Question

Find the equation of the line perpendicular line to 3x-4y=7 and passing through (-1,4)

Answer 1

As the line is perpendicular to the equation 3x - 4y = 7
then,
m1 × m2 = -1
by the equation
4y = 3x -7
y = 3/ 4 x -7/4
so m1 = 3/4
so m2 will be - 4/ 3
so equation of line is
y - 4 = -4 /3 ( x - (-1))
that is
y-4 = -4/3 ( x +1 )
so equation is 4x + 3y = 8
_____ hope it helps

Answer 2

3x-4y=7
4y = 3x - 7
y = 3/4x -7/4
Slope = 3/4 (y= mx + c )
Let slope of line perpendicular to this line be m.
So
m * 3/4 = -1
m = -4/3
Given it passes through (-1,4)
Line passing through (x1,y1) and slope m is
y-y1 = m(x-x1)
Here
x1= -1
y1= 4
m= -4/3
y - 4 = -4/3(x+1)
3y -12 = -4x -4
3y + 4x = 8 ans