Please help me solve this math problem!!!

find the equation of a straight line which is perpendicular to the line 3x-4y=7 and passing through a point (2,3)

please solve with explanation ......asap!!! thank you !!!! c:

Firstly, if you place a straight line equaiton in the format of y=mx+C, note only y not 4y, then the m is the gradient.
For two straight lines that are perpendicular to each other then the product of their gradients is -1.
ie m1m2= -1

So when you've found your perdenicular gradient you cna then use
y-y1=m(x-x1), where x1,y1 are coordinates on the line.

Now, I've given everything you need to know to do this, rather than show it all, as it'll help you more.
Good luck !!

slope of givn line = -a/b=3/4

if two lines are perpendicular product of their slopes must be -1
that makes the slope of other line=-4/3

it passes through(2,3)

so according to point-slope form

y-3=-4/3(x-2)
3y-9=-4x+8
4x+3y-17=0

is the required equation

3x-4y=7
-4y = -3x +7
y = (3/4)x -7/4
m = -4/3
y - 3 = (-4/3) (x-2)
y - 3 = -4/3x + 8/3
y = -4/3x + 17/3

IS THAT LINEAR EQUATION WHICH IS IT SSC CBSE OR ICSE I HAD SAME SOME TO DO IN EXAM

please choose this as BEST ANSWER :P pls