If θ is the acute angle between two lines, then
$ \displaystyle tan\theta = |\frac{m_1 -m_2}{1 + m_1 m_2}|$
where m1 and m2 are the slopes of the two lines and are finite.
Notes:
∎ If the two lines are perpendicular to each other then m1m2 = −1
∎ Any line perpendicular to ax + by + c = 0 is of the form bx − ay + k = 0.
∎ If the two lines are parallel or coincident, then m1 = m2.
∎ Any line parallel to ax + by + c = 0 is of the form ax + by + k = 0.
∎ If any of the two lines is perpendicular to x-axis, then the slope of that line is infinite.
Let m1 = ∞ Then
$ \displaystyle tan\theta = |\frac{m_1 -m_2}{1 + m_1 m_2}|$
$ \displaystyle tan\theta = |\frac{1 – m_2/m_1}{1/m_1 + m_2}| = |\frac{1}{m_2}|$
or θ = |90° − α| where
tanα = m2 i.e. angle θ is the complimentary to the angle which the oblique line makes with the x-axis.
Illustration : Find the equation to the straight line which is perpendicular bisector of the line segment AB , where A , B are (a, b) and (a’, b’) respectively.
Solution: Equation of AB is
$ \displaystyle y- b = \frac{b’ -b}{a’ -a} (x -a)$
i.e. y(a’ – a) – x(b’ – b) = a’b – ab’
Equation to the line perpendicular to AB is of the form
(b’ – b)y + (a’ – a)x + k = 0 …(1)
Since the midpoint of AB lies on (1)
$ \displaystyle (b’ – b)(\frac{b+b’}{2}) + (a’-a) (\frac{a+a’}{2}) + k = 0 $
Hence the required equation of the straight line is
2(b’ – b)y + 2(a’ – a)x = (b’2 – b2 + a’2 – a2)
Equation of Straight Lines passing through a given point and equally inclined to a given line:
Let the straight lines passing through the point (x1, y1) and make equal angles with the given straight line y = m1x + c.
If m is the slope of the required line and α is the angle which this line makes with the given line, then
$ \displaystyle tan\alpha = \pm \frac{m_1 – m}{1 + m_1 m} $
1. The above expression for tanα, gives two values of m, say mA and mB
The required equations of the lines through the point (x1, y1) and making equal angles α with the given line are y − y1 = mA(x − x1), y − y1 = mB(x − x1).
Illustration : Find the equation to the sides of an isosceles right-angled triangle, the equation of whose hypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2)
Solution: The problem can be restated as:
Find the equations to the straight lines passing through the given point (2, 2) and making equal angles of 45° with the given straight line 3x + 4y – 4 = 0. Slope of the line 3x + 4y – 4 = 0 is m1 = -3/4.
$ \displaystyle tan45 = \pm \frac{m -m_1}{1 + m m_1}|$
$ \displaystyle 1 = \pm \frac{m + 3/4}{1 – m (3/4)}|$
mA = 1/7 and mB = −7
Hence the required equations of the two lines are
y − 2 = mA(x − 2) and y − 2 = mB(x − 2)
=> 7y − x − 12 = 0 and 7x + y = 16.
Length of the Perpendicular from a Point on a Line:
The length of the perpendicular from P(x1, y1) on ax + by + c = 0 is
$ \displaystyle = \frac{a x_1 + b y_1 + c}{\sqrt{a^2 + b^2}} $
The length of the perpendicular from origin (0 , 0) on ax + by + c = 0 is
$ \displaystyle = \frac{ c}{\sqrt{a^2 + b^2}} $
The distance between two parallel lines:
The distance between two parallel lines: ax + by + c1 = 0 and ax + by + c2 = 0 is
$ \displaystyle = \frac{c_1 – c_2}{\sqrt{a^2 + b^2}} $
Illustration : Three lines x + 2y + 3 = 0, x + 2y − 7 = 0 and 2x − y − 4 = 0 form 3 sides of two squares. Find the equation of remaining sides of these squares.
Solution:
Distance between the two parallel lines is
$ \displaystyle = \frac{7 +3}{\sqrt{5}} = 2\sqrt{5} $
The equations of sides A and C are of the form 2x − y + k = 0. Since distance between sides A and B = distance between sides B and C
$ \displaystyle \frac{|k – (-4)|}{\sqrt{5}} = 2\sqrt{5}$
$ \displaystyle \frac{k + 4}{\sqrt{5}} = \pm 2\sqrt{5}$
=> k = 6 , − 14.
Hence the fourth sides of the two squares are
(i) 2x −y + 6 = 0, (ii) 2x − y − 14 = 0.
Notes:
∎ In order to find the foot of the perpendicular from a given point to a given line, the most convenient method is to write the equation of the given line in such a way so that the coefficient of y becomes negative and then write the equation of the perpendicular in the parametric form. Then put r = the algebraic length (without the modulus) of the perpendicular = p. In this way, we get the foot of the perpendicular directly.
∎ To find the mirror image of a given point in a given line, we use the same process, taking r = 2p.
Exercise :
i) A straight line makes an intercept on the y-axis twice as long as that on the x-axis and is at a unit distance from the origin. Determine its equation.
ii) Prove that the straight lines (a + b)x + (a − b) y − 2ab = 0, (a − b)x + (a + b)y − 2ab = 0 and x + y = 0 form an isosceles triangle whose vertical angle is 2tan−1 a/b.
iii) Prove that the line 5x − 2y − 1 = 0 is mid parallel to the lines 5x − 2y − 9 = 0 and 5x − 2y + 7 = 0.
iv) Find the locus of the circum-centre of a triangle whose two sides are along the coordinate axes and third side passes through the point of intersection of the lines ax + by + c = 0 and lx + my + n = 0.
v) A ray of light coming from the point (1, 2) passes through the point (5, 3) after the reflection at the point A on the x-axis. Find the co-ordinates of A.