Question

1. Can you frame 3 equations in such a way that when it is paired with the 1st equation you should get a pair of
a) Parallel lines b) Coincident lines c) intersecting lines

Answer 1

Given: a) Parallel lines b) Coincident lines c) intersecting lines

To find:  frame 3 equations.

Solution:

• Consider the first equation as : 2x + 3y - 8 = 0

• Let second equation be: ax + by + c = 0

         Answer a:

         For parallel lines, a1/a2 = b1/b2 ≠ c1/c2

         Here a1 = 2, b1 = 3, a2 = a, b2 = b, c1 = -8, c2 = c

         Now putting values, we get:

         2/a = 3/b ≠ -8/c

         So putting any value of a, b and c according to condition, we get equations as:

         2x + 3y + 4 = 0 and 2x + 3y - 9 = 0

         Answer b:

         For coincident lines: a1/a2 = b1/b2 = c1/c2

         Here a1 = 2, b1 = 3, a2 = a, b2 = b, c1 = -8, c2 = c

         Now putting values, we get:

         2/a = 3/b = -8/c

         So putting any value of a, b and c according to condition, we get equations as:

         4x + 6y - 16 = 0 and 8x + 12y - 32 = 0

         Answer c:

         For intersecting lines: a1/a2 ≠ b1/b2

         Here a1 = 2, b1 = 3, a2 = a, b2 = b

         Now putting values, we get:

         2/a ≠ 3/b

         So putting any value of a and b except 2 and 3, we get equations as:

         4x + 5y = 0 and 5x + 9y = 4

Answer:

So the equations formed are:

a)Parallel lines:  2x + 3y - 8 = 0 and 2x + 3y + 4 = 0 and 2x + 3y - 9 = 0

b)Coincident lines: 2x + 3y - 8 = 0 and 4x + 6y - 16=0 and 8x + 12y - 32 = 0

c)Intersecting lines: 2x + 3y - 8 = 0 and 4x + 5y = 0 and 5x + 9y = 4