given - equation of hypotenuse
vertex containing right angle
to find- equation of other two legs
kvnmurty:
is it a square ? rectangle ? Rhombus ? parallelogram?
ofc it is a trianglr i said na hypotenuse sir
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Let A (a, b) be the vertex containing the right angle.
Let the equation of hypotenuse BC be y = m x + c
we are given a, b, m, and c.
Let equation of AB be : y = p x + q and point A is on AB.
b = a p + q => q = b - a p
So equation of AB : y = p (x - a) + b , p is a parameter that is the slope of AB. There are infinitely many lines AB possible.
equation of AC : y = -1/p ( x - a) + b , as A(a, b) on AC and its slope is -1/p as AC is perpendicular to AB.
Now, the point B will be the intersection of BC : y = m x + c and AB :
y = m x + c = p(x - a) + b
x (m -p) = b - c - a p
x = (b - c- a p) / (m - p)
y = m (b - c - a p) / (m - p)
Now, the point of intersection of BC and AC is C:
y = m x + c = -1/p * (x - a) + b
x (m + 1/ p) = b - c + a/p
x = (p b - p c + a) / (pm + 1)
y = m (p b - p c + a) / (p m + 1)
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The solution:
Given hypotenuse BC: y = m x + c and A (a, b) , the vertex with right angle, we find the equations of AB, AC and the coordinates of B and C with a parameter p. There are infinitely many such triangles.
AB: y = p (x - a) + b
AC : y = -1/p (x - a) + b
B = [ (b - c- a p) / (m - p), m (b - c - a p) / (m - p) ]
C = [ (p b - p c + a) / (pm + 1) , m (p b - p c + a) / (p m + 1) ]
Let the equation of hypotenuse BC be y = m x + c
we are given a, b, m, and c.
Let equation of AB be : y = p x + q and point A is on AB.
b = a p + q => q = b - a p
So equation of AB : y = p (x - a) + b , p is a parameter that is the slope of AB. There are infinitely many lines AB possible.
equation of AC : y = -1/p ( x - a) + b , as A(a, b) on AC and its slope is -1/p as AC is perpendicular to AB.
Now, the point B will be the intersection of BC : y = m x + c and AB :
y = m x + c = p(x - a) + b
x (m -p) = b - c - a p
x = (b - c- a p) / (m - p)
y = m (b - c - a p) / (m - p)
Now, the point of intersection of BC and AC is C:
y = m x + c = -1/p * (x - a) + b
x (m + 1/ p) = b - c + a/p
x = (p b - p c + a) / (pm + 1)
y = m (p b - p c + a) / (p m + 1)
================
The solution:
Given hypotenuse BC: y = m x + c and A (a, b) , the vertex with right angle, we find the equations of AB, AC and the coordinates of B and C with a parameter p. There are infinitely many such triangles.
AB: y = p (x - a) + b
AC : y = -1/p (x - a) + b
B = [ (b - c- a p) / (m - p), m (b - c - a p) / (m - p) ]
C = [ (p b - p c + a) / (pm + 1) , m (p b - p c + a) / (p m + 1) ]
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Length of hypotenuse will be:
BC^2 = [ (pm+1) (b-c-ap) - (m-p)(pb-pc+a) ]^2 + [ m (pm+1)(b-c-ap) - m (m-p)(pb - pc +a) ]^2
= (1+m^2) [ pmb - pmc - ap^2m + b - c - ap - mpb + mpc - ma + p^2b - p^2c+pa ]^2
= (1+m^2) [ - am p^2 + b - c - ma + b p^2 - c p^2 ]^2
BC^2 = (1+m^2) (1+ p^2) (b - c - a m)
BC^2 = [ (pm+1) (b-c-ap) - (m-p)(pb-pc+a) ]^2 + [ m (pm+1)(b-c-ap) - m (m-p)(pb - pc +a) ]^2
= (1+m^2) [ pmb - pmc - ap^2m + b - c - ap - mpb + mpc - ma + p^2b - p^2c+pa ]^2
= (1+m^2) [ - am p^2 + b - c - ma + b p^2 - c p^2 ]^2
BC^2 = (1+m^2) (1+ p^2) (b - c - a m)
So length of the hypotenuse of the triangle ABC depends on parameter p, with minimum value when p = 0.
length of side AB² = [ (b - am)² (1 + p²) - 2 c (b - am) (1 + pm) + c² (1+ m²) ] / (m-p)²
OK
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