{2}+y^{2}+2ax+a^{2}-(r^{'})^{2}=0$ so that $\frac{}{} + \frac{}{}=0 \Longrightarrow Sr^{'}+S^{'}r^{'}=0$, that is,

$\Longrightarrow (r+r^{'})(x^{2}+y^{2}+a^{2})-2ax(r^{'}-r)-rr^{'}(r+r^{'})=0$

$\Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}-r}{r^{'}+r}x + a^{2}-rr^{'}=0$…call this I.

and $\frac{S}{r} - \frac{S^{'}}{r^{'}}=0$ and in turn $\Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}+r}{r^{'}-r}+a^{2}+rr^{'}=0$…call this II.

Now, since $2(\{ -a\frac{r^{'}-r}{r^{'}+r}\})(\{-a\frac{r^{'}+r}{r^{'}-r} \})+ 2\times 0 \times 0 =2a^{2}=(a^{2}-rr^{'}) +(a^{2}+rr^{'})$.

The circles I and II intersect orthogonally.

Question 5:

Let P, Q, R, S be the centres of the four circles each of which is cut by a fixed circle orthogonally. If $t_{1}^{2}$, $t_{2}^{2}$, $t_{3}^{2}$, $t_{4}^{2}$ be the squares of the lengths of the tangents to the four circles from a point in their plane, then prove that

$t_{1}^{2}\Delta QRS +t_{2}^{2}\Delta RSP + t_{3}^{2}\Delta SPQ + t_{4}^{2}\Delta PQR=0$

Solution 5:

Let the equations of the four circles be

$x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$, $i=1,2,3,4$, then centres of these circles are as follows:

$P(-g_{1},-f_{1})$, $Q(-g_{2},-f_{2})$, $R(-g_{3},-f_{3})$, and $S(-g_{4},-f_{4})$

Let the fixed point in the plane be taken as the origin, then $t_{1}^{2}=c_{1}$, $t_{2}^{2}=c_{2}$, $t_{3}^{2}=c_{3}$ and $t_{4}^{2}=c_{4}$. Let the equation of the fixed circle cutting the four circles orthogonally be

$x^{2}+y^{2}+2gx + 2fy +c=0$, then $2gg_{1}+2ff_{1}=c+c_{1}=c+t_{1}^{2}$, or

we get the following:

$2gg_{i}+2ff_{i}-c-t_{i}^{2}=0$, for $i=1,2,3,4$.

Eliminating the unknowns g, f, c we get

$\left| \begin{array}{cccc} 2g_{1} & 2f_{1} & -1 & -t_{1}^{2} \\ 2g_{2} & 2f_{2} & -1 & -t_{2}^{2}\\ 2g_{3} & 2f_{3} & -1 & -t_{3}^{2}\\ 2g_{4} & 2f_{4} & -1 & -t_{4}^{2} \end{array} \right|$

or, $t_{1}^{2}|D_{1}|-t_{2}^{2}|D_{2}|+t_{3}^{2}|D_{3}|-t_{4}^{2}|D_{4}|=0$

where $|D_{1}|= \left| \begin{array}{ccc} g_{2} & f_{2} & 1\\ g_{1} & f_{1} & 1 \\ g_{4} & f_{4} & 1 \end{array} \right |=2\Delta QRS$,

and $|D_{2}|=2\Delta PRS$, $|D_{3}|=2\Delta PQS$ and $|D_{4}|=2\Delta PQR$

Hence, we get the following:

$t_{1}^{2}\Delta QRS + t_{2}^{2}\Delta RSP + t_{3}^{2}SPQ + t_{4}^{2}PQR=0$.

Homework Quiz Coordinate Geometry:

1. OAB is any chord of a circle which passes through O, a point in the plane of the circle and meets it in points A and B. A point P is taken on this chord such that OP is (i) arithmetic mean (ii) geometric mean of OA and OB. Prove that the locus of P in either case is a circle. Determine the circle.
2. Let $2x^{2}+y^{2}-3xy=0$ be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length OA.
3. Let P, Q and R be the centres and $r_{1}, r_{2}, r_{3}$ are the radii respectively of three coaxial circles. Show that $r_{1}^{2}QR + r_{2}^{2}RP + r_{3}^{2}PQ=-PQ. QR. RP$
4. If ABC be any triangle and $A^{'}B^{'}C^{'}$ be the triangle formed by the polars of the points A, B, C with respect to a circle, so that $B^{'}C^{'}$ is the polar of A; $C^{'}A^{'}$ is the polar of B and $A^{'}B^{'}$ is the polar of C. Prove that the lines $AA^{'}$, $BB^{'}$ and $CC^{'}$ meet in a point.

That’s all, folks !

Nalin Pithwa.

## Listening math : Gaurish Korpal way :-) :-) :-)

https://gaurish4math.wordpress.com/2018/02/17/listening-maths/

Thanks Gaurish —

From — Nalin Pithwa.

## Some random sample problems-solutions in Coordinate Geometry: I: IITJEE Mains Maths tutorials

Question I:

The point $(4,1)$ undergoes the following transformations, successively:

a) reflection about the line $y=x$.

b) translation through a distance 2 units along the positive directions of the x-axis.

c) rotation through an angle of $\pi/4$ about the origin in the anticlockwise direction.

d) reflection about $x=0$.

Hint: draw the diagrams at very step!

Ans: $(1/\sqrt{2}, 7/\sqrt{2})$

Question 2:

$A_{1}, A_{2}, A_{3}, \ldots, A_{n}$ are n points in a plane whose co-ordinates are $(x_{1}, y_{1})$, $(x_{2},y_{2})$, $\ldots$, $(x_{n},y_{n})$ respectively. $A_{1}$, $A_{2}$ is bisected at the point $G_{1}$, $G_{1}A_{3}$ is divided in the ratio 1:2 at $G_{2}$, $G_{2}A_{4}$ is divided in the ratio $1:3$ at $G_{3}$, $G_{3}A_{3}$ is divided in the ratio $1:4$ at $G_{4}$ and so on until all n points are exhausted. Show that the co-ordinates of the final point so obtained are

$(\frac{1}{n}(x_{1}+x_{2}+ \ldots + x_{n}) , \frac{1}{n}(y_{1}+y_{2}+ \ldots + y_{n}) )$.

Solution 2:

The co-ordinates of $G_{1}$ are $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$.

Now, $G_{2}$ divides $G_{1}A_{3}$ in the ratio $1:2$. Hence, the co-ordinates of $G_{2}$ are

$( \frac{1}{3}(\frac{2(x_{1}+x_{2})}{2}+x_{3}), \frac{1}{3}(\frac{3(y_{1}+y_{2})}{2}+y_{3}))$, or $(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3})$.

Again, $G_{3}$ divides $G_{2}A_{4}$ in the ratio $1:4$. Therefore, the co-ordinates of $G_{3}$ are $(\frac{1}{4}(\frac{3(x_{1}+x_{2}+x_{3})}{3}+x_{4}) ,\frac{1}{4}(\frac{3(y_{1}+y_{2}+y_{3})}{3}+y_{4}) )$, or

$( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4} )$.

Proceeding in this manner,we can show that the coordinates of the final point obtained will be

$(\frac{1}{n}(x_{1}+x_{2}+x_{3}+\ldots + x_{n}),\frac{1}{n}(y_{1}+y_{2}+y_{3}+\ldots + y_{n}))$.

Remark: For a rigorous proof, prove the above by mathematical induction.

Question 3:

A line L intersects the three sides BC, CA, and AB of a triangle ABC at P, Q and R, respectively. Show that $\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1$

Solution 3:

Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, and $C(x_{3},y_{3})$ be the vertices of $\triangle ABC$, and let $lx+my+n=0$ be equation of the line L. If P divides BC in the ratio $\lambda:1$, then the coordinates of P are $(\frac{\lambda x_{3}+x_{2}}{\lambda + 1} ,\frac{\lambda y_{3}+y_{2}}{\lambda + 1})$.

Also, as P lies on L, we have $l(\frac{\lambda x_{3}+x_{2}}{\lambda + 1})+m(\frac{\lambda y_{3}+y_{2}}{\lambda + 1})+n=0$

$\Longrightarrow \frac{lx_{2}+my_{2}+n}{lx_{3}+my_{3}+n}=\lambda=\frac{BP}{PC}$…..call this relation I.

Similarly, we can obtain $\frac{CQ}{QA}=-\frac{lx_{3}+my_{3}+n}{lx_{1}+my_{1}+n}$….call this relation II.

and so, also, we can prove that $\frac{AR}{RB}=-\frac{lx_{1}+my_{1}+n}{lx_{2}+my_{2}+n}$…call this III.

Multiplying, I, II and III, we get the desired result.

The above is the famous Menelaus’s theorem of plane geometry proved with elementary tools of co-ordinate geometry. As a homework quiz, try proving the equally famous Ceva’s theorem of plane geometry with elementary tools of co-ordinate geometry.

Question 4:

A triangle has the lines $y=m_{1}x$ and $y=m_{2}x$ as two of its sides, with $m_{1}$ and $m_{2}$ being roots of the equation $bx^{2}+2hx+a=0$. If $H(a,b)$ is the orthocentre of the triangle, show that the equation of the third side is $(a+b)(ax+by)=ab(a+b-2h)$.

Solution 4:

Since the given lines intersect at the origin, one of the triangle lies at the origin O(0,0). Let OA and OB be the given lines $y=m_{1}x$ and $y=m_{2}x$, respectively. Let the equation of AB be $lx+my=1$. Now, as OH is perpendicular to AB, we have

$\frac{b}{a}=\frac{m}{l}$, $\Longrightarrow \frac{l}{a}=\frac{m}{b}=k$, say…call this equation I

Also, the coordinates of A and B are respectively,

$(\frac{1}{l+mm_{1}}, \frac{m_{1}}{l+mm_{1}})$ and $(\frac{1}{l+mm_{2}} , \frac{m_{2}}{l+mm_{2}})$

Therefore, the equation of AB is

$(y-\frac{m_{1}}{l+mm_{1}})=-\frac{1}{m_{2}}(x-\frac{1}{l+mm_{1}})$

or $x+m_{2}y=\frac{1+m_{1}m_{2}}{1+mm_{1}}$…call this II.

Similarly, the equation of BH is $x+m_{1}y=\frac{1+m_{1}m_{2}}{1+mm_{2}}$….call this III.

Solving II and III, we get the coordinates of H. Subtracting III from II, we get

$y=\frac{(1+m_{1}m_{2})m}{l^{2}+lm(m_{1}+m_{2})+m^{2}m_{1}m_{2}}$

Since $m_{1}$ and $m_{2}$ are the roots of the equation $bx^{2}+2hx+a=0$, we have $m_{1}+m_{2}=-\frac{2h}{b}$ and $m_{1}m_{2}=a/b$.

$\Longrightarrow y=\frac{(a+b)m}{bl^{2}-2hlm+am^{2}} \Longrightarrow \frac{m}{b}=\frac{bl^{2}-2hlm+am^{2}}{a+b}$ because y=b for H.

$\Longrightarrow k=\frac{k^{2}(ba^{2}-2hab+ab^{2})}{a+b} \Longrightarrow k=\frac{a+b}{ab(a-2h+b)}$.

Hence, the equation of AB is

$ax+by=\frac{1}{k}=\frac{ab(a+b-2h)}{a+b}$

$\Longrightarrow (a+b)(ax+by)=ab(a+b-2h)$

More later,

Nalin Pithwa.

## World Maths day : Mar 7 competitions

Thanks to Ms. Colleen Young:

https://colleenyoung.wordpress.com/2018/02/15/world-maths-day-2018/