Question

The line ${(k+1)}^{2}x+ky-{2k}^2-25=0$ passes through a point regardless of the value k. Which of the following is the line slope $2$ passing through the point?

• A. $y=2x-8$
• B. $y=2x-5$
• C. $y=2x-4$
• D. $y=2x+8$

Answer 1

The correct option is A $y=2x-8$

$(k+1{)}^{2}x+ky-2k^2-25=0$

$y=mx+c$

$ky=-(k+1{)}^{2}x+(2k^2+25)$

$y=-\frac{(k+1{)}^2}{k}x+\frac{(2k^2+25)}{k}$

$k=-2+\sqrt{3}k=-2-\sqrt{3}$

$y=mx+c$

$m={\left(\frac{k+1}{k}\right)}^2=2$

$-(k+1{)}^2=2k$

$k=\frac{-4\pm \sqrt{2}}{2\times 1}$

$=-2\pm \sqrt{3}$

$=\frac{2k^2+25}{k}=$

$c=2[-2+\sqrt{3}+\frac{25}{-2+\sqrt{3}}\times \frac{-2-\sqrt{3}}{-2-\sqrt{3}}]$

$=2[-2\sqrt{3}\pm 5(-2-\sqrt{3}\left)\right]$

$=2[-2\sqrt{3}+10+5\sqrt{3}]=2[10+3\sqrt{3}]$

$c=8$

$y=2x-8$