The equation of a straight line l1 is 3y+4x=12.Find the equation of another line l2 which is perpendicular to l1 and passes through [1, -2].

Answer 1

First we need to find the slope of the line l1. We can rearrange the equation to solve for y:

3y + 4x = 12

3y = -4x + 12

y = (-4/3)x + 4

The slope of l1 is therefore -4/3.

To find the slope of the line l2, we know that it must be the negative reciprocal of the slope of l1.

Slope of l2 = 3/4

Now we can use the point-slope form of a linear equation:

y - y1 = m(x - x1)

Where m is the slope and (x1, y1) is a point on the line.

We know that (x1, y1) = (1, -2) and m = 3/4, so:

y - (-2) = (3/4)(x - 1)

y + 2 = (3/4)x - 3/4

y = (3/4)x - 11/4

So the equation of line l2 is y = (3/4)x - 11/4.

Answer 2

Just a note:
Since the first line was expressed in standard form, maybe the second should also be in standard form:
3x-4y = 11

Answer 3

Yes, you are correct. Thank you for bringing this to my attention.

To convert the equation of line l2 to standard form, we can rearrange it as follows:

y = (3/4)x - 11/4

-3/4 x + y = -11/4

Multiplying both sides by -4 to eliminate the fraction:

3x - 4y = 11

So the equation of line l2 in standard form is 3x - 4y = 11, which is perpendicular to line l1 and passes through the point (1, -2).