Pair of Straight Lines
The equations of two or more lines can be expressed together by an equation of degree higher than one. As we see that a linear equation in x and y represents a straight line, the product of two linear equations represent two straight lines, that is a pair of straight lines. Hence we study pair of straight lines as a quadratic equations in x and y.
Let L1 ≡ a1x + b1y + c1 = 0 and L2 ≡ a2x + b2y + c2 = 0, be separate equations of two straight lines. If P (x1, y1) is a point on L1, then it satisfies the equaiton L1 = 0. Similarly, if P (x1, y1) is on L2 then L2 = 0. If P (x1, y1) lies either on L1 = 0 or L2 = 0, then P (x1, y1) satisfies the equation (L1 ) (L2) = 0, and no other point satisfies L1 · L2 = 0. Therefore the equation L1 · L2 = 0 represents the pair of straight lines L1 = 0 and L2 = 0.
We first consider a simple case. Both the lines in this pair pass through the origin. Thus, their equations can be written as
The above equation suggests that the general equation of a pair of straight lines passing through the origin with slopes m1 and m2, ax2 + 2hxy + by2 = 0 is a homogenous equation of degree two, implying that the degree of each term is 2.
As a consequence of this formula, we can conclude that
· The lines are real and distinct, if m1 and m2 are real and distinct, that is if h2 − ab > 0
· The lines are real and coincident, if m1 and m2 are real and equal, that is if h2 − ab = 0
· The lines are not real (imaginary), if m1 and m2 are not real, that is if h2 − ab < 0
Also, we see that the lines represented by (6.30), are parallel (since both pass through the origin, the lines are coincident lines) if tan θ = 0, that is h2 − ab = 0, and perpendicular if cot θ = 0 that is a + b = 0
ax2 + 2hxy + by2 = 0 Let the equations of the two straight lines be y − m1x = 0 and y − m2x = 0
We know that the equation of bisectors is the locus of points from which the perpendicular drawn to the two straight lines are equal.
Let P (p, q) be any point on the locus of bisectors.
The perpendiculars from P (p, q) to the line y − m1x = 0 is equal to the perpendicular from P (p, q) to y − m2x = 0
Consider the equations of two arbitrary lines l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0 The combined equation of the two lines is
(l1x + m1y + n1) (l2x + m2y + n2) = 0
If we multiply the above two factors together, we get a more general equation to a pair of straight lines has the form
ax2 + 2hxy + by2 + 2gx + 2f y + c = 0 ......................6.40
An equation of the form (6.40) will always represent a pair of straight lines, provided it must able to be factorized into two linear factors of the form l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0.
Condition that the general second degree equation ax2 + 2hxy + by2 + 2gx + 2f y + c = 0 should represent a pair of straight lines
Let us rearrange the equation of the pair of straight lines ax2 + 2hxy + by2 + 2gx + 2f y + c = 0 as a quadratic in x, we have
Since each of the above equations represents a straight line, they must be of the first degree in x and y. Therefore the expression under the radical sign should be a perfect square and the condition for this is
Related Topics
Privacy Policy, Terms and Conditions, DMCA Policy and Compliant
Copyright © 2018-2024 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.