How do you find the length of the line $x = A t + B, y = C t + D, a \leq t \leq b$ $x=At+B, y=Ct+D, a<=t<=b$ ?

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How do you find the length of the line $x = A t + B, y = C t + D, a \leq t \leq b$ $x=At+B, y=Ct+D, a<=t<=b$ ?

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Answer 1 · 2018-03-04T11:45:18

$l = | a - b | \sqrt{A^{2} + C^{2}}$ $l = abs(a-b)sqrt(A^2+C^2)$

Explanation:

Given the line

$L \to p = p_{0} + \to v t$ $L -> p = p_0 + vec v t$

with

$p = (x, y)$ $p = (x,y)$
$p_{0} = (B, D)$ $p_0 =(B,D)$
$\to v = (A, C)$ $vec v = (A,C)$

we have

$p_{a} = p_{0} + \to v a$ $p_a = p_0 + vec v a$
$p_{b} = p_{0} + \to v b$ $p_b = p_0 + vec v b$

and then

$l = ∥ p_{a} - p_{b} ∥ = ∥ ∥ p_{0} + (\to v) a - p_{0} - (\to v) b ∥ ∥ = ∥ ∥ \to v (a - b) ∥ ∥ = | a - b | ∥ ∥ \to v ∥ ∥ = | a - b | \sqrt{A^{2} + C^{2}}$ $l = norm(p_a-p_b)= norm(p_0+ (vec v) a - p_0 - (vec v) b) = norm(vec v(a-b)) = abs(a-b)norm (vec v) = abs(a-b)sqrt(A^2+C^2)$

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How do you find the length of the line $x = A t + B, y = C t + D, a \leq t \leq b$ $x=At+B, y=Ct+D, a<=t<=b$ ?

1 Answer

Explanation:

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How do you find the length of the line x=At+B, y=Ct+D, a<=t<=bx=At+B,y=Ct+D,a≤t≤bx=At+B, y=Ct+D, a<=t<=b?

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Explanation:

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How do you find the length of the line $x = A t + B, y = C t + D, a \leq t \leq b$ $x=At+B, y=Ct+D, a<=t<=b$ ?