Question

If the slope of the line (x/a+y/b-1)+k ( x/b + y/a-1)=0 is -1 , then the value of 'k' is ​

Answer 1

$Given \: \: Question \: Is \: \\ \\ ( \frac{x}{a} + \frac{y}{b} - 1) + k( \frac{x}{b} + \frac{y}{a} - 1) = 0 \\ \\ if \: \: slope \: is \: \: \: - 1 \: \: \: find \: k \\ \\ \\ Answer \: \\ \\ The \:given \: equation \: can \: be \: rewrite \: as \\ \\ ( \frac{x}{a} + \frac{kx}{b} + \frac{y}{b} + \frac{ky}{a} - k - 1) = 0 \\ \\ ( \frac{1}{a} + \frac{k}{b} )x + ( \frac{1}{b} + \frac{k}{a} )y - (k + 1) = 0 \\ \\ now \: \: slope \: of \: this \: straight \: line \: is \\ \\ \frac{ - ( \frac{1}{a} + \frac{k}{b}) }{( \frac{1}{b} + \frac{k}{a} )} = - 1 \\ \\ becoz \: \: slope \: of \: straight \: line \: having \: equation \: \\ \\ \alpha x + \beta y + c = 0 \\ is \: \: \: \frac{ - \alpha }{ \beta } \\ => \\ \frac{ (\frac{1}{a} + \frac{k}{b}) }{ (\frac{1}{b} + \frac{k}{a}) } = 1 \\ \\ \frac{(b + ak)}{(a + bk)} = 1 \\ \\ b + ak = a+ bk \\ \\ (ak - bk) = (a - b) \\ \\ k(a - b) = (a - b) \\ \\ k = \frac{(a - b)}{(a - b)} \\ \\ k = 1 \\ \\ Therefore \: \: slope \: of \: given \: line \: is \: - 1 \: \: for \: \: k = 1$

Answer 2

Answer:

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