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Nitella [24]
3 years ago
13

Give the slope intercept form of the equation of the line that is perpendicular to 5x-3y=8

Mathematics
1 answer:
Stels [109]3 years ago
5 0
Hello : 
let the slope : a    and ; b the slope of the line : 5x - 3y = 8
<span>the line that is perpendicular to 5x-3y=8  so : a×b = - 1 
but : 
5x - 3y = 8...      y = (5/3)x - 8/3  when : b = 5/3
</span><span> the slope intercept form of the equation of the line that is perpendicular
 to 5x-3y=8 is : y = (5/3)x +c </span>
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Since a_1,a_2,a_3,\cdots are in arithmetic progression,

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b_2 = 2b_1

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\cdots\implies b_n=2^{n-1}b_1

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\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n

\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2

It follows that

a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n +  n(n-1)

so the left side is

2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n

Also recall that

\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}

so that the right side is

b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)

Solve for c.

2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}

Now, the numerator increases more slowly than the denominator, since

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This means we only need to check if the claim is true for any n\in\{1,2,3,4\}.

n=1 doesn't work, since that makes c=0.

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c = \dfrac{12}{2^3 - 6 - 1} = 12

If n=4, then

c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N

There is only one value for which the claim is true, c=12.

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2 years ago
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