Answer:
Either a can be 0 or 10/3
Step-by-step explanation:
Given that a right triangle has vertices as (6,2) and we have to find the third vertex on the y axis
Since the third vertex lie on y axis, it would have coordinates as (0,a)
Using slope formula we can say one pair of sides will be perpendicular
Let A (6,2) B (5,5) and C (0,a)
Slope of AB = change in y coordinate /change in x coordinate = -3
Slope of BC = [tex]\frac{a-5}{-5}[/tex]
Slope of CA = [tex]\frac{a-2}{-6}[/tex]
If AB and BC are perpendicular then
[tex]\frac{a-5}{-5}*-3 = -1\\\\3a-15 = -5\\a=10/3[/tex]... i
OR if BC and CA are perpendicular then,
(a-5)(a-2)= -30
[tex]a^2-7a+40 =0\\[/tex]
No real roots
OR if CA and AB are perpendicular
(a-2) =-2
a=0
Either a can be 0 or 10/3
