Question 6761
The equation of the line that passes through the point (2, -3) and is perpendicular to the line whose equation is: 5x + 3y = 9 can be found by recalling that perpendicular lines have slopes that are the negative reciprocal of each other.
So, you need to determine the slope of the given line and then find the negative reciprocal of it to get the slope of the new line.

Put the given equation into the slope-intercept form: y = mx + b

5x + 3y = 9 Subtract 5x from both sides.
3y = -5x + 9 Divide both sides by 3.
y = (-5/3) + 3 The slope of the given line is: m = -5/3  The negative reciprocal of -5/3 is 3/5 and this is the slope of the new line.

So, for the new line you can write: y = (3/5)x + b  Now you need to find b, the y-intercept.
This done by substituting the x and y coordinates from the given point (2, -3) into y = (3/5)x + b and solving for b.

-3 = (3/5)2 + b Solve for b.
-3 = 6/5 + b Subtract 6/5 from both sides.
(-15/5)-(6/5) = b
-21/5 = b

Now you can write the equation of the new line:  y = (3/5)x - 21/5

You can check this by seeing that the given point, (2, -3) satisfies this equation and by seeng that the slope of this line is the negative reciprocal of the slope of the given line.
This is left as an exercise for the student.  If you have any further issues with this, please call again.