170k views
5 votes
Find the equation of the straight line parallel to q) 4x+3y=1 and passes through point (0,1)​

User Tiendv
by
7.0k points

1 Answer

5 votes

Answer: To find the equation of the straight line parallel to the given line 4x+3y=1 and passing through the point (0,1), we can use the fact that parallel lines have the same slope.

First, we need to rearrange the given line into slope-intercept form, y = mx + b, where m is the slope of the line:

4x + 3y = 1

3y = -4x + 1

y = (-4/3)x + 1/3

Therefore, the slope of the given line is -4/3. Now, we can use this slope and the point (0,1) to find the equation of the line we're looking for:

y - y1 = m(x - x1) (point-slope form)

y - 1 = (-4/3)(x - 0)

y - 1 = (-4/3)x

y = (-4/3)x + 1

So the equation of the straight line parallel to 4x+3y=1 and passing through point (0,1) is y = (-4/3)x + 1.

Explanation:

User Evgeny Goldin
by
7.1k points