Coordinates

I answered the same question here: https://web2.0calc.com/questions/coordinates_83752#r2

The area of the triangle is:  A  =  ½ · (base) · (height)  =  ½ · ( 7 ) · ( 3 )   =  21/2

The equation of line(BA) is:  y  =  (3/10)(x + 6).

The line  x = k  is a vertical line that passes through the x-intercept  (k, 0).

It also passes through the point  ( k, (3/10)(k + 6) )  on the line(BA).

The region of the triangle to the left of the line  x = k  is a triangle with base length = k + 6

and whose height is  (3/10)(k  + 6).

The area of this triangle is:  ½ · (k + 6) · (3/10)(k + 6)

Since this area = one-half of the total area of the triangle:

     ½ · (k + 6) · (3/10)(k + 6)  =  ½ · (21/2)

                        (3/10)(k + 6)²  =  21/2

                                  (k + 6)²  =  35

                                    k + 6  =  sqrt(35)                       (you can ignore the other answer)

                                          k  =  sqrt(35) - 6                  (a point slightly to the left of the axis)