Posted by **Jake** on Friday, January 21, 2011 at 3:18pm.

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).

- Math -
**MathMate**, Friday, January 21, 2011 at 9:54pm
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

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