Question

What is the equation of the line in slope-intercept that is perpendicular to the line `4y - 2 = 3x` and passes through the point (6,1)?

Answer 1

Let,the equation of the line required is `y=mx+c` where, `m` is the slope and `c` is the `Y` intercept.

Given equation of line is `4y-2=3x`

or, `y=3/4 x +1/2`

Now,for these two lines to be perpendicular product of their slope has to be `-1`

i.e `m(3/4)=-1`

so, `m=-4/3`

Hence,the equation becomes, `y=-4/3x+c`

Given,that this line passes through `(6,1)`,putting the values in our equation we get,

`1=(-4/3)*6 +c`

or, `c=9`

So,the required equation becomes, `y=-4/3 x+9`

or, `3y+4x=27` graph{3y+4x=27 [-10, 10, -5, 5]}