Question 150632
Here is your original line and point (-10,9).
{{{drawing( 300, 300, -15, 5, -5, 15,grid( 1 ),circle( -10,9, .3 ), graph( 300, 300, -15, 5, -5, 15, (-13/4)x-1.3)) }}} 
Perpendicular lines have slopes that are negative reciprocals of each other.
{{{m[1]m[2]=-1}}}
{{{(-13/4)m[2]=-1}}}
{{{m[2]=4/13}}}
Point-slope form using the point (-10,9) and the slope, just calculated, is
{{{y-y[1]=m(x-x[1])}}}
{{{y-9=(4/13)(x-(-10))}}}
Make it into slope-intercept form,
{{{y-9=(4/13)(x+10)}}}
{{{y=(4/13)x+(40/13)+9}}}
{{{y=(4/13)x+(40/13)+117/13}}}
{{{y=(4/13)x+157/13}}}
{{{drawing( 300, 300, -15, 5, -5, 15,grid( 1 ),circle( -10,9, .3 ), graph( 300, 300, -15, 5, -5, 15, (-13/4)x-1.3, (4/13)x+(157/13))) }}}