MHB Solve $ac+bd-ad-bc$ Given $\dfrac{3}{4}$, $\sqrt{a^2+c^2}=15$

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Given the ratios of positive integers \( \frac{a}{c} = \frac{3}{4} \) and \( \frac{b}{d} = \frac{3}{4} \), the values of \( a \), \( b \), \( c \), and \( d \) can be expressed as \( a = 3k \), \( b = 3m \), \( c = 4k \), and \( d = 4m \) for some integers \( k \) and \( m \). The equation \( \sqrt{a^2+c^2} - \sqrt{b^2+d^2} = 15 \) can be simplified using these expressions, leading to \( \sqrt{25k^2} - \sqrt{25m^2} = 15 \). This results in \( 5k - 5m = 15 \), giving \( k - m = 3 \). Finally, substituting back into the expression \( ac + bd - ad - bc \) yields a specific value based on the integers derived from \( k \) and \( m \).
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Assume that $a, b, c, d$ are positive integers and $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3}{4}$, and $$\sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15$$, find $ac+bd-ad-bc$.
 
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Re: Find ac+bd-ad-bc

anemone said:
Assume that $a, b, c, d$ are positive integers and $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3}{4}$, and $$\sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15$$, find $ac+bd-ad-bc$.

Sqrt(a^2+c^2) = 5/4 c
Sqrt(b^2+d^2) = 5/4 d
so 5/4(c-d) = 15or c-d = 12
ac + bd – ad –bc
= ¾ c^2 + ¾ d^2 – ¾ cd – ¾ cd
= ¾(c-d)^2 = ¾ * 12^2 = 108
 
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