Pair Of Straight Lines

Pair Of Straight Lines

BOOST YOUR CONCEPTS
1.The point of intersection of the pair of straight lins H = ax² + 2hxy + by² = 0 is (0, 0).


2.If \( S \equiv ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \), represents a pair of lines and \( h^2 > ab \), then the point of intersection of the pair of lines is \( \left( {\frac{{hf - bg}} {{ab - h^2 }},\frac{{gh - af}} {{ab - h^2 }}} \right) \)
Note : (i) The point of intersection of the pair of lines \( S \equiv 0 \) is also obtained by solving \( \frac{{\partial s}} {{\partial x}} = 0 \) and \( \frac{{\partial s}} {{\partial y}} = 0 \) . Here, \( \frac{{\partial s}} {{\partial x}} \) and \( \frac{{\partial s}} {{\partial y}} \) represents the partial differentiation of s w.r.t x and y respectively.
(ii) If the equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) represents a pair of intersecting lines, then the square of the distance of their point of intersection from the origin is  \( \frac{{c\left( {a + b} \right) - f^2 - g^2 }} {{ab - h^2 }} \) .
(iii) If the equation \( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) represents a pair of perpendicular lines, then the square of the distance of their points of intersection from the origin is \( \frac{{f^2 + g^2 }} {{h^2 - ab}}\text{ }or\text{ }\frac{{f^2 + g^2 }} {{a^2 + h^2 }}\text{ }or\text{ }\frac{{f^2 + g^2 }} {{b^2 + h^2 }} \)

3.The equation of the pair of lines passing through the origin and forming an isosceles triangle with the line ax + by + c = 0 is \( \left( {ax + by} \right)^2 - k\left( {bx - ay} \right)^2 = 0 \).
a)If k = 1, then the triangle is right angled isosceles triangle 
b)If k = 3, then the triangle is equilateral triangle.
c) If \( k = \frac{1} {3} \), then the triangle is an isosceles and obtuse angled triangle.

4.The triangle formed by the pair of lines \( S \equiv ax^2 + 2hxy + by^2 = 0 \) and the line lx + my + n = 0 is 
a)Equilateral if \( ax^2 + 2hxy + by^2 = \left( {lx + my} \right)^2 - 3\left( {mx - ly} \right)^2 \)
b)Isosceles if \( h\left( {l^2 - m^2 } \right) = \left( {a - b} \right)lm \)
c)Right angled if a + b = 0 or s(l, m) = 0.

5.Theorem : The product of the perpendiculars from \( \left( {\alpha ,\beta } \right) \) to the pair of lines \( ax^2 + 2hxy + by^2 = 0 \) is \( \frac{{\left| {a\alpha ^2 + 2h\alpha \beta + b\beta ^2 } \right|}} {{\sqrt {\left( {a - b} \right)^2 + 4h^2 } }} \)
Proof:

Given \( H \equiv ax^2 + 2hxy + by^2 = 0 \). Let \( l_1 x + m_1 y = 0 \) and \( l_2 x + m_2 y = 0 \) be the two lines represented by \( H \equiv 0 \).
We have,
\( ax^2 + 2hxy + by^2 = \left( {l_1 x + m_1 y} \right)\left( {l_2 x + m_2 y} \right) \)
\( = l_1 l_2 x^2 + \left( {l_1 m_2 + l_2 m_1 } \right)xy + m_1 m_2 y^2 \)
Comparing the coefficients of like terms on both sides, we get
\( a = l_1 l_2 \left| {2h = l_1 m_2 + l_2 m_1 } \right|b = m_1 m_2 \to \left( 1 \right) \)
Let d₁ and d₂ be the lengths of the perpendicular distances from the point \( \left( {\alpha ,\beta } \right) \) to the lines \( l_1 x + m_1 y = 0 \) and \( l_2 x + m_2 y = 0 \) respectively.
We have, \( d_1 = \frac{{\left| {l_1 \alpha + m_1 \beta } \right|}} {{\sqrt {l_1^2 + m_1^2 } }} \)
and \( d_2 = \frac{{\left| {l_2 \alpha + m_2 \beta } \right|}} {{\sqrt {l_2^2 + m_2^2 } }} \)
Now, the product of the perpendicular distances.
\( d_1 d_2 = \frac{{\left| {l_1 \alpha + m_1 \beta } \right|}} {{\sqrt {l_1^2 + m_1^2 } }} \cdot \frac{{\left| {l_2 \alpha + m_2 \beta } \right|}} {{\sqrt {l_2^2 + m_2^2 } }} \)
\( = \frac{{\left| {\left( {l_1 \alpha + m_1 \beta } \right)\left( {l_2 \alpha + m_2 \beta } \right)} \right|}} {{\sqrt {\left( {l_1^2 + m_1^2 } \right)\left( {l_2^2 + m_2^2 } \right)} }} \)
\( \begin{gathered} = \frac{{\left| {l_1 l_2 \alpha ^2 + \left( {l_1 m_2 + l_2 m_1 } \right)\alpha \beta + m_1 m_2 \beta ^2 } \right|}} {{\sqrt {l_1^2 l_2^2 + l_1^2 m_2^2 + l_2^2 m_1^2 + m_1^2 m_2^2 } }} \hfill \\ = \frac{{\left| {l_1 l_2 \alpha ^2 + \left( {l_1 m_2 + l_2 m_1 } \right)\alpha \beta + m_1 m_2 \beta ^2 } \right|}} {{\sqrt {\left( {l_1 l_2 - m_1 m_2 } \right)^2 + \left( {l_1 m_2 + l_2 m_1 } \right)^2 } }} \hfill \\ \end{gathered} \)
\( = \frac{{\left| {a\alpha ^2 + 2h\alpha \beta + b\beta ^2 } \right|}} {{\sqrt {\left( {a - b} \right)^2 + \left( {2h} \right)^2 } }} \)  (from (1))
\( = \frac{{\left| {a\alpha ^2 + 2h\alpha \beta + b\beta ^2 } \right|}} {{\sqrt {\left( {a - b} \right)^2 + 4h^2 } }} \)

6.The product of perpendiculars from \( \left( {\alpha ,\beta } \right) \) to the pair of lines
\( S \equiv ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \) is \( \frac{{\left| {S_{11} } \right|}} {{\sqrt {\left( {a - b} \right)^2 + 4h^2 } }} \)
i.e.,\( \frac{{\left| {a\alpha ^2 + 2h\alpha \beta + b\beta ^2 + 2g\alpha + 2f\beta + c} \right|}} {{\sqrt {\left( {a - b} \right)^2 + 4h^2 } }} \)

7.The product of perpendiculars from O(0, 0) to the pair of lines represented by \( S \equiv 0 \) is \( d = \frac{{\left| c \right|}} {{\sqrt {\left( {a - b} \right)^2 + 4h^2 } }} \)

8.The area of an equilateral triangle formed by the line ax + by + c = 0 with the pair of lines \( \left( {ax + by} \right)^2 - 3\left( {bx - ay} \right)^2 = 0 \) is \( \frac{{c^2 }} {{\sqrt 3 \left( {a^2 + b^2 } \right)}}or\frac{{p^2 }} {{\sqrt 3 }} \) where p is the perpendicular distance from the origin to the line ax + by + c = 0.