does euqation of pair of lines have a xy term ? a second degree equation has xy term what about a pair of lines or pair of circles?
Answers
Answered by
0
A pair of lines represented by :
L1 : ax + by + c = 0 , slope = m1 = -a/b
and L2 = d x + e y + f = 0 slope m2 = - d/e
L1L2: (a x + b y + c) (d x + e y + f) = 0
L1L2: a d x² + b e y² + (b d + a e) x y + (a f + c d) x + (b f + c e) y + c f = 0
So a pair of straight lines has a XY term.
= (ad/be) x² + y² + (d/e + a/b) x y + (af + cd)/be x + (f/e + c/b) y + cf/ be = 0
= m1 m2 x² + y² - (m1 + m2) x y + ( af + cd)/be x + (f/e + c/b) y + cf/be = 0
in this form, the coefficient of x y is the sum of slopes of the lines.
=====================
a circles:
generally for a circle, the equation is : (x - a)² + (y - b)² = r²
This does not contain a term in XY.
L1 : ax + by + c = 0 , slope = m1 = -a/b
and L2 = d x + e y + f = 0 slope m2 = - d/e
L1L2: (a x + b y + c) (d x + e y + f) = 0
L1L2: a d x² + b e y² + (b d + a e) x y + (a f + c d) x + (b f + c e) y + c f = 0
So a pair of straight lines has a XY term.
= (ad/be) x² + y² + (d/e + a/b) x y + (af + cd)/be x + (f/e + c/b) y + cf/ be = 0
= m1 m2 x² + y² - (m1 + m2) x y + ( af + cd)/be x + (f/e + c/b) y + cf/be = 0
in this form, the coefficient of x y is the sum of slopes of the lines.
=====================
a circles:
generally for a circle, the equation is : (x - a)² + (y - b)² = r²
This does not contain a term in XY.
Similar questions