SOLUTION: A triangle has vertices A(-1,k) B(6,k-1) and C(2,-1) where k is a positive number. (a) Calculate the gradient of AB (My answer was -1/7) (b) Find expressions for the gradient of:

Algebra ->  Points-lines-and-rays -> SOLUTION: A triangle has vertices A(-1,k) B(6,k-1) and C(2,-1) where k is a positive number. (a) Calculate the gradient of AB (My answer was -1/7) (b) Find expressions for the gradient of:      Log On


   



Question 23753: A triangle has vertices A(-1,k) B(6,k-1) and C(2,-1) where k is a positive number.
(a) Calculate the gradient of AB (My answer was -1/7)
(b) Find expressions for the gradient of:
(i) AC - I got (-1-k)/3
(ii) BC - I got k/4
(c)Find the value of k for which angle ACB + 90 degrees
( I know that m1m2 = -1, but I keep getting a negative answer for k)
Your help is very much appreciated!

Answer by venugopalramana(3286) About Me  (Show Source):
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A triangle has vertices A(-1,k) B(6,k-1) and C(2,-1) where k is a positive number.
(a) Calculate the gradient of AB (My answer was -1/7)
(K-1-K)/(6+1)=-1/7..........VERY GOOD
(b) Find expressions for the gradient of:
(i) AC - I got (-1-k)/3...VERY GOOD
(ii) BC - I got k/4.....VERY GOOD..
(c)Find the value of k for which angle ACB + 90 degrees
( I know that m1m2 = -1,EXCELLENT....OK LET US SEE.
(-1-K)*K/(3*4)= -1..MULTIPLYING BY -12 BOTH SIDES
K(K+1)=12..
K^2+K-12=0
FIND 2 FACTORS OF -12 WHOSE SUM IS +1...THEY ARE +4 AND -3..SO SPLIT K AS 4K-3K
K^2+4K-3K-12=0
K(K+4)-3(K+4)=0
(K-3)(K+4)=0
K-3=0....OR....K+4=0..
SO K=3 OR -4...
(but I keep getting a negative answer for k)YOU CAN HAVE + VE OR - VE...ANSWER FOR K ..NO PROBLEMS...
Your help is very much appreciated!IN FACT YOU DONT NEED ANY HELP..YOU ARE GOOD BY YOUR SELF AND YOU ARE SURE TO DO WELL.YOU ONLY NEED A LITTLE ENCOURAGEMENT AND SUPPORT...KEEP IT UP...