Question

The line passing through (-4,-2) and (2, -3) is perpendicular
to line through (a, 5) and (2,-1). Find a​

Answer 1

First we find the distance of the coordinates of (-4,2); (2,-3)
root(x2-x1)^2+(y2-y1)^2
Root(2+4)^2+(-3+2)^2
root(6)^2+(-1)^2
36+1=37=root37

Now we find the distance of (a,5) and (2,-1)
Root (x2-x1)^2+(y2-y1)^2
Root (a-2)^2+(-1-5)^2
Root a^2-4a+4+(-6)^2
Root a^2-4a+4+36
Root a^2-4a+40

Equating both equations

Root a^2-4a+40=root 37
Root and root gets canceled

a^2-4a+40-37=0
a^2-4a+3=0
Now we factorise this equation
a^2-a-3a+3=0
a(a-1)-3(a-1)=0
(a-3)(a-1)=0
a-3=0
a=3
a-1=0
a=1

There are both numbers which satisfies the value of a that is 1,3