Question

if the line (x-y+1) + k(y-2k+4) =0 makes equal intercepts on the axes . Then what is the value of k ?​

Answer 1

$\textbf{Given:}$

$\text{Equation of the line is (x-y+1)+k(y-2k+4)=0}$

$\textbf{To find:}$

$\text{The value of k}$

$\textbf{Solution:}$

$\text{Consider,}$

$(x-y+1)+k(y-2k+4)=0$

$x+(k-1)y-2k^2+4k+1=0$

$x+(k-1)y=2k^2-4k-1$

$\text{Divide both sides by 2k^2-4k-1 }$

$\dfrac{x}{2k^2-4k-1}+\dfrac{(k-1)y}{2k^2-4k-1}=1$

$\dfrac{x}{2k^2-4k-1}+\dfrac{y}{\dfrac{2k^2-4k-1}{k-1}}=1$

$\text{Comparing this with}$

$\bf\,\dfrac{x}{a}+\dfrac{y}{b}=1$

$\text{we get}$

$\text{x-intercept,}\;a=2k^2-4k-1$

$\text{y-intercept,}\;b=\dfrac{2k^2-4k-1}{k-1}$

$\text{But,}\;a=b$

$2k^2-4k-1=\dfrac{2k^2-4k-1}{k-1}$

$1=\dfrac{1}{k-1}$

$k-1=1$

$\implies\boxed{\bf\,k=2}$

$\therefore\textbf{The value of k is 2}$

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