Hi! i'm not sure of the answer to this.

write the equation of a line that is perpendicular to the given line and passes through the given point y-2= 7/3 (x+5) ; (-5,2)
answer options:
y+2= -3/7 (x-5)
y-5= - 3/7 (x+2)
y-2= -3/7 (x+5)
y+2=3/7 (x-5)

y - 2 = (-3/7)(x + 5)

Step-by-step explanation:

Recall that the slopes of a line and another line perpendicular to the first are negative reciprocals.

The slope of the given line is 7/3.  The negative reciprocal of that is -3/7.  Thus, we immediately eliminate the fourth answer choice.

We must now find the equation of the perpendicular line, with slope -3/7, that passes through (-5, 2).  We'll do that using the point-slope formula for the equation of a straight line.

That formula is    y - k = m(x - h), where m is the slope and (h, k) is the given point.

Substituting 2 for k, -5 for h and -3/7 for m, we get:

y - 2 = (-3/7)(x + 5).  This matches the 3rd given possible answer.

idk man good luck man