Let the equation x(x + 2)(12–k) = 2 have equal roots. Then the distance of the point (k, k/2) from the line 3x+4y+5=0 is

Question

Let the equation \(x(x+2)(12-k)=2\) have equal roots. Then the distance of the point \(\left(\mathrm{k}, \frac{\mathrm{k}}{2}\right)\) from the line 3x + 4y + 5 = 0 is

(1) 15

(2) \(5 \sqrt{3}\)

(3) \(15 \sqrt{5}\)

(4) 12  

Answer 1

Correct option is: (1) 15 

\(\left(x^{2}+2 x\right)(12-k)=2\)

\(\lambda \mathrm{x}^{2}+2 \lambda \mathrm{x}-2=0\)

\(\mathrm{k} \neq 12\) Let \(12-\mathrm{k}=\lambda\)

\(\mathrm{D}=0\)

\(4 \lambda^{2}+8 \lambda=0\)

\(\lambda=0\) or \(\lambda=-2\)

\(\Rightarrow 12-\mathrm{k}=-2\)

\(\mathrm{k}=14\)

So \(P\left(k, \frac{k}{2}\right)=(14,7)\)

\(d=\left|\frac{3 \times 14+4 \times 7+5}{5}\right|=15\)