Let `P=(x_0,y_0)` be point and `l` line with equation `Ax+By+C=0` , then the distance between point `P` and line `l` is

`d(P,l)=(|Ax_0+By_0+C|)/(sqrt(A^2+B^2))`

In your case `x_0=2,` `y_0=-1,` `A=3,` `B=4` and `C=0.` Hence we have

`d=(|3cdot2+4cdot(-1)|)/(sqrt(3^2+4^2))=(|6-8|)/(sqrt(9+16))=|-2|/sqrt25=2/5`

**The distance is** `2/5.`

Another way of doing this is to find the point of intersection Q of the perpendicular drawn from the point P(2, -1) to the line 3x + 4y = 0 and then determine the distance between the two points P and Q.

Write 3x + 4y = 0 in slope intercept form

4y = -3x

y = (-3/4)x

The slope of a line perpendicular to this is 4/3. The equation of a a line with slope 4/3 and passing through (2, -1) is

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

4x - 8 = 3y + 3

4x = 3y + 11

The point of intersection of 4x = 3y + 11 and 3x + 4y = 0 can be determined by solving the two equations.

Substitute x = (-4/3)y in 4x = 3y + 11

4*(-4/3)y = 3y + 11

-16y = 9y + 33

25y = -33

y = -33/25

x = 44/25

The distance between (2, -1) and (44/25, -33/25) is

D = `sqrt((2 - 44/25)^2 + (-1 + 33/25)^2)`

= `(1/25)sqrt((50 - 44)^2 + (-25 + 33)^2)`

= `(1/25)sqrt(6^2 + 8^2)`

= `10/25`

= 0.4

The distance between the point (-2, 1) and the line 3x + 4y = 0 is 0.4