The combined equation
of a pair of straight lines L_{1 }= a_{1}x + b_{1}y + c_{1}
= 0 and

L_{2 }= a_{2}x + b_{2}y + c_{2} = 0 is (a_{1}x
+ b_{1}y + c_{1}) (a_{2}x + b_{2}y + c_{2})
= 0 i.e. L_{1}L_{2} = 0. Opening the brackets and comparing the
terms with the terms of general equation of 2^{nd} degree

ax^{2} + 2hxy + by^{2} + 2gx +2fy + c= 0, we can get all the
following results for pair of

straight lines. .

The general equation of
second degree ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0
represents a pair of straight lines if and h^{2}
≥
ab.

&⇒
abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0 and h^{2}
≥
ab.

The homogeneous second
degree equation ax^{2} + 2hxy + by^{2} = 0 represents a pair of
straight lines through the origin if h^{2} ≥
ab.

If the lines through
the origin whose joint equation is ax^{2} + 2hxy + by^{2} = 0,
are y = m_{1}x and y = m_{2}x, then

y^{2}
- (m_{1} + m_{2})xy + m_{1}m_{2}x^{2} =
0 and y^{2} + xy + = 0
are identical , so that

.

If q be the angle between two lines, through the origin, then

= ±.

The lines are
perpendicular if a + b = 0 and coincident if h^{2} = ab.

In the more general case, the lines
represented by ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 will
be perpendicular if a + b = 0, parallel if the terms of second degree make a
perfect square i.e. ax^{2} + 2hxy + by^{2} gets converted into
(l_{1}x ± m_{1}y)^{2},
coincident if the whole equation makes a perfect square i.e. ax^{2} +
2hxy +by^{2} + 2gx +2fy + c can be written as (lx + my + n)^{2 }.
.

Note:

Point of intersection
of the two lines represented by ax^{2}+2hxy+by^{2}+2gx+2fy+c=0
is obtained by solving the equations =ax +
hy + g = 0 and =
hx + by + f = 0 where denotes
the derivative of f with respect to x, keeping y constant anddenotes
the derivative of f with respect to y, keeping x constant. The fact can be used
in splitting ax^{2} + by^{2} + 2hxy + 2gx + 2fy + c = 0 into
equations of two straight lines. With the above method, the point of
intersection can be found. Now only the slopes need to be determined.

It should be noted that
the line ax + hy + g = 0 and hx + by + f = 0 are not the lines represented by
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0. These are the lines
concurrent with the lines represented by given equation.

If the line lx + my +
n = 0, ((n > 0) i.e. the line does not pass
through origin) cuts the curve ax ax is the equation of the lines OA and OB |

Find the
orthocentre of a triangle the coordinates of whose vertices A, B, C are (at_{1}t_{2},
a(t_{1} + t_{2})), (at_{2}t_{3}, a(t_{2}
+ t_{3})) and (at_{3}t_{1}, a(t_{3} + t_{1}))
respectively.

Slope of BC =
Similarly equation of CF is y
- a(t Subtracting (1) from (2) we get x = -a Hence the orthocentre of the triangle is (-a,
a(t |

If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0 (a, b and c being distinct and different from 1) are concurrent, then prove that .

Since the given lines are concurrent

Operating C_{2} →
C_{2} – C_{1} and C_{3} → C_{3}
– C_{1}, we get

&⇒
a (b –1) (c–1) – (b–1) (1–a) – (c–1) (1–a) = 0

&⇒

&⇒
.

Prove that the
straight lines joining the origin to the points of intersection of the straight
line hx + ky = 2hk and the curve

(x – k)^{2} + (y – h)^{2} = c^{2} are at right angles
if h^{2} + k^{2} = c^{2}

Making the equation of the curve homogeneous with the help of the line, we get

or 4h^{2}k^{2}x^{2}
+ 4h^{2}k^{2}y^{2} – 4hk^{2}x (hx + ky) – 4h^{2}ky
(hx + ky) + (h^{2}+k^{2}-c^{2}) (h^{2}x^{2}
+ k^{2}y^{2} + 2hkxy) = 0

This is the equation of
the pair of lines joining the origin to the points of intersection of the given
line and the curve. They will be at right angles if coefficient of x^{2}
+ coefficient of y^{2} = 0.

(h^{2} + k^{2})
(h^{2} + k^{2} – c^{2}) = 0 since [h^{2} + k^{2}
>
0]

&⇒ h^{2} + k^{2} = c^{2}