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.
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:
10x - 4y - 19 = 0
check:
slope = 10/4 = 5/2 <----- same as original
let x = 2
20 - 4y -19 = 0
-4y = -1
y = 1/4
All is good!
To find the equation of a line parallel to L1 and passing through a given point, we need to use the slope-intercept form of a line: y = mx + b.
1. First, we need to find the slope of line L1. The equation of L1 is given as 2y - 5x - 1 = 0. Rewriting it in slope-intercept form, we get:
2y = 5x + 1
y = (5/2)x + 1/2
From this equation, we can see that the slope of line L1, m1, is 5/2.
2. Since line L2 is parallel to L1, it will have the same slope. Therefore, the slope of line L2, m2, is also 5/2.
3. We know that the equation of L2 passes through the point (2, 1/4). We can use this point, along with the slope, to find the y-intercept, b2, of the line L2.
Using the point-slope form of a line, we have:
y - y1 = m(x - x1)
Substituting the given point (2, 1/4) and the slope (5/2) into the equation, we get:
y - 1/4 = (5/2)(x - 2)
Expanding and simplifying the equation:
y - 1/4 = (5/2)x - 5
y = (5/2)x - 5 + 1/4
y = (5/2)x - 20/4 + 1/4
y = (5/2)x - 19/4
Therefore, the equation of line L2 is 2y - 5x - 19 = 0.