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Expert replies
by sud21 » Sat Jan 28, 2012 11:44 pm
Three points (0,3), (-2,0), and (b,0) composed a right triangle. b=?
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Source: — Problem Solving |

by LalaB » Sun Jan 29, 2012 1:00 am
b(0;0)
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by neelgandham » Sun Jan 29, 2012 9:04 am
Lala,
I think your solution is incomplete. Do correct me if I am wrong!

Q: Three points (0,3), (-2,0), and (b,0) composed a right triangle. b=?

Solution :
Let point (0,3) be labelled A
Let point (-2,0) be labelled B
Let point (b,0) be labelled C

Case 1: If line segments AC(on Y axis), BC(on X axis) are perpendicular(Red triangle in the attachment) and line segment AB is the hypotenuse, then line segments AC and BC meet at the origin (0,0). The value of b is 0

Case 2: If line segments AB(on X axis), BC(On X axis) are perpendicular and line segment AC is the hypotenuse. This is impossible as points B and C are on the same line(X axis) and angle between line segments AB and BC is a constant = Inverse Tan(3/2) !=90 degress

Case 3: If line segments AB,AC are perpendicular (Yellow triangle in the attachment) and line segment AB is the hypotenuse, then the product of slopes of AB and AC should be equal to -1.i.e. Slope of AB * Slope of AC = -1
(3/2)*(3/-b)=-1
b = 4.5

So the value of b can be 0 or 4.5
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by ronnie1985 » Sun Jan 29, 2012 10:25 am
By pythagoras theorem
(b+2)^2 = (2^2+3^2)+(b^2+3^2)
b = 4.5
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by LalaB » Sun Jan 29, 2012 9:45 pm
neelgandham, yep u r right.
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