Asked by Jake
Find the value of the constant t such that the triangle formed by A (2,2) ,B (4,-5) ,C (6,t) is right angled at B(4,-5).
Answers
Answered by
MathMate
Line passing through two given points P1(x1,y1) and P2(x2,y2)
L: (y-y1)/(y2-y1) = (x-x1)/(x2-x1)
or (x2-x1)(y-y1) = (y2-y1)(x-x1)
For line AB:
AB: (4-2)(y-2) = (-5-2)(x-2)
=> AB: 2(y-2) = (-7)(x-2)
=> AB: y = -3.5x + 9
Similarly, line BC is:
BC: (6-4)*(y+5)=(t+5)*(x-4)
=> BC: y=((t+5)/2)x - 2t-15
If the two lines are to be perpendicular to each other, the product of the slopes must be -1, or
-3.5*(t+5)/2 = -1
Solve for t to get t=-31/7
Check:
Slope of AB: (-5-2)/(4-2)=-3.5
Slope of BC: (-31/7+5)/(6-4)= 2/7
Product of slopes : -3.5*2/7=-1 OK
L: (y-y1)/(y2-y1) = (x-x1)/(x2-x1)
or (x2-x1)(y-y1) = (y2-y1)(x-x1)
For line AB:
AB: (4-2)(y-2) = (-5-2)(x-2)
=> AB: 2(y-2) = (-7)(x-2)
=> AB: y = -3.5x + 9
Similarly, line BC is:
BC: (6-4)*(y+5)=(t+5)*(x-4)
=> BC: y=((t+5)/2)x - 2t-15
If the two lines are to be perpendicular to each other, the product of the slopes must be -1, or
-3.5*(t+5)/2 = -1
Solve for t to get t=-31/7
Check:
Slope of AB: (-5-2)/(4-2)=-3.5
Slope of BC: (-31/7+5)/(6-4)= 2/7
Product of slopes : -3.5*2/7=-1 OK
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.