**Problem I:**

If , and are the vertices of a triangle ABC, then prove that the **equation of the median** through A is given by:

**Solution I:**

If D is the mid-point of BC, its co-ordinates are

Therefore, **equation of the median AD** is , which in turn, implies that,

Now apply the row transformation to the previous determinant. So, we get

, using the sum property of determinants.

Hence, the proof.

**Problem 2:**

If is the **area of the triangle** with vertices , and is the **area of the triangle** with vertices , , and , and is the **area of the triangle** with vertices , , . Then, prove that there is no value of for which the areas of triangles, , and are in GP.

**Solution 2:**

We have , and

.

Applying the following column transformations to the above determinant, and , we get

and

so that .

Now, , and are in GP, if

, that is,

, where . But, for this value of , the vertices of the given triangles are not defined. Hence, , and and cannot be in GP for any value of .

**Problem 3:**

Two points P and Q are taken on the line joining the points and such that . Circles are drawn on AP, PQ, and QB as diameters. The locus of the point S, the sum of the squares of the length of the tangents from which to the three circles is equal to , is

(a)

(b)

(c)

(d) .

**Ans. b.**

**Solution 3:**

Since , the co-ordinates of P are and of Q are , equations of the circles on AP, PQ, and QB as diameters are respectively.

*Please draw the diagram.*

So, we get

So, if be any point of the locus, then .

So, the required locus of is .

More later,

Nalin Pithwa.