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.