**Problem 1:**

The line cuts the circle at P and Q. The line cuts the circle at R and S. If P, Q, R and S are concyclic, prove that

.

**Solution I;**

An equation of a circle through P and Q is …call this equation I.

And, an equation of a circle through R and S is …call this equation II.

If P, Q, R and S are concyclic, then I and II represent the same circle for same values of and .

or

so also,

or

or .

Eliminating and , we get the following:

, that is,

**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 where be n fixed points. Let be the given line. Thus, as per given hypthesis, we have

where and

which shows that the given line passes through the fixed point .

**Problem III:**

The straight lines and are intersecting. Find the straight line such that L is the bisector of the angle between and .

**Solution III:**

Let the equation of the line be where the slopes of are respectively

.

Since L is the bisector of the angle between and we have

Hence, the equation of the required line is .

**Problem IV:**

If a, b are real numbers and , find the locus represented by .

PS: Please draw a right angled triangle PMA, with right angle at M, and P being and A being .

**Solution IV:**

Let and , then the given equation becomes .

where and which is the slope of , which in turn implies

, or . The given equation now becomes

….call this as relation I.

If M is the foot of the perpendicular from a point P(x,y) on the line and A is the point which clearly lies on this line, then from relation I, we have

. Hence, the locus of P is a straight line through the point inclined at an angle with the line .

**Problem V:**

Find the co-ordinates of the orthocentre of the triangle formed by the lines and and , where , and show that for all values of t and u, the orthocentre lies on the line .

**Solution V:**

Let the equation of the side BC be . Then, the coordinates of B and C are and , respectively, where and 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 …let us call this equaiton I.

And, equation of CF is …let us call this equation II.

Solving I and II, we get the following:

, which in turn implies that

and , so that the orthocentre is the point which lies on the line .

Cheers,

Nalin Pithwa