Problem 1:
The line joining
and
is produced to the point
so that
, then find the value of
.
Solution 1:
As M divides AB externally in the ratio
, we have
and
which in turn



Problem 2:
If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points
and
, then where does the orthocentre lie?
Solution 2:
From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre
and the centroid
, that is,
, or
. That is, the orthocentre lies on this line.
Problem 3:
If a, b, c are unequal and different from 1 such that the points
,
and
are collinear, then which of the following option is true?
a: 
b: 
c: 
d: 
Solution 3:
Suppose the given points lie on the line
then a, b, c are the roots of the equation :
, or

and
, that is, 
Eliminating l, m, n, we get 
, that is, option (d) is the answer.
Problem 4:
If
and
are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points
with
lie?
Solution 4:
Note: Centre of Mean Position is
.
Let the coordinates of the centre of mean position of the points
,
be
then
and 
, 
and 
, and 
, that is, the CM lies on this line.
Problem 5:
The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of
?
Solution 5:
Equation of the line L in the two coordinate systems is
, and
where
are the new coordinate of a point
when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.


or
. So, the value is zero.
Problem 6:
Let O be the origin,
and
and
are points such that
and
, then which of the following options is true:
a: P lies either inside the triangle OAB or in the third quadrant
b: P cannot lie inside the triangle OAB
c: P lies inside the triangle OAB
d: P lies in the first quadrant only.
Solution 6:
Since
, P either lies in the first quadrant or in the third quadrant. The inequality
represents all points below the line
. So that
and
imply that either P lies inside the triangle OAB or in the third quadrant.
Problem 7:
An equation of a line through the point
whose distance from the point
has the greatest value is :
option i: 
option ii: 
option iii: 
option iv: 
Solution 7:
Let the equation of the line through
be
. If p denotes the length of the perpendicular from
on this line, then 
, say
then
is greatest if and only if s is greatest.
Now, 
so that
. Also,
, if
, and
, if 
and
, if
. So s is greatest for
. And, thus, the equation of the required line is
.
Problem 8:
The points
,
, Slatex C(4,0)$ and
are the vertices of a :
option a: parallelogram
option b: rectangle
option c: rhombus
option d: square.
Note: more than one option may be right. Please mark all that are right.
Solution 8:
Mid-point of AC = 
Mid-point of BD = 
the diagonals AC and BD bisect each other.
ABCD is a parallelogram.
Next,
and
and since the diagonals are also equal, it is a rectangle.
As
and
, the adjacent sides are not equal and hence, it is neither a rhombus nor a square.
Problem 9:
Equations
and
will represent the same line if
option i: 
option ii: 
option iii: 
option iv: 
Solution 9:
The two lines will be identical if there exists some real number k, such that
, and
, and
.
or 
or
, and
or 
That is,
or
, or
.
Next,
. Hence,
, or
.
Problem 10:
The circumcentre of a triangle with vertices
,
and
lies at the origin, where
and
. Show that it’s orthocentre lies on the line 
Solution 10:
As the circumcentre of the triangle is at the origin O, we have
, where r is the radius of the circumcircle.
Hence, 
Therefore, the coordinates of A are
. Similarly, the coordinates of B are
and those of C are
. Thus, the coordinates of the centroid G of
are
.
Now, if
is the orthocentre of
, then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.
Hence, 

because
.
Hence, the orthocentre
lies on the line
.
Hope this gives an assorted flavour. More stuff later,
Nalin Pithwa.