Pairs of straight lines
General equation of second degree
The general equation of first degree in x and y always represents a straight line.
Let A_{1}x + B_{1}y + C_{1} = 0.......................(i)
Let A_{2}x + B_{2}y + C_{2} = 0.......................(ii)
be the equations of two straight lines.
Now,
Combining these equations we have,
(A_{1}x + B_{1}y + C_{1}) (A_{2}x + B_{2}y + C_{2}) = 0........................(iii)
The coordinates of any point, which satisfy the equation (iii), will also satisfy either equation (i) or equation (ii).
Similarly,
the coordinates of any point, which satisfy any one of the equation (i) or (ii) will also satisfy the equation (iii).
Therefore, equation (iii) represents two separate straight lines (i) and (ii). In other words, equation (iii) represents a pair of the straight lines given by (i) and (ii). So, (iii) is the equation of a pair of lines.
Now,
Expanding the left hand side of equation (iii) we get,
A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} + (A_{1}C_{2} + A_{2}C_{1})x + (B_{1}C_{2} + B_{2}C_{1})y + C_{1}C_{2} = 0
If we put A_{1}A_{2} = a,A_{1}B_{2} + A_{2}B_{1} = 2h, B_{1}B_{2} = b, A_{1}C_{2} + A_{2}C_{1} = 2g, B_{1}C_{2} + B_{2}C_{1 }= 2f and C_{1}C_{2} = c, then the above equation becomes,
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0..........................(iv)
This equation is called the general equation of the second degree in x and y. Thus we see that the equation of a pair of lines is a second degree equation. But the converse of this statement is not always true. It means every second degree equations in x and y may not represent a pair of straight lines. Equations of second degree will represent a pair of straight lines only if the left hand side can be resolved into two linear factors.
Consider an equation y^{2}  3xy + 2x^{2} = 0. This equation is equivalent to (y  x) (y  2x) = 0.
So, the equation y^{2}  3xy + 2x^{2} = 0 represents the two straight lines y  x = 0 and y  2x = 0.
Similarly, the equation xy = 0 represents the two straight lines x = 0 and y = 0.
Again,
Consider an equation x^{2}  5x + 6 = 0 represents two straight lines x  2 = 0 and x  3 = 0.
And the equation x^{2}  y^{2} = 0 represents the two straight lines x + y = 0 and x  y = 0.
Condition that the general equation of second degree may represent a line pair
The general equation of second degree in x and y is
ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0
or, ax^{2} + (2hy + 2g)x + (by^{2} + 2fy + c) = 0
This is quadratic equation in x.
So, x = \(\frac {(2hy + 2g) ± \sqrt {(2hy + 2g)^2  4a(by^2 + 2fy + c)}}{2a}\)
or, x = \(\frac {(hy + g) ± \sqrt {(hy + g)^2  a(by^2 + 2fy + c)}}{a}\)
These two equations will be linear if
(hy + g)^{2}  a (by^{2} + 2fy + c) is a perfect square.
or, h^{2}y^{2} + 2ghy + g^{2}  aby^{2}  2afy  ac is a perfect square.
i.e. (h^{2}  ab) y^{2} + (2gh  2af) y + (g^{2}  ac) is a perfect square.
i.e. (2gh  2af)^{2}  4 (h^{2}  ab) (g^{2}  ac) = 0
i.e. g^{2}h^{2}  2ghaf + a^{2}f^{2}  g^{2}h^{2} + h^{2}ac + abg^{2}  a^{2}bc = 0
i.e. a (af^{2} + bg^{2} + ch^{2}  2fgh  abc) = 0
i.e. abc + 2fgh  af^{2}  bg^{2}  ch^{2} = 0
Hence, the general equation of second degree ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 may represent a line pair if abc + 2fgh  af^{2}  bg^{2}  ch^{2} = 0.
Quadratic Equation
Any equation in the form of ax^{2} + bx + c = 0 is called quadratic equation in x.
Multiplying both sides of this equation by 4a we get,
4a (ax^{2} + bx + c) = 0
or, 4a^{2}x^{2} + 4abx + 4ac = 0
or, (2ax)^{2} + 2 . 2ax . b + b^{2}  b^{2} + 4ac = 0
or, (2ax + b)^{2} = b^{2}  4ac
or, (2ax + b)^{2} = (\(\sqrt {b^2  4ac}\))^{2}
or, 2ax + b = ± \(\sqrt {b^2  4ac}\)
or, 2ax =  b ± \(\sqrt {b^2  4ac}\)
or, x = \(\frac { b ± \sqrt {b^2  4ac}}{2a}\)
Let α = \(\frac { b + \sqrt {b^2  4ac}}{2a}\) and β = \(\frac { b  \sqrt {b^2  4ac}}{2a}\).
Then α and β are called roots of the quadratic equation ax^{2} + bx + c = 0.
Now,
\begin{align*} \text{Sum of the roots (α + β)} &= \frac { b + \sqrt {b^2  4ac}}{2a} + \frac { b  \sqrt {b^2  4ac}}{2a}\\ &=  \frac ba\\ &= \frac {coefficient\;of\;x}{coefficient\;of\;x^2}\\ \end{align*}
\begin{align*} \text{Sum of the roots (αβ)} &= (\frac { b + \sqrt {b^2  4ac}}{2a}) (\frac { b  \sqrt {b^2  4ac}}{2a})\\ &= \frac ca\\ &= \frac {constant\;term}{coefficient\;of\;x^2}\\ \end{align*}
Homogeneous equation of second degree
An equation in x and y in which the sum of the power of x and y in every term is the same, is called homogenous equation. If this sum is two, then the equation is called a homogenous equation of second degree. The equation ax^{2} + 2hxy + by^{2} = 0 is the general homogenous equation of the second degree.
A homogeneous equation of the second degree represents a pair of straight lines which pass through the origin.
Proof: Consider the homogenous equation of the second degree.
ax^{2 }+ 2hxy + by^{2} = 0...........................(i)
or, by^{2} + 2hxy + ax^{2} = 0
If b ≠ 0, the equation can be written as y^{2} + \(\frac {2h}b\)xy + \(\frac ab\)x^{2} = 0
or, (\(\frac yx\))^{2} + \(\frac {2h}b\)(\(\frac yx\)) + \(\frac ab\) = 0
This is quadratic equation in \(\frac yx\). So it has two roots. Let these roots be m_{1} and m_{2}.
Then,
\(\frac yx\) = m_{1} and \(\frac yx\) = m_{2}
or, y = m_{1}x and y = m_{2}x.
These two equations are the equations of straight lines passing through the origin.
If b = 0, then equation (i) becomes
ax^{2} + 2hxy = 0
or, x(ax + 2hy) = 0 which represents two straight lines x = 0 and ax + 2hy = 0.
These two lines pass through the origin.
Hence, the homogenous equation of second degree always represents a pair of straight lines passing through the origin.
Angle between the line pair represented by ax^{2} + 2hxy + by^{2} = 0
Homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0
or, by^{2} + 2hxy + ax^{2} = 0
or, y^{2} + \(\frac{2h}{b}\)(\(\frac yx\)) + \(\frac ab\) = 0.........................(i)
This is quadratic equation in \(\frac yx\). So, it has two roots. Let these two roots be m_{1} and m_{2}.
Then,
\(\frac yx\) = m_{1} and \(\frac yx\) = m_{2}
or, y = m_{1}x and y = m_{2}x which are two seperate equations represente by the given equation ax^{2} + 2hxy + by^{2} = 0.
Now,
From the quadratic equation (i)
m_{1} + m_{2} = \(\frac {2h}b\) and m_{1}m_{2} = \(\frac ab\)
Now,
\begin{align*} m_1 m_2 &= \sqrt {(m_1 + m_2)^2  4m_1m_2}\\ &= \sqrt {\frac{4h^2}{b^2}  4\frac ab}\\ &= \sqrt {\frac {4h^2  4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2  ab}\\ \end{align*}
Let \(\theta\) be the angle between the lines y = m_{1}x and y = m_{2}x. Then,
\begin{align*} tan\theta &= ± \frac {m_1  m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2  ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2  ab}}{a + b}\\ \end{align*}
∴ \(\theta\) = tan^{1 }(± \(\frac {2\sqrt {h^2  ab}}{a + b}\))
Second Method:
We have ax^{2} + 2hxy + by^{2} = 0............................................(i)
Let the seperate equations represented by this equation be y = m_{1}x and y = m_{2}x.
Now,
The combined equation of theses equation is:
(y  m_{1}x) (y  m_{2}x) = 0
or, y^{2}  (m_{1} + m_{2})xy + m_{1}m_{2}x^{2} = 0
or, m_{1}m_{2}x^{2} (m_{1} + m_{2})xy + y^{2} = 0.......................................(ii)
Comparing (i) and (ii) we have,
\(\frac {m_1m_2}{a}\) = \(\frac { (m_1 + m_2)}{2h}\) = \(\frac 1b\)
Taking 1^{st} and last, m_{1}m_{2} = \(\frac ab\)
Taking 2^{nd} and last, m_{1} + m_{2} = \(\frac {2h}b\)
Now,
\begin{align*} m_1  m_2 &= \sqrt {(m_1 + m_2)^2  4m_1m_2}\\ &= \sqrt {\frac {4h^2}{b^2}  4\frac ab}\\ &= \sqrt {\frac {4h^2  4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2  ab}\\ \end{align*}
Let \(\theta\) be the angle between the lines y = m_{1}x and y = m_{2}x.
Then,
\begin{align*} tan\theta &=± \frac {m_1  m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2  ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2  ab}}{a + b}\\ \end{align*}
∴ \(\theta\) = tan^{1} (± \(\frac {2\sqrt {h^2  ab}}{a + b}\))
Condition that the straight lines given by the equation ax^{2} + 2hxy + by^{2} = 0 may be (1) Perpendicular and (2) Coincident.
 If a + b = 0 the value of tan\(\theta\) is∞ and hence \(\theta\) is 90°. Hence two straight lines represented by the equation ax^{2} + 2hxy + by^{2} = 0 are perpendicular to each other if coefficient of x^{2} + coefficient of y^{2} = 0.
For example: the equations x^{2}  y^{2} = 0 and 6x^{2} + 11xy  6y^{2}= 0 both represent pairs of straight lines at right angles.
Similarly, whatever be the value of h, the equation x^{2} + 2hxy  y^{2} = 0 represents a pair of straight lines at right angles.  If h^{2} = ab, the value of tan\(\theta\) is zero and hence \(\theta\) is zero. But both the straight lines represented by ax^{2} + 2hxy + by^{2} = 0 pass through the origin. So, the two lines are coincident.
Hence, two straight lines represented by the equation ax^{2} + 2hxy + by^{2} = 0 are coincident ifh^{2} = ab.
This may be seen directly from the original equation ax^{2} + 2hxy + by^{2} = 0. If h^{2} = ab i.e. h = \(\sqrt {ab}\), then the original equation becomes,
ax^{2} + 2\(\sqrt {ab}\)xy + by^{2} = 0
or, (\(\sqrt a\)x)^{2} + 2 \(\sqrt a\) \(\sqrt b\) xy + (\(\sqrt b\)y)^{2} = 0
or, (\(\sqrt a\)x + \(\sqrt b\)y)^{2} = 0 which gives two coincident straight lines
i.e.\(\sqrt a\)x + \(\sqrt b\)y = 0 and\(\sqrt a\)x + \(\sqrt b\)y = 0.
If the equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of lines, then ax^{2} + 2hxy + by^{2} = 0 represents a pair of lines through the origin parallel to the above pair.
Proof:
Let ax^{2}+ 2hxy + by^{2} + 2gx + 2fy + c = 0............................................................(i)
represent a pair of straight lines. Then the lefthand side can be resolved into two linear factors. Let these factors be A_{1}x + B_{1}y + C_{1}and A_{2}x + B_{2}y + C_{2} = 0.
Now
Combining equation of these equations is:
(A_{1}x + B_{1}y + C_{1}) (A_{2}x + B_{2}y + C_{2}) = 0
or, A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} + (A_{1}C_{2} + A_{2}C_{1})x + (B_{1}C_{2} + B_{2}C_{1})y + C_{1}C_{2} = 0..............................................(ii)
Equating the coefficients of like terms in equation (i) and (ii) we have,
A_{1}A_{2} = a^{2}, A_{1}B_{2} + A_{2}B_{1} = 2h, B_{1}B_{2} = b^{2}, A_{1}C_{2} + A_{2}C_{1} = 2g, B_{1}C_{2} + B_{2}C_{1} = 2f, C_{1}C_{2} = c^{2}
Now,
Equation of the straight line parallel to A_{1}x + B_{1}y + C_{1} = 0 and passing through the origin is:
A_{1}x+ B_{1}y = 0.........................................(iii)
Again,
Equation of straight line parallel to A_{2}x + B_{2}y + C_{2} = 0 and passing through the origin is:
A_{2}x + B_{2}y = 0.........................................(iv)
Now,
Combining (iii) and (iv) we have,
(A_{1}x + B_{1}y) (A_{2}x + B_{2}y) = 0
or, A_{1}A_{2}x^{2} + (A_{1}B_{2} + A_{2}B_{1})xy + B_{1}B_{2}y^{2} = 0
or, ax^{2} + 2hxy + by^{2} = 0. This completes the proof.
Note:Angles between the line pair ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 are same as the angles between the pair ax^{2} + 2hxy + by^{2} = 0.
The angles are given by,
tan^{1} (± \(\frac {2\sqrt {h^2  ab}}{a + b}\))
Hence, the two lines will be perpendicular to each other if a + b = 0 and they will beparallel if h^{2}= ab.
Here,
3x^{2}  5xy  2y^{2}  x + 2y = 0
or, 3x^{2}  6xy + xy  2y^{2}  x + 2y = 0
or, 3x(x  2y) + y(x  2y)  1(x  2y) = 0
or, (x  2y) (3x + y  1) = 0
Either: x  2y = 0
Or: 3x + y  1 = 0
∴ The required equations are: x  2y = 0 and 3x + y  1 = 0_{Ans}
Here,
x^{2}  3xy + 2y^{2} = 0
or, x^{2}  2xy  xy + 2y^{2} = 0
or, x(x  2y)  y(x  2y) = 0
or, (x  2y) (x  y) = 0
Either: x  2y = 0
Or: x  y = 0
∴ The required seperate equations are:x  2y = 0 andx  y = 0 _{Ans}
Here,
The given equation is:6x^{2} + 11xy  6y^{2} = 0..................(1)
The homogeneous equation of second is: ax^{2} + 2hxy + by^{2} = 0.................(2)
Comparing (1) and (2)
a = 6
2h = 11 i.e. h = \(\frac {11}2\)
b = 6
Now.
a + b = 0
or, 6  6 = 0
∴ 0 = 0
Hence, a + b = 0 is satisfied by the given equation so the equation6x^{2} + 11xy  6y^{2} = 0 are perpendicular to each other. _{Proved}
Here,
Given equation is:x^{2}  4xy + 4y^{2} = 0......................(1)
The homogenous equation of second degree is:
ax^{2}+ 2hxy + by^{2} = 0...........................(2)
Comparing (1) and (2)
a = 1
2h = 4 i.e. h = \(\frac {4}2\) = 2
b = 4
Now,
h^{2} = ab
or, (2)^{2} = 1× 4
∴ 4 = 4
Hence, h^{2} = ab is satisfied by the given equation so the equationx^{2}  4xy + 4y^{2} = 0 are coincident to each other. _{Proved}
Here,
ax^{2} + 2hxy + by^{2} = 0
When: h^{2} = ab then straight lines are coincident each other.
When: a + b = 0 then straight lines are perpendicular each other.
Given eq^{n} of line are:
x = 2y
i.e. x  2y = 0..........................(1)
2x = y
i.e. 2x  y = 0..........................(2)
Combined eq^{n} of (1) and (2) is:
(x  2y) (2x  y) = 0
or, 2x^{2} xy  4xy + 2y^{2} = 0
or, 2x^{2}  5xy + 2y^{2} = 0
∴ The required eq^{n} is: 2x^{2}  5xy + 2y^{2} = 0 _{Ans}
Here,
Given eq^{n} is: y^{2} = x^{2}
or, x^{2}  y^{2} = 0
or, (x  y) (x + y) = 0
either: x + y = 0
Or, x  y = 0
Slope of eq^{n} x + y = 0 is: m_{1} = \(\frac 11\) = 1
Slope of eq^{n} x y = 0 is: m_{2} = \(\frac 1{1}\) = 1
m_{1}× m_{2} = 1× 1 = 1
Hence, they are perpendicular to each other. _{Proved}
Here,
Given eq^{n} is:
6x^{2}  5xy  6y^{2} = 0...............................(1)
The eq^{n}of homogenous is:
ax^{2} + 2hxy + by^{2} = 0..........................(2)
Comparing eq^{n} (1) and (2)
a = 6
b = 6
Now,
a + b = 0
or, 6  6 = 0
∴ 0 = 0
Hence, the angle between two lines is 90°. _{Proved}
Here,
Given eq^{n} is:
x^{2}  2xy sec\(\alpha\) + y^{2} = 0...........................(1)
The homogenous eq^{n} is:
ax^{2} + 2hxy + by^{2} = 0..............................(2)
Comparing eq^{n} (1) and (2)
a = 1
h =  sec\(\alpha\)
b = 1
Now,
tan\(\theta\) =± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {(sec\alpha)^2  1 × 1}}{1 + 1}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {sec^2\alpha  1}}2\)
or, tan\(\theta\) =± \(\sqrt {sec^2\alpha  1}\)
or, tan\(\theta\) =± \(\sqrt {tan^2\alpha}\)
or, tan\(\theta\) =± tan\(\alpha\)
∴ \(\theta\) = \(\alpha\) _{Ans}
Here,
Given eq^{n} are:
x cos\(\alpha\) + y sin\(\alpha\) = 0...................(1)
x sin\(\alpha\) + y cos\(\alpha\) = 0...................(2)
Combined eq^{n} of (1) and (2)
(x cos\(\alpha\) + y sin\(\alpha\)) (x sin\(\alpha\) + y cos\(\alpha\)) = 0
or, x^{2} cos\(\alpha\).sin\(\alpha\) +xy cos^{2}\(\alpha\) + xy sin^{2}\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) = 0
or, x^{2} cos\(\alpha\).sin\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) + xy (cos^{2}\(\alpha\) + sin^{2}\(\alpha\)) = 0
or, sin\(\alpha\).cos\(\alpha\) (x^{2} + y^{2}) + xy× 1 = 0
∴(x^{2} + y^{2}) sin\(\alpha\).cos\(\alpha\) + xy = 0 _{Ans}
Given equation is:
x^{2} + 4xy + y^{2} = 0.............................(1)
The homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0.....................(2)
Comparing (1) and (2)
a = 1
h =2
b = 1
Now,
tan\(\theta\) = ± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\theta\) = ± \(\frac {2\sqrt {2^2  1 × 1}}{1 + 1}\)
or, tan\(\theta\) = ± \(\frac {2\sqrt {4  1}}2\)
or, tan\(\theta\) = ± \(\sqrt 3\)
∴ obtuse angle (\(\theta\) = tan^{1} (\(\sqrt 3\))
∴ \(\theta\) = (90 + 30)° = 120°_{Ans}
Here,
Given equation is:
3x^{2} + 2y^{2}  5xy = 0
The homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0.....................(2)
Comparing (1) and (2)
a = 1
h =2
b = 1
Now,
tan\(\theta\) = \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\theta\) = \(\frac {2\sqrt {(\frac {5}2)^2  3 × 2}}{3 + 2}\)
or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25}4  \frac 61}}5\)
or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25  24}4}}5\)
or, tan\(\theta\) = \(\frac {2 × \frac 12}{\frac 51}\)
or, tan\(\theta\) = \(\frac 15\)
or, \(\theta\) = tan^{1}(\(\frac 15\))
∴ \(\theta\) = 11.31° _{Ans}
Here,
The given equation is:
12x^{2}  23xy + 5y^{2} = 0.....................(1)
The homogeneous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0.....................(2)
Comparing (1) and (2)
a =12
h = \(\frac {23}2\)
b = 5
Now,
tan\(\theta\) = ±\(\frac {2\sqrt {h^2  ab}}{a + b}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {23}2)^2  12 × 5}}{12 + 5}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529}4  60}}{17}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529  240}4}}{17}\)
or,tan\(\theta\) = ± \(\frac {2\sqrt {289}}2\)× \(\frac 1{17}\)
or,tan\(\theta\) = ± \(\frac {17}{17}\)
∴tan\(\theta\) = ± 1
Taking +ve sign,
tan\(\theta\) = 1
or, tan\(\theta\) = tan 45°
∴ \(\theta\) = 45°
Taking ve sign,
tan\(\theta\) = 1
or, tan\(\theta\) = tan (180  35)
or, tan\(\theta\) = tan 135°
∴ \(\theta\) = 135°
Hence, the obtuse angle between the pair of equation is: 135°. _{Ans}
Here,
The homogenous equation of second degree is:
ax^{2} + 2hxy +by^{2} = 0................................(1)
The given equation is:
2x^{2}  5xy + 2y^{2} = 0...................................(2)
Comparing (1) and (2)
a = 2
2h = 5 i.e. h = \(\frac 52\)
b = 2
If \(\theta\) be the angle between pair of lines:
tan\(\theta\) =± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {(\frac {5}2)^2  2 × 2}}{2 + 2}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4  \frac 41}}4\)
or, tan\(\theta\) =± \(\frac {\sqrt {\frac {25  16}4}}2\)
or, tan\(\theta\) =± \(\sqrt {\frac 94}\)× \(\frac 12\)
or, tan\(\theta\) =± \(\frac 32\)× \(\frac 12\)
or, tan\(\theta\) =± \(\frac 34\)
∴ \(\theta\) = tan^{1}(± \(\frac 34\))
Hence, the acute angle (\(\theta\)) =tan^{1}(± \(\frac 34\)) _{Ans}
Here,
Given equation is:
6x^{2} + 5xy  3x  2y  6y^{2} = 0
or, 6x^{2} + 9xy  4xy  6y^{2}  3x  2y = 0
or, 3x (2x + 3y)  2y (2x + 3y)  1 (2x + 3y) = 0
or, (2x + 3y) (3x  2y  1) = 0
Either: 2x + 3y = 0..........................(1)
Or: 3x  2y  1 = 0............................(2)
Slope of equation (1), m_{1} = \(\frac 23\)
Slope of equation (2), m_{2} = \(\frac 32\)
Again,
m_{1}× m_{2} = \(\frac {2}3\)× \(\frac 32\) = 1
The product of two slopes = 1
Hence, these equations are perpendicular each other. _{Hence, Proved}
Here,
Let: the two equation of the lines through origin be:
a_{1}x + b_{1}y = 0..............................(1)
a_{2}x + b_{2}y = 0..............................(2)
Combined equation of (1) and (2) is:
(a_{1}x + b_{1}y) (a_{2}x + b_{2}y) = 0
or, a_{1}a_{2}x^{2} + a_{1}b_{2}xy + a_{2}b_{1}xy + b_{1}b_{2}y^{2} = 0
or,a_{1}a_{2}x^{2} + (a_{1}b_{2}+ a_{2}b_{1})_{}xy + b_{1}b_{2}y^{2} = 0
Let:
a_{1}a_{2} = a
b_{1}b_{2} = b
(a_{1}b_{2}+ a_{2}b_{1}) = 2h
Now,
ax^{2} + 2hxy + by^{2} = 0
Thus,ax^{2} + 2hxy + by^{2} = 0 is homogenous equation of second degree. _{Ans}
Here,
The given equation is: ax^{2} + 2hxy + by^{2} = 0 if a≠ 0, then:
The given equation is multiplied by 'a' on both sides:
a^{2}x^{2} + 2ahxy + aby^{2} = 0
or (ax)^{2} + 2 (ax) (hy) + (hy)^{2}  h^{2}y^{2} + aby^{2} = 0
or, (ax + hy)^{2}  (h^{2}  ab) y^{2} = 0
or, (ax + hy)^{2}  {(\(\sqrt {(h^2  ab)y})}^{2} = 0
or, (ax + hy + \(\sqrt {(h^2  ab)y}\)) (ax + hy  \(\sqrt {(h^2  ab)y}\)) = 0
Either:(ax + hy + \(\sqrt {(h^2  ab)y}\)) = 0.................(1)
Or:(ax + hy  \(\sqrt {(h^2  ab)y}\)) = 0..........................(2)
Equation (1) and (2) are satisfied (0, 0) so both straight lines will passes through origin.
Again,
If a = 0
The equation ax^{2}+ 2hxy + by^{2} = 0 will be
2hxy + by^{2} = 0
or, y(2hx + by) = 0
Either: y = 0.......................(3)
Or: 2hx + by = 0..............(4)
Equation (3) and (4) are satisfied by the coordinates (0, 0).
Hence, the second degree homogeneous equation: ax^{2} + 2hxy + by^{2} = 0 always represents a pair of straight lines through the origin. _{Proved}
Here,
Given equationax^{2} + 2hxy + by^{2}= 0 represents two straight lines that passes through the origin.
Let: y = m_{1}x and y = m_{2}x are the two lines.
y m_{1}x = 0............................(1)
y  m_{2}x = 0............................(2)
Product of the equation (1) and (2) is:
(y  m_{1}x) (y  m_{2}x) = 0
or, m_{1}m_{2}x^{2}  m_{1}xy  m_{2}xy + y^{2} = 0
or, m_{1}m_{2}x^{2}  (m_{1 }+m_{2})xy + y^{2} = 0...................(3)
Given equationax^{2} + 2hxy + by^{2}= 0
\(\frac ab\)x^{2} + \(\frac {2h}b\) xy + y^{2} = 0...................(4)
Comparing equation (3) and (4), we get:
m_{1} + m_{2} = \(\frac {2h}b\) and m_{1}m_{2} = \(\frac ab\)
Let \(\theta\) be the angle between the lines:
tan\(\theta\) =± \(\frac {m_1  m_2}{1 + m_1m_2}\)
or, tan\(\theta\) =± \(\frac {\sqrt {(m_1 + m_2)^2  4m_1m_2}}{1 + m_1m_2}\) [\(\because\) a  b = \(\sqrt {(a + b)^2  4ab}\)]
or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2}{b^2}  \frac {4a}b}}{1 + \frac ab}\)
or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2  4ab}{b^2}}}{\frac {b + a}b}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {h^2  ab}}{a + b}\)× \(\frac bb\)
or, tan\(\theta\) =± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
∴\(\theta\) = tan^{1}(± \(\frac {2\sqrt {h^2  ab}}{a + b}\))_{ Ans}
Here,
x^{2}  5xy + 4y^{2} = 0
or, x^{2}  xy  4xy + 4y^{2} = 0
or, x(x  y)  4y(x  y) = 0
or, (x  y) (x  4y) = 0
Either: x  y = 0.................(1)
Or: x  4y = 0......................(2)
The eq^{n} (1) changes into parallel form is:
x  y + k_{1} = 0......................(3)
The point (1, 1) passes through eq^{n} (1)
1  1 + k_{1} = 0
∴ k_{1} = 0
Putting the value of k_{1} in eq^{n} (3)
x  y + 0 = 0
x  y = 0............................(4)
The eq^{n}(2) changes in parallel form is:
x  4y + k_{2} = 0..................(5)
The point (1, 1) passes through eq^{n} (5)
1  4× 1 + k_{2} = 0
or, 3 + k_{2} = 0
∴ k_{2} = 3
Putting the value of k_{2} in eq^{n} (5)
x  4y + 3 = 0.......................(6)
The eq^{n} of pair of line is:
(x  y) (x  4y + 3) = 0
or, x^{2}  4xy + 3x  xy  4y^{2}  3y = 0
∴x^{2}  5xy  4y^{2} + 3x  3y = 0_{Ans}
Here,
Given equation is:
x^{2}  xy  2y^{2} = 0
or, x^{2}  2xy + xy  2y^{2} = 0
or, x (x  2y) + y (x  2y) = 0
or, (x  2y) (x + y) = 0
Either: x  2y = 0.........................(1)
Or: x + y = 0.................................(2)
The eq^{n} (1) changes in perpendicular form is:
2x  y + k_{1} = 0
2x + y  k_{1} = 0..........................(3)
The eq^{n} (3) passes through origin (0, 0):
2× 0 + 0  k_{1} = 0
∴ k_{1} = 0
Putting the value of k_{1} in eq^{n} (3)
2x + y  0 = 0
2x + y = 0........................(4)
The eq^{n} (2) change in perpendicular form is:
x  y + k_{2} = 0........................(5)
The eq^{n} (5) passes through origin (0, 0)
0  0 + k_{2} = 0
∴ k_{2} = 0
Putting the value of k_{2} in eq^{n} (5)
x  y + 0 = 0
x  y = 0...........................(6)
The equation of the pairs of lines is:
(2x + y) (x  y) = 0
or, 2x^{2}  2xy + xy  y^{2} = 0
∴2x^{2}  xy  y^{2} = 0_{Ans}
Given eq^{n} of line is:
3x^{2}  8xy + 5y^{2}^{}= 0
or, 3x^{2}  3xy  5xy + 5y^{2}^{}= 0
or, 3x (x  y)  5y (x  y) = 0
or, (x  y) (3x  5y) = 0
Either: x  y = 0.......................(1)
Or: 3x  5y = 0.........................(2)
The eq^{n} (1) changes in perpendicular form is:
x + y + k_{1} = 0...........................(3)
The point (2, 3) passes through eq^{n} (3)
2 + 3 + k_{1} = 0
or, 5 + k_{1} = 0
∴ k_{1} = 5
Putting the value of k_{1} in eq^{n} (3)
x + y  5 = 0..........................(4)
The eq^{n} (2) change in perpendicular form is:
5x + 3y + k_{2} = 0...................(5)
The point (2, 3) passes through eq^{n} (5)
5× 2 + 3× 3 + k_{2} = 0
or, 10 + 9 + k_{2} = 0
or, 19 + k_{2} = 0
∴ k_{2} =  19
Putting the value of k_{2} in eq^{n} (5)
5x + 3y  19 = 0......................(6)
The eq^{n} of pairs of lines is:
(x + y  5) (5x + 3y  19) = 0
or, 5x^{2} + 3xy  19x + 5xy + 3y^{2}  19y  25x  15y + 95 = 0
∴ 5x^{2} + 8xy + 3y^{2}  44x  34y + 95 = 0_{Ans}
Here,
x^{2}  xy  2y^{2} = 0
or, x^{2}  2xy + xy  2y^{2} = 0
or, x(x  2y) + y(x  2y) = 0
or, (x  2y) (x + y) = 0
Either: x  2y = 0..................(1)
Or: x + y = 0...........................(2)
The eq^{n} (1) change in perpendicular form is:
2x + y + k_{1} = 0.......................(3)
The point (3, 1) passes through eq^{n} (3)
2× 3  1 + k_{1} = 0
or, 6  1 + k_{1} = 0
∴ k_{1}= 5
Putting the value of k_{1} in eq^{n} (3)
2x + y  5 = 0..........................(4)
The eq^{n} (2) change in perpendicular form is:
x  y + k_{2} = 0.........................(5)
The point (3, 1) passes througheq^{n} (5)
3 + 1 + k_{2} = 0
∴ k_{2} = 4
Putting the value of k_{2} in eq^{n} (5)
x  y  4 = 0.........................(6)
The eq^{n} of pair of lines is:
(2x + y  5) (x  y  4) = 0
or, 2x^{2}  2xy  8x + xy  y^{2}  4y  5x + 5y + 20 = 0
∴ 2x^{2}  xy y^{2}  13x + y + 20 = 0_{Ans}
Here,
2x^{2} + 5xy + 3y^{2} = 0
or, 2x^{2} + 3xy + 2xy + 3y^{2} = 0
or, x(2x + 3y) + y(2x + 3y) = 0
or, (2x + 3y) (x + y) = 0
Either: 2x + 3y = 0
Or: x + y = 0
Given eq^{n} is: 2x^{2} + 5xy + 3y^{2} = 0......................(1)
Homogenous eq^{n} is: ax^{2} + 2hxy + by^{2} = 0...................(2)
Comparing eq^{n} (1) and (2)
a = 2
h = \(\frac 52\)
b = 3
We know that:
tan\(\theta\) =± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or,tan\(\theta\) =± \(\frac {2\sqrt {(\frac 52)^2  2 × 3}}{2 + 3}\)
or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4  6}}5\)
or,tan\(\theta\) =± \(\frac {2\sqrt {\frac {25  24}4}}5\)
or,tan\(\theta\) =± \(\frac {2\sqrt {\frac 14}}5\)
or,tan\(\theta\) =± \(\frac {2 × \frac 12}5\)
or,tan\(\theta\) =± \(\frac 15\)
Taking +ve sign,
\(\theta\) = tan^{1} (\(\frac 15\)) = 11.31°
Taking vesign,
\(\theta\) = (180°  11.31°) = 168.69°
∴ Required eq^{n} are: 2x + 3y = 0 and x + y = 0 and angle between them are: 11.31° and 168.69°. _{Ans}
Given eq^{n} is:x^{2} + 2xy sec\(\theta\) + y^{2} = 0........................(1)
Homogenous eq^{n} is: ax^{2} + 2hxy + by^{2} = 0..........(2)
Comparingeq^{n}(1) and (2)
a = 1
h = sec\(\theta\)
b = 1
If \(\alpha\) be the angle between pairs of lines then,
tan\(\alpha\) = \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta  1 × 1}}{1 + 1}\)
or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta  1}}2\)
or, tan\(\alpha\) = \(\sqrt {tan^2\theta}\)
or, tan\(\alpha\) = tan\(\theta\)
∴\(\alpha\) = \(\theta\) _{Proved}
Again,
The given eq^{n} is:
y^{2} + 2xy sec\(\theta\) + x^{2} = 0..........................(3)
The quadratic eq^{n} is:
ax^{2} + bx + c = 0.............................(4)
Comparing (3) and (4)
a = 1
b = 2x sec\(\theta\)
c = x^{2}
We know,
x = \(\frac {b ± \sqrt {b^2  4ac}}{2a}\)
or, y = \(\frac {(2x sec\theta) ± \sqrt {(2x sec\theta)^2  4 × 1 × x^2}}{2 × 1}\)
or, y = \(\frac {2x sec\theta ± \sqrt {4x^2 sec^2\theta  4x^2}}2\)
or, y = \(\frac {2x sec\theta ± \sqrt {4x^2 (sec^2\theta  1)}}2\)
or, y = \(\frac {2x sec\theta ± 2x\sqrt {tan^2\theta}}2\)
or, y = \(\frac {2x sec\theta ± 2x tan\theta}2\)
or, y = \(\frac {2(x sec\theta ± xtan\theta)}2\)
or, y =  xsec\(\theta\)± x tan\(\theta\)
Taking +ve sign:
y = xsec\(\theta\) + xtan\(\theta\)
Taking ve sign:
y = xsec\(\theta\)  xtan\(\theta\)
∴ The required eq^{n} are:y = xsec\(\theta\) + xtan\(\theta\) andy = xsec\(\theta\)  xtan\(\theta\) _{Ans}
Given equation is:x^{2}  2xy cosec\(\theta\) + y^{2} = 0.....................(1)
Homogenous equation is: ax^{2} + 2hxy + by^{2} = 0......................(2)
Comparing eq^{n} (1) and (2)
a = 1
h = cosec\(\theta\)
b = 1
If \(\alpha\) be the angle between pair of lines then:
tan\(\alpha\) =± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\alpha\) =± \(\frac {2\sqrt {(cosec\theta)^2  1 × 1}}{1 + 1}\)
or, tan\(\alpha\) =± \(\frac {2\sqrt {(cosec\theta)^2  1}}2\)
or, tan\(\alpha\) =± \(\frac {2\sqrt {cosec^2\theta  1}}2\)
or, tan\(\alpha\) =± \(\sqrt {cot^2\theta}\)
∴ tan\(\alpha\) =± cot\(\theta\)
Taking +ve sign,
tan\(\alpha\) = cot\(\theta\)
tan\(\alpha\) = tan(\(\frac p2\)  \(\theta\))
∴\(\alpha\) = (\(\frac p2\)  \(\theta\))_{Hence, Proved}
Here,
Given equation is:
2x^{2} + 7xy + 3y^{2} = 0
or, 2x^{2} + 6xy + xy + 3y^{2} = 0
or, 2x(x + 3y) + y(x + 3y) = 0
or, (x + 3y)(2x + y) = 0
The two equation represented by2x^{2} + 7xy + 3y^{2} = 0 are:
2x + y = 0......................(1)
x + 3y = 0......................(2)
Now,
Slope of equation (1), m_{1} = \(\frac {xcoefficient}{ycoefficient}\) = 2
Slope of equation (2),m_{2} =\(\frac {xcoefficient}{ycoefficient}\) = \(\frac 13\)
If the angle between the lines be \(\theta\) then:
tan\(\theta\) =± \(\frac {m_1  m_2}{1 + m_1m_2}\)
or, tan\(\theta\) =± \(\frac {2 + \frac 13}{1 + (2) (\frac 13)}\)
or, tan\(\theta\) = ± \(\frac {\frac {6 + 1}3}{\frac {3 + 2}3}\)
or, tan\(\theta\) = ± \(\frac {5}3\)× \(\frac 35\)
or, tan\(\theta\) = ± 1
Taking +ve sign,
tan\(\theta\) = tan 45°
∴ \(\theta\) = 45°
Taking ve sign,
tan\(\theta\) = tan (180  45)° = tan 135°
∴ \(\theta\) = 135°
∴ Required angles (\(\theta\)) = 45° and 135°_{Ans}
The given equation is:
2x^{2} + 3xy  2y^{2} = 0
or, 2x^{2} + 4xy  xy  2y^{2} = 0
or, 2x(x + 2y)  y(x + 2y) = 0
or, (x + 2y) (2x  y) = 0
∴ Equation are: x + 2y = 0 and 2x  y = 0
Again,
2x^{2} + 3xy  2y^{2} = 0............................(1)
The homogenous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0........................(2)
Comparingequation (1) and (2)
a = 2
h = \(\frac 32\)
b = 2
tan\(\theta\) = \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\theta\) = \(\frac {2\sqrt {(\frac 32)^2  2 × (2)}}{2 + (2)}\)
or, tan\(\theta\) = \(\frac {2\sqrt {\frac 94 + \frac 41}}{2  2}\)
or, tan\(\theta\) = \(\frac {2\sqrt {\frac {9 + 16}4}}0\)
or, tan\(\theta\) =∞
∴ \(\theta\) = tan^{1}∞ = 90°
∴ The required equations are: x + 2y = 0 and 2x  y = 0 and angle between the pairs of straight lines is 90°._{Ans}
Here,
Given equation is:
2x^{2} 3xy + y^{2} = 0......................(1)
The homogenous equation of second degree is:
ax^{2} + 2hxy + by^{2} = 0..................(2)
Comparing equation (1) and (2)
a = 2
h = \(\frac 32\)
b =1
Now,
tan\(\theta\) = ± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {3}2)^2  2 × 1}}{2 + 1}\)
or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac 94  \frac 21}}3\)
or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac {9  8}4}}3\)
or, tan\(\theta\) = ± \(\frac {2 × \frac 12}3\)
or, tan\(\theta\) = ± \(\frac 13\)
For the obtuse angle,
tan\(\theta\) = \(\frac 13\)
\(\theta\) = tan^{1} (\(\frac 13\))
Now,
2x^{2}  3xy + y^{2} = 0
or, 2x^{2}  2xy  xy + y^{2} = 0
or, 2x(x  y)  y(x  y) = 0
or, (x  y) (2x  y) = 0
∴ The required pairs of lines are: x  y = 0 and 2x  y = 0. _{Ans}
Here,
6x^{2} + 5xy  3x + 2y  6y^{2} = 0
or, 6x^{2} + 5xy  6y^{2}  3x + 2y = 0
or, 6x^{2} + 9xy  4xy  6y^{2}  3x + 2y = 0
or, 3x(2x + 3y)  2y(2x + 3y)  1(3x  2y) = 0
or, (2x + 3y  1) (3x  2y) = 0
Either: 2x + 3y  1 = 0...........(1)
Or: 3x  2y = 0.........................(2)
Slope of equation (1), m_{1} = \(\frac {3}{2}\) = \(\frac 32\)
Slope of equation (2),m_{2} = \(\frac 23\)
∴ m_{1}× m_{2} = \(\frac 32\)× \(\frac {2}3\) = 1
Hence, the product of two slopes = 1 so these equations are perpendicular to each other. _{Proved}
The given equation is:
x^{2}  2xy cosec\(\theta\) + y^{2} = 0
or, (y)^{2}  (2x cosec\(\theta\)) y + (x)^{2} = 0..................(1)
The quadratic equation is:
ax^{2} + bx + c = 0......................(2)
Comparing (1) and (2)
a = 1
b = 2x cosec\(\theta\)
c = x^{2}
x = \(\frac {b ± \sqrt {b^2  4ac}}{2a}\)
y = (2x cosec\(\theta\))± \(\frac {\sqrt {(2x cosec\theta)^2  4(1) (x^2)}}{2× 1}\)
y = \(\frac {2x cosec\theta ± \sqrt {4x^2 cosec^2\theta  4x^2}}2\)
y = \(\frac {2x cosec\theta ± 2x \sqrt {cosec^2\theta  1}}2\)
y = \(\frac {2(x cosec\theta ± x cot\theta)}2\)
y = x cosec\(\theta\)± x cot\(\theta\)
The pair of lines are:
y  x (cosec\(\theta\)  cot\(\theta\)) = 0 and
y  x (cosec\(\theta\) + cot\(\theta\)) = 0
Comparing x^{2} 2xy cosec\(\theta\) + y^{2} = 0 and ax^{2} + 2hxy + by^{2} = 0
a = 1
h =  cosec\(\theta\)
b = 1
We know that:
tanA =± \(\frac {2\sqrt {h^2  ab}}{a + b}\)
or, tanA = ± \(\frac {2\sqrt {(cosec\theta)^2  1 × 1}}{1 + 1}\)
or, tanA =± \(\frac {2\sqrt {cosec^2\theta  1}}2\)
or, tanA =± cot\(\theta\)
∴ A = tan^{1} (± cot\(\theta\))
∴ The required equations are: y  x (cosec\(\theta\)  cot\(\theta\)) = 0 and y  x (cosec\(\theta\) + cot\(\theta\)) = 0 and angle (A) = tan^{1} (± cot\(\theta\)). _{Ans}

2x^{2} + 7xy + 3y^{2 } = 0
2xy =0 , x3y = 0
2x+y =0 , x+3y = 0
2x+3 =0 , x+2y = 0
2x3 =0 , x3y = 0

2x^{2 } 5xy + 3y^{2 }= 0.
x+y =0 , 3x2y =0
xy =0 , 2x0y =3
xy =0 , 2x3y =0
x+y =0 , 2x+3y =0

2x^{2 }+ xy 6y^{2} = 0 .
2x3y = 0 , x+2y = 0.
2x+3y = 0 , x+3y = 0.
2x+3y = 0 , x2y = 0.
1x2y = 0 , x+3y = 0.

X^{2 }+ x  y  y^{2 }= 0
x  y = 0 , x + 2y = 0
x + y = 0 , x + 3y = 0
x  y = 2 , x + 3y = 0
x + y = 0 , y + 2x = 0

3x^{2}  5xy  2y^{2 } x + 2y = 0
x = 2y , 1x+y = 3
x = 2y , 3x+y = 1
x = 2y , 3x+y = 1
x = 1y , 3x+y = 2

Find the single equation which represents the pair of straight lines x=2 and y = 3
xy  3x  2y + 6 = 0
xy  3x  2y = 6 + 0
xy  3x  3y  6 = 0
xy  3x + 2y  6 = 0

Find the homogeneous equation of 2nd degree from the pair of lines x = 2y&2x = y.
2x^{2 } 4xy + 3y^{2 } = 0.
2x^{2 } 5xy + 2y^{2 } = 0.
5x^{2 } 2xy + 2y^{2 } = 0.
2x^{2 } 2xy  2y^{2 } = 0

6x^{2 } xy  y^{2 }= 0
40(^0) , 145(^0)
45(^0) , 135(^0)
90(^0) , 135(^0)
45(^0) , 125(^0)

x^{2 }^{}  3xy  4y^{2}^{} = 0
99.04(^0) , 100.96(^0)
09.54(^0) , 126.96(^0)
θ = tan^{1 }(± (frac{5}{3}) ) or 59.04(^0) , 120.96(^0)
θ = tan^{1 }(± (frac{3}{5}) ) , 110.86(^0)

X^{2 } 2xy cosec α + y^{2 } = 0
90(^o)  α
45(^o)  α
180(^o)  α
360(^o)  α

x^{2 }^{}^{}^{}^{}^{} 4xy + y^{2 }= 0
360(^o)
120(^o)
180(^o)
145(^o)

3x^{2 }^{} 7xy + 2y^{2 } = 0
190(^o)
360(^o)
135(^o)
180(^o)

x^{2}  4xy + y^{2}^{} = 0
360(^o)
90(^o)
180(^o)
60(^o)

(frac{1}{2}) x^{2} + 2xy + (frac{1}{2}) y^{2 }
60(^o)
120(^o)
390(^o)
360(^o)

If the pair of lines represented by an equation mx^{2 } 5xy  6y^{2} = 0 are perpendicular to each other find the value of m.
6
12
5
4

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