Answer to Question #142374 in Analytic Geometry for me

Question

the coordinates of point P and Q are (-2,3) and (1,2) respectively .A straight line QR is perpendicular ti the straight line PQ.find the equation of the straight line QR

Answer 1

$a(x-x_0)+b(y-y_0)=0\text{ where } \vec n\text{ normal line vector}$

$(x_0,y_0)=Q(1,2)$

$\vec n=\vec {PQ}=(1-(-2),2-3)=(3,-1)$

$3(x-1)-1*(y-2)=0$

$3x-y-1=0$

Answer: $3x-y-1=0$