Answer:
<em>x=-7</em>
<em>y=11</em>
Step-by-step explanation:
<u>Perpendicular Vectors</u>
Two vectors defined as their endpoints
and
are perpendicular if their dot product a.b is zero. The dot product is
![\vec u.\vec v=ac+bd](https://tex.z-dn.net/?f=%5Cvec%20u.%5Cvec%20v%3Dac%2Bbd)
In other words
ac+bd=0
Let's treat all the points as the extremes of vectors, so we can easily find the missing coordinates
B (3,5) and C (7,15) define a segment, the vector
![\overrightarrow{BC}==](https://tex.z-dn.net/?f=%5Coverrightarrow%7BBC%7D%3D%3C7-3%2C15-5%3E%3D%3C4%2C10%3E)
The point A is A (x,9), we need to form a vector with B
![\overrightarrow{AB}==](https://tex.z-dn.net/?f=%5Coverrightarrow%7BAB%7D%3D%3Cx-3%2C9-5%3E%3D%3Cx-3%2C4%3E)
this vector must be perpendicular to BC, so, applying the dot product we have
![4(x-3)+40=0](https://tex.z-dn.net/?f=4%28x-3%29%2B40%3D0)
![4x-12=-40](https://tex.z-dn.net/?f=4x-12%3D-40)
![x=-7](https://tex.z-dn.net/?f=x%3D-7)
The point D is D(17,y), we need to form a vector with C
![\overrightarrow{CD}==](https://tex.z-dn.net/?f=%5Coverrightarrow%7BCD%7D%3D%3C17-7%2Cy-15%3E%3D%3C10%2Cy-15%3E)
this vector must be perpendicular to BC, so, applying the dot product we have
![(4)(10)+(10)(y-15)=0](https://tex.z-dn.net/?f=%284%29%2810%29%2B%2810%29%28y-15%29%3D0)
![40+10y-150=0](https://tex.z-dn.net/?f=40%2B10y-150%3D0)
![10y=110](https://tex.z-dn.net/?f=10y%3D110)
![y=11](https://tex.z-dn.net/?f=y%3D11)
The points are
![\boxed{A(-7,9), D(17,11)}](https://tex.z-dn.net/?f=%5Cboxed%7BA%28-7%2C9%29%2C%20D%2817%2C11%29%7D)