Asked by Jared
The points A(1,5) and B(9,3) are part of a triangle ΔABC. The triangle has a right angle at A and and sides satisfy AB=2AC. Find a point C such that C lies above the line AB.
Answers
Answered by
Reiny
Did you make your sketch ?
Two concepts are needed here:
1. the slope of perpendicular lines are negative reciprocals of each other
2. You have to know how to find the distance between two points.
1.
slope of AB = (9-1)/(3-5) = -4
so slope of AC = + 1/4
but slope of AC = (y-5)/(x-1)
so:
(y-5)/(x-1) = 1/4
x-1 = 4y - 20
x = 4y - 19
2.
AB = √68 , leaving it up to you to check that.
AC = √((x-1)^2 + (y-5)^2 )
but AB = 2AC
√68 = 2√((x-1)^2 + (y-5)^2 )
square both sides and expand
68 = x^2 - 2x + 1 + y^2 - 10y + 25
x^2 - 2x + y^2 - 10y -42 = 0
subbing in x = 4y-19
(4y-19)^2 - 2(4y-19) + y^2 - 10y - 42 = 0
leaving it up to you to solve for y, then find the matching x
you will get 2 points, but you want the one above line AB
check my arithmetic
Two concepts are needed here:
1. the slope of perpendicular lines are negative reciprocals of each other
2. You have to know how to find the distance between two points.
1.
slope of AB = (9-1)/(3-5) = -4
so slope of AC = + 1/4
but slope of AC = (y-5)/(x-1)
so:
(y-5)/(x-1) = 1/4
x-1 = 4y - 20
x = 4y - 19
2.
AB = √68 , leaving it up to you to check that.
AC = √((x-1)^2 + (y-5)^2 )
but AB = 2AC
√68 = 2√((x-1)^2 + (y-5)^2 )
square both sides and expand
68 = x^2 - 2x + 1 + y^2 - 10y + 25
x^2 - 2x + y^2 - 10y -42 = 0
subbing in x = 4y-19
(4y-19)^2 - 2(4y-19) + y^2 - 10y - 42 = 0
leaving it up to you to solve for y, then find the matching x
you will get 2 points, but you want the one above line AB
check my arithmetic
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