SOLUTION: Find b such that the points A(2,b),B(5,5)and C(-6,0) are the vertices of right angled {{{highlight(triangle)}}} with angle < BAC = 90 degree.

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Question 1148023: Find b such that the points A(2,b),B(5,5)and C(-6,0) are the vertices of right angled with angle < BAC = 90 degree.
Found 3 solutions by greenestamps, Boreal, ikleyn:
Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


corrected response.... the slopes are negative reciprocals, not equal!

The right angle is at A, so segments BA and CA are perpendicular; so the product of their slopes is -1.

AB slope: (5-b)/3
AC slope: b/8






or


Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
AB distance is sqrt (diff x^2+diff y^2) or sqrt((b-5)^2+9) or sqrt(b^2-10b+34)
BC distance is sqrt(11^2+5^2)=sqrt(146). I will call that the hypotenuse
CA distance is sqrt(8^2+b^2)
AB^2+CA^2=BC^2
or b^2-10b+34+b^2+64=146
or 2b^2-10b+98=146
or b^2-5b-24=0
(b-8)(b+3)=0
b=8 only positive root
The points are (2, 8) (5, 5) and (-6, 0)
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

            The solution by @Boreal is uncompleted.

            The solution by @greenestamps is incorrect.

            I came to bring the correct solution. 


The right angle is at A, so segments BA and CA are perpendicular. Hence, their slopes are inversely opposite.


AB slope: 

AC slope: 


The condition of perpendicularity is THIS equation

     = 

saying that the slope AC is inversely opposite to the slope AB.


Simplify and solve for "b" :


     = -24

     = 0.


Factor the quadratic polynomial in the left side

    (b-8)*(b+3) = 0.


The two roots are  b= 8  and  b= -3.


ANSWER.  The problem has two solutions  b= 8  and  b= -3.

Solved and completed.


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