Pair of straight lines
The combined equation
of a pair of straight lines L1 = a1x + b1y + c1
= 0 and
L2 = a2x + b2y + c2 = 0 is (a1x
+ b1y + c1) (a2x + b2y + c2)
= 0 i.e. L1L2 = 0. Opening the brackets and comparing the
terms with the terms of general equation of 2nd degree
ax2 + 2hxy + by2 + 2gx +2fy + c= 0, we can get all the
following results for pair of
straight lines. .
The general equation of
second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
represents a pair of straight lines if and h2
≥
ab.
&⇒
abc + 2fgh - af2 - bg2 - ch2 = 0 and h2
≥
ab.
The homogeneous second
degree equation ax2 + 2hxy + by2 = 0 represents a pair of
straight lines through the origin if h2 ≥
ab.
If the lines through
the origin whose joint equation is ax2 + 2hxy + by2 = 0,
are y = m1x and y = m2x, then
y2
- (m1 + m2)xy + m1m2x2 =
0 and y2 + 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 h2 = ab.
In the more general case, the lines
represented by ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will
be perpendicular if a + b = 0, parallel if the terms of second degree make a
perfect square i.e. ax2 + 2hxy + by2 gets converted into
(l1x ± m1y)2,
coincident if the whole equation makes a perfect square i.e. ax2 +
2hxy +by2 + 2gx +2fy + c can be written as (lx + my + n)2 .
.
Note:
Point of intersection
of the two lines represented by ax2+2hxy+by2+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 ax2 + by2 + 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
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. These are the lines
concurrent with the lines represented by given equation.
Joint equation of pair of lines
joining the origin and the points of intersection of a curve and a line:
If the line lx + my +
n = 0, ((n > 0) i.e. the line does not pass
through origin) cuts the curve ax2 + 2hxy + by2 + 2gx +
2fy + c = 0 at two points A and B, then the joint equation of straight lines
passing through A and B and the origin is given by homogenizing the equation
of the curve by the equation of the line. i.e. .
ax2 + 2hxy
+ by2 + (2gx + 2fy)
is the equation of
the lines OA and OB
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Example 1
Find the
orthocentre of a triangle the coordinates of whose vertices A, B, C are (at1t2,
a(t1 + t2)), (at2t3, a(t2
+ t3)) and (at3t1, a(t3 + t1))
respectively.
Solution:
Slope
of BC =
&⇒ Slope of AD = -t3.
&⇒
Equation of AD is y - a(t1 + t2) = -t3(x -
at1t2)..(1).
Similarly
equation of CF is
y
- a(t3 + t1) = -t2(x - at3t1) (2)
Subtracting
(1) from (2) we get
x = -a &⇒
y = a(t1 + t2 + t3 + t1t2t3).
Hence the
orthocentre of the triangle is
(-a,
a(t1 + t2 + t3 + t1t2t3)).
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Example 2
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 .
Solution:
Since the given lines are concurrent
Operating C2 →
C2 – C1 and C3 → C3
– C1, we get
&⇒
a (b –1) (c–1) – (b–1) (1–a) – (c–1) (1–a) = 0
&⇒
&⇒
.
Example 3
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 = c2 are at right angles
if h2 + k2 = c2
Solution:
Making the equation of the curve
homogeneous with the help of the line, we get
or 4h2k2x2
+ 4h2k2y2 – 4hk2x (hx + ky) – 4h2ky
(hx + ky) + (h2+k2-c2) (h2x2
+ k2y2 + 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 x2
+ coefficient of y2 = 0.
(h2 + k2)
(h2 + k2 – c2) = 0 since [h2 + k2
>
0]
&⇒ h2 + k2 = c2