Question

What is the equation of a line perpendicular to 3x+ 4y =12and passes thru the point (3,2)? Write the equation in slope-intercept form (y=mx+b)

Answer 1

y=4/3X -2

Explanation:

4y=12-3X

divide both sides by4

y=-3x/4 +3

m1×m2= -1

m1= -3/4

-3/4× m2=-1

m2=4/3

Gradient is now=4/3

m= y -y1/x-x1

4/3= y-2/x-3

3y - 6= 4x - 12

3y= 4x - 12 +6

3y= 4x -6

divide both sides by 3

y =4/3x - 2

Answer 2

Answer: 3y= 4x-6

Explanation:

given line = 4y= -3x+12

slope= -3/4

so the perpendicular lines slope will be 4/3

new line = y = (4/3)x+c

==> 2 = (4/3)3+c

==> 2 = 4+c

==> c= -2

3y= 4x-6