Example 1
The
equations of the sides of a triangle are y = m_{1}x + c_{1 }, y
= m_{2}x + c_{2} and
x = 0. Prove that the area of the triangle is
Solution:
The sides of the triangle are y = m_{1}x
+c_{1}, . . . . . (1).
y = m_{2}x +c_{2} .
. . . . (2).
and x = 0 . .
. . . . (3).
On solving (1) and (2), we get one
vertex as .
Similarly other vertices are (0, c_{1})
and ( 0, c_{2})
Area of triangle = = .
Example 2
A
variable line through the point (6/5, 6/5) cuts the coordinate axes at the
points A and B. If the point P divides AB internally in the ratio 2:1, show
that the locus of P is 5xy = 2 (2x + y).
Solution:
Let the equation of
the variable line be
This meets the coordinate axes at A (a, 0) and B (0, b). Let P (h, k) be the
point which divides AB in the ratio 2:1. Then the coordinates of P are
&⇒
h = &⇒
a = 3h


and b = . Here, a and b are the
variables.
Since (1) passes
through , … (ii)
Putting the value of a
and b in (ii), we get
&⇒
&⇒
2k + 4h = 5hk
&⇒ 5hk = 2(2h + k)
Hence the locus of (h,
k) is 5xy = 2(2x+ y)
Example 3
Find the condition
in a and b, such that the portion of the line
ax + by – 1 = 0, intercepted between the lines ax + y + 1 = 0 and x + by = 0 subtends
a right angle at the origin.
Solution:
Given lines are ax + y + 1 = 0
. . . (1).
x+ by = 0 .
. . (2).
ax+ by = 1
. . . (3).
Joint equation of (1) and (2) is (ax
+ y + 1) ( x+ by ) = 0
&⇒
ax^{2} + by^{2} + (ab + 1) xy + x + by = 0
Making (4) homogeneous with the help
of (1) we have
ax^{2} + by^{2} +
(ab+ 1) xy + x(ax + by) + by (ax +by) =0
since angle between these two lines
is 90°
\Coefficient
of x^{2} + coeff. of y^{2} = 0.
2a + b + b^{2} = 0 is the
required condition. .
Example 4
Find
the locus of the circumcentre of a triangle whose two sides are along the
coordinate axes and third side passes through the point of intersection of the
lines ax + by + c = 0 and lx + my + n = 0.
Solution:
Let the equation of the third line be
(ax + by + c) + l(lx
+ my + n) = 0 where l is a parameter.
It meets the xaxis at A where y = 0
and x = .
Also it meets the yaxis at B, where
x = 0 and y = .
The triangle OAB is a right angled
triangle. Its circumcentre is the midpoint of the hypotenuse. Let it be(a,
b).
&⇒
2a
= and
&⇒
Hence the locus of (a,b)
is
&⇒
2xy (ma – bl) + x (an – lc)+ y(mc – bn) = 0.
Example 5
Let the sides of
a parallelogram be U =a, U = b, V = a¢
and V = b¢ where U = lx
+my + n, V =l¢x = m¢y + n¢. Show that the equation of the diagonal through
the point of intersection of U = a and V =a¢
and of U = b and V = b¢ is given by the
equation: = 0 .
Solution:
Given that diagonal
AC passes through the points of intersection of
U = a and V = a¢, and U = b and V = b¢.
The equation of the line AC can be written as .


eliminating l
and m from these equations, we have
&⇒