Example 1
Prove that (4, -1), (6, 0), (7,2)
and (5, 1) are the vertices of rhombus. Is it a square?.
Solution:
Let the given points be A, B, C and D
respectively. .
Then, coordinates of mid-point of AC
are 
Coordinates of the mid-points of
BD are 
Thus, AC and BD have the same mid-point. Hence, ABCD is
a parallelogram. .
Now, AB = 
BC =
\
AB = BC. .
So, ABCD is a parallelogram whose
adjacent sides are equal
Hence, ABCD is a rhombus. .
Now, AC = 
and BD =
.
Clearly, AC >
BD. So, ABCD is not a square. .
Eaxmple 2
Let P (sinq, cosq) (0 £
q £ 2p)
be a point and let OAB be a triangle with vertices (0, 0),
and 
Find q if P lies inside the DOAB.
Solution:
Equations of lines
along OA, OB and AB are y = 0, x = 0, and x+y =
respectively. Now P and
B will lie on the same side of y = 0 if cosq > 0.
Similarly P and A will lie on the same side of x = 0 if sin q
> 0 and P and O will lie on the same side of x + y = if sin q
+ cosq < .
Hence P will lie inside the D ABC,
if sinq
> 0, cosq > 0 and sinq
+ cosq <.
|
Now sinq
+ cosq <
&⇒
sin ( q + p /4) < 
i.e. 0 < q
+ p/4 < p/3 .
or  < q
+ p/4 < p
Since sinq>0
and cos q > 0, so
0< q<
|
|
Example 3
A
line intersects the straight lines 5x – y – 4 = 0 and 3x – 4y – 4 = 0 at A and
B respectively. If a point P (1, 5) on the line AB is such that
AP: PB = 2 : 1 (internally), find the point A.
Solution:
|
Let
AP = 2r. Then PB = r.
Let
the slope of AB be tanq. .
The
equation of the line AB is
 = r, so that
A
º (1 + 2rcosq, 5 + 2rsinq)
and
Bº (1 – rcosq, 5 – rsinq)
or
Aº (1 - 2rcosq, 5 - 2rsinq)
and
Bº (1 + rcosq, 5 + rsinq)
The
point ‘A’ lies on the line
5x
– y – 4 = 0, So that
|
|
5 (1 + 2rcosq)
- (5 +2rsinq) - 4 = 0 ….(i).
The Point ‘B’ lies on
the line 3x – 4y – 4 0,
So that 3 (1–rcosq)
+ 4(5–rsinq) – 4 = 0 ….(ii).
Let rcosq
= a and rsinq = b. .
&⇒
10a – 2b – 4 = 0 and –3a
– 4b + 19 = 0
On solving for a
and b, we get 
Hence the point A is 
Similarly when A º
(1 - 2rcosq, 5 - 2rsinq)
and B º
(1 + rcosq, 5 + rsinq),
we get
- 10a
+ 2b - 4 = 0 and 3a +4b +19 = 0
&⇒
a
= 
, b
= 
Hence the point A is .
Example 4
Four points A(a, 0), B(b, 0), C(g, 0) and D(d, 0), with a < b < g
< d or
a > b > g > d
, are such that a and b are the roots of ax2
+2hx + b = 0 and g and d are the roots of a¢x2 +2h¢x + b¢ = 0. If 
= l, 
= m and
ab¢ + ba¢ = 2hh¢, show that l
+ m =0.
Solution:
We have a
+ b = -2h/a, ab
= b/a. Also, g + d = 
, gd = 
&⇒ l
= 
. Also, m
= 
, l
+ m = +
= 
= 
= 
= 
= 
= 0.
Example 5
Find
the points farthest and nearest on the curve 5x2 + 5y2 +
6xy– 8 = 0 from origin. Also determine the equations of the lines through the
origin on which the points are occurring. .
Solution:
Any line through the
origin is 
Any point on this line
is
(r cosq, r sinq). If this point lies on the given
curve,
then 5r2 + 6r2sinq
cosq = 8
&⇒
r2 = 
&⇒
rmax = 2 (for q= 
) and rmin =
1 (for q = p/4).
The required points are
(
) and (
). The lines are
y = – x and y = x.
&⇒
