Asked by khanyee
The equation of a line L1 is 2y-5x-1=0. Line L2 is parallel to L1 and passes through (2,1/4). What is the equation of L2, please? help me.
Answers
Answered by
Reiny
2y - 5x - 1 = 0
better form:
5x - 2y + 1 = 0
since your new line is parallel to the above, it would differ only in the constant, since it must have the same slope
new line:
5x - 2y + c = 0
but the point (2, 1/4) lies on it, so
10 - 1/2 + c = 0
c = -19/2
5x - 2y - 19/2 = 0
or in better form without fractions:
<b>10x - 4y - 19 = 0</b>
check:
slope = 10/4 = 5/2 <----- same as original
let x = 2
20 - 4y -19 = 0
-4y = -1
y = 1/4
All is good!
better form:
5x - 2y + 1 = 0
since your new line is parallel to the above, it would differ only in the constant, since it must have the same slope
new line:
5x - 2y + c = 0
but the point (2, 1/4) lies on it, so
10 - 1/2 + c = 0
c = -19/2
5x - 2y - 19/2 = 0
or in better form without fractions:
<b>10x - 4y - 19 = 0</b>
check:
slope = 10/4 = 5/2 <----- same as original
let x = 2
20 - 4y -19 = 0
-4y = -1
y = 1/4
All is good!
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