3x + 5y = 7;( -1 , 2) write an eqautian thats perpidicaluar to the given line that passes through the given point

5x -3y + 11 = 0 .

Step-by-step explanation:

Firstly let's find the slope of the line by converting the equation into slope intercept form as ,

=> 3x +5y = 7

=> 5y = 7 - 3x

=> y = (7-3x)/5

=> y = -3/5x + 7/5

=> m = -⅗

Now the slope of the line perpendicular to this line will be , say m'

=> m' =- 1/m

=> m' = -1/-⅗

=> m = 5/3

=> ( x - x1) m = ( y - y1)

=> ( x - (-1)) × 5/3 = y - 2

=> ( x +1)×5 = 3( y -2)

=> 5x + 5 = 3y -6

=> 5x - 3y +5 +6=0

=> 5x -3y +11 = 0

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

practice. practice. practice.