Question 123320
Equation of the point passing through 2 points is given by: 


{{{(y - y1)/(y2 - y1)}}} = {{{(x - x1)/(x2 - x1) }}}


The two points are: (4, 2) and (-4, 6) 


plugging in these points: 


{{{(y - 2)/(6 - 2)}}} = {{{(x - 4)/(-4 - 4)}}} 


{{{(y - 2)/4 }}} = {{{(x - 4)/(-8)}}} 


==> (y - 2) = {{{(x - 4)(-1/2)}}} 


This can be written as: 


==> 2(y - 2) = (-)(x - 4)


==> 2y - 4 = -x + 4 


==> 2y + x - 4 - 4 = 0


==> 2y + x - 8 = 0 


Or ==> y = -x/2 + 4 -------------------(1)


This is the equation of the line. 


So now to check whether the above equation is perpendicular to the given line 

y = 2x + 3   --------------------------(2)


For thsi condition to satisfy the product of their slopes should be equal to negative 1.


So first lets find the slope of eqn (1) ==> m1 = -1/2 


Slope of eqn (2) ==> m2 = 2 


So now the product of m1 & m2 = (-1/2) * (2)  = -1 


Hence both the lines are penpendicular. 


Regards