Relation between Two Lines
Let L1 and L2 be the two lines as
L1 : a1x + b1y + c1 = 0
L2 : a2x + b2y + c2 = 0
Two lines are said to be parallel if the below condition is satisfied,
\(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}} \neq \frac{{{c}_{1}}}{{{c}_{2}}}\end{array} \)
Two lines intersect at a point if
\(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}} \neq \frac{{{b}_{1}}}{{{b}_{2}}}\end{array} \)
Two lines coincide if
\(\begin{array}{l}\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\end{array} \)
Angle between Straight Lines
\(\begin{array}{l}Let\,\,\,\,{{L}_{1}}\,\,\,\,\equiv \,\,\,y={{m}_{1}}x+{{c}_{1}}\end{array} \)
and
\(\begin{array}{l}{{L}_{2}}\,\,\,\equiv \,\,\,y={{m}_{2}}x+{{c}_{2}}\end{array} \)
\(\begin{array}{l}\text{Angle} = \theta ={{\tan }^{-1}}\left| \left( \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}} \right) \right|\end{array} \)
Special Cases:
\(\begin{array}{l}\Rightarrow {{m}_{2}}={{m}_{1}}\,\,\,\,\,\,\,\to \,\,\,\,\,lines\,are\,parallel\\\end{array} \)
\(\begin{array}{l}\Rightarrow \,\,{{m}_{1}}{{m}_{2}}=-1,\,\,\,\,\,\,\,\,\,\,lines\,L1\,\And L2\,are\,perpendicular\,to\,each\,other\end{array} \)
Length of Perpendicular from a Point on a Line
The length of the perpendicular from P(x1, y1) on ax + by + c = 0 is
\(\begin{array}{l}\ell =\left| \frac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\end{array} \)
B (x, y) is the foot of perpendicular is given by
\(\begin{array}{l}\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{-(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\end{array} \)
A’(h, k) is mirror image, given by
\(\begin{array}{l}\frac{h-{{x}_{1}}}{a}=\frac{k-{{y}_{1}}}{b}=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{\left( {{a}^{2}}+{{b}^{2}} \right)}\end{array} \)
Angular Bisector of Straight lines
An angle bisector has an equal perpendicular distance from the two given lines.
The equation of line L can be given as
\(\begin{array}{l}\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{\sqrt{{{a}_{1}}^{2}+{{b}_{1}}^{2}}}=\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}{\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}}\end{array} \)
Family of Lines:
The general equation of the family of lines through the point of intersection of two given lines, L1 & L2, is given by L1 +λ L2 = 0
Where λ is a parameter.
Concurrency of Three Lines
Let the lines be
\(\begin{array}{l}{{L}_{1}}\equiv {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\end{array} \)
\(\begin{array}{l}{{L}_{2}}\equiv {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\end{array} \)
and
\(\begin{array}{l}{{L}_{3}}\equiv {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}=0\end{array} \)
So, the condition for the concurrency of lines is
\(\begin{array}{l}\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=0\end{array} \)
Pair of Straight Lines
Join equation of lines L1 & L2 represents P. S. L (a1x + b1y+c1) (a2x+b2y+c2) = 0
i.e. f(x,y) . g(x,y) = 0
Let’s defines a standard form of the equation:-
ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 represent conics curve equation
Condition for curve of being P.O.S.L Δ = abc + 2fgh – af2 – bg2 – ch2 = 0
If Δ ≠ 0, (i) parabola h2 = ab
(ii) hyperbola h2 < ab
(iii) circle h2 = 0, a = b
(iv) ellipse h2 > ab
Now, let’s see how did we get Δ = 0
General equation ax2 + 2gx + 2hxy + by2 + 2fy + c = 0
ax2 + (2g+2hy)x + (by2 + 2fy + c) = 0
we can consider the above equation as a quadratic equation in x, keeping y constant.
\(\begin{array}{l}x=\frac{-(2g+2hy)\pm \sqrt{{{(2g+2hy)}^{2}}-4a(b{{y}^{2}}+2fy+c)}}{2a}\end{array} \)
, so
\(\begin{array}{l}x=\frac{-(2g+2hy)\pm \sqrt{Q(y)}}{2a}\end{array} \)
Now, Q(y) has to be a perfect square, and only then, we can get two different line equations Q(y) in the perfect square for that Δ value of Q(y) should be zero.
From there D = 0
abc + 2fgh – bg2 – af2 – ch2 = 0
Or
\(\begin{array}{l}\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{matrix} \right|=0\end{array} \)
Note:
1. Point of intersection
To find point of intersection of two lines (P.O.S.L), solve the P.O.S.L, factorize it in (L1).(L2) = 0 or f(x, y) . g(u,y) = 0
2. Angle between the lines
\(\begin{array}{l}\tan \theta =\left( \left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right| \right)\end{array} \)
Special cases:
h2 = ab → lines are either parallel or coincident
h2 < ab → imaginary line
h2 > ab → Two distinct lines
a + b = 0 ⇒ perpendicular line
3. P.O.S.L passing through the origin, then
⇒ (y – m1x) (y – m2x) = 0
y2 – m2yx – m1xy – m1m2x2 = 0
y2 – (m1 + m2) xy – m1m2x2 = 0
⇒ ax2 + 2hxy + by2 = 0
\(\begin{array}{l}\Rightarrow {{y}^{2}}+\frac{2h}{b}xy+\frac{ab}{b}{{x}^{2}}=0\end{array} \)
\(\begin{array}{l}\Rightarrow m1 + m2 = \frac{2h}{b}\end{array} \)
\(\begin{array}{l}m_1 m_2=\frac{a}{b}\end{array} \)
\(\begin{array}{l}\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{\sqrt{{{({{m}_{1}}+{{m}_{2}})}^{2}}4{{m}_{1}}{{m}_{2}}}}{1+{{m}_{1}}{{m}_{2}}} \right|=\left| \frac{2\sqrt{{{h}^{2}}-ab}}{a+b} \right|\end{array} \)