slope of AB = 1/2 , you had that
slope of BC = 2/(k-4)
since angle B is 90°, the above slopes must be negative reciprocals of each other, that is ....
2/(k-4) = -2
solve for k
point A(-1,-2) B(7,2) C(k,4) K is a constant. The vertices of ABC. Angle ABC is a right angle.
Ive worked out the gradient of AB to be 1/2.
How do you calculate the value of k??
thanks for the help
5 answers
sorry but confused
confused about what ?
The property of slopes of perpendicular lines , or
the procedure to solve the equation ?
The property of slopes of perpendicular lines , or
the procedure to solve the equation ?
where you get slope of BC = 2/(k-4) from??
As BC is perpendicular to AB, being a right angled triangle. The gradient of BC would be -2.
For the equation of BC,
y-2=-2(x-7)
y-2=-2x+14
2x+y=16
Then put C(k,4) as x and y,
2(k)+(4)=16
2k+4=16
2k=16-4
2k=12
k=6
For the equation of BC,
y-2=-2(x-7)
y-2=-2x+14
2x+y=16
Then put C(k,4) as x and y,
2(k)+(4)=16
2k+4=16
2k=16-4
2k=12
k=6