SOLUTION: find the eq of the legs of a right isosceles triangle if the of its hypotenuse is x-2y-3=0 and the vertex at the right angle is (1,6)

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Question 984277: find the eq of the legs of a right isosceles triangle if the of its hypotenuse is x-2y-3=0 and the vertex at the right angle is (1,6)
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
-2y=-x%2B3
y=x%2F2-3%2F2

The equation of the line through (1,6) and perpendicular to the given equation is y=-2%28x-1%29%2B6. Find the intersection point of these two lines. You will use information about 45-45-90 triangles. Note that the intersection point distance from (1,6) is also the distance ON THE GIVEN LINE that you want from the intersection point. Use the Distance Formula.

FINDING THE INTERSECTION POINT
x%2F2-3%2F2=-2%28x-1%29%2B6
x%2F2%2B2%28x-1%29=6%2B3%2F2
x%2B4%28x-1%29=12%2B3
x%2B4x-4=15
5x=19
x=19%2F5
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y=-2%2819%2F5-1%29%2B6
y=38%2F5-5%2F5%2B30%2F5
y=63%2F5
-
Intersection point on the given line, (19/5,63/5).

Distance from (1,6) to intersection point (19/5,63/5):
sqrt%28%2819%2F5-1%29%5E2%2B%2863%2F5-6%29%5E2%29
sqrt%28%2814%2F5%29%5E2%2B%2833%2F5%29%5E2%29
%281%2F5%29sqrt%2814%5E2%2B33%5E2%29
highlight_green%28%281%2F5%29sqrt%281285%29%29
%281%2F5%29%28sqrt%285%2A257%29%29-----5 and 257 both prime numbers

Next:
Whare are the points on the line y=x/2-3/2 which are (1/5)sqrt(1285) length from point (19/5,63/5) ?

Can you take care of this from here?