Question

Write the equation of a line that is perpendicular to 2x-7y=13 and which passes through the point ( -3,8).

Answer 1

Given parameters:

Equation of the line given  2x - 7y = 13

Coordinates of points = -3, 8

Unknown:

Equation of a line perpendicular = ?

Solution;

 To find the equation of this line, simply find the negative inverse of the given line.

         m  = $-\frac{1}{m}$

          2x - 7y = 13 , this is an equation of a straight line

  let us write it in the form y = mx + c so as to derive the slope

        y and x are coordinates

         m is the slope

        c is the y intercept

  2x - 7y = 13;

        -7y = -2x + 13

  divide through by -7;

             y = $\frac{2}{7}x$ + $\frac{13}{7}$

The slope of this line is $\frac{2}{7}$;

           Now the slope of a line perpendicular to it will be $-\frac{7}{2}$;

  the equation of the line will be;

        y  = $-\frac{7}{2} x + C$

Since the coordinate of this line is (-3,8) , let us find C;

     y  = 8 and x = -3;

     8  = $-\frac{7}{2}$ (-3) + C

         multiply through by 2;

 

    16 = 21 + 2C

    16 - 21  = 2C

      -5 = 2C

        C  = $\frac{-5}{2}$

Now the equation of the line is;

              y  = $-\frac{7}{2} x + \frac{5}{2}$