The slopes of perpendicular line segments are 1/2 and d/3 What is the value of d?
2 answers:
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Remark
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.
Givens
a = x
b = x + 4
c = 20
Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2
Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20 Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0
2x^2 + 8x - 4 = 0 Divide through by 2
x^2 + 4x - 2 = 0
Use the quadratic formula
a = 1
b = 4
c = - 2
From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6) This is an extraneous root. No line can be minus.
x1 = + 0.4495
x2 = x + 4 = 4.4495
It's a right triangle so the Pythagorean Theorem applies. All you have to do is put the right things in the right places of the formula.
Givens
a = x
b = x + 4
c = 20
Formula and Substitution.
a^2 + b^2 = c^2
x^2 + (x + 4)^2 = 20^2
Solution
x^2 + x^2 + 8x + 16 = 20 Collect the like terms on the left.
2x^2 + 8x + 16 = 20 Subtract 20 from both sides.
2x^2 + 8x + 16 - 20 = 0
2x^2 + 8x - 4 = 0 Divide through by 2
x^2 + 4x - 2 = 0
Use the quadratic formula
a = 1
b = 4
c = - 2
From which x = (-4 +/- sqrt(24) ) / 2
x1 = (- 4 +/- sqrt(4*6) ) / 2
x1 = (- 4 +/- 2 sqrt(6) ) / 2
x1 = -2 + sqrt(6)
x2 = -2 - sqrt(6) This is an extraneous root. No line can be minus.
x1 = + 0.4495
x2 = x + 4 = 4.4495
Complete the square.
Use de Moivre's theorem to compute the square roots of the right side.
Now, taking square roots on both sides, we have
Use de Moivre's theorem again to take square roots on both sides.