perpendicular to 3x+4y=13 and through (2,7)

Question

Summer Math packet question. write the equation in slope intercept and standard form

Answer 1

Equation of Line L₂= m₂X + b₂ Perpendicular to L_{1 : 3X + 4y = 13} passing through ( 2, 7)

 

 

    L1 :

             y = -3/4 X + 13/4

 

 

         then, m_{2 =} 4/3

 

 

       L_2:  Y ₌ 4/3 X ₊ b₂

              7 = 4/3(2) + b₂

 

              b_{2 = 7 - 8/3 = 13/3}

 

        L₂ : Y = 4/3 X + 13/3    

 

Answer 2

3x+4y=13

 

first we want to write this in standard form which is: y=mx+b       (where the slope is m and y-intercept is b)

 

so 4y=13-3x  and so y = (13-3x)/4 = 13/4 - 3/4 x  or  Y =(-3/4) X + (13/4)

 

The slope of this line is -3/4

 

For a line to be perpendicular to that, obviously it has to have the opposite sign. It turns out that a the slope of a perpendicular line is negative reciprocal  (inverse) of the slope of the original line. So we can say:

 

Y' = (4/3) X +b'          wher b' is the y intercept of the perpendicular line.

 

Since the line passes through (2,7), we have:

 

7 = (4/3) X + 2  

 

So  (4/3)X=5     and X = 5*(4/3) =20/3

 

So Y' = (4/3) X + (20/3)

 

Note: I have denoted the perpendicular line as Y' since I have already defined the original line as Y.