Consider the lines \(\mathrm{x}(3\lambda +1)+\mathrm{y}(7\lambda +2)=17\lambda +5\), \(\lambda\) being a parameter, all …

Rearrange the given equation by grouping terms with \(\lambda\):

\(x(3\lambda +1)+y(7\lambda +2)−17\lambda −5=0\)

\(\lambda (3x+7y−17)+(x+2y−5)=0\)

1. \(x+2y−5=0\Rightarrow x=5−2y\)
2. \(3x+7y−17=0\)

Substitute (1) into (2): \(3(5−2y)+7y−17=0\) \(15−6y+7y−17=0\) \(y−2=0\Rightarrow y=2\)

Then, \(x=5−2(2)=1\). So, the fixed point is \(P(1,2)\).

• Slope of \(OP=\frac{2−0}{1−0}=2\).
• Therefore, the slope of line \(L\) is perpendicular, which is \(−\frac{1}{2}\).

Equation of line \(L\) passing through \((1,2)\) with slope \(−\frac{1}{2}\)

: \(y−2=−\frac{1}{2}(x−1)\) \(2y−4=−x+1\) \(x+2y−5=0\)

Find the perpendicular distance \(d\) from point \((3,6)\) to line \(x+2y−5=0\):

\(d=\frac{\mathrm{∣}1(3)+2(6)−5\mathrm{∣}}{\sqrt{{1}^{2}+{2}^{2}}}\) \(d=\frac{\mathrm{∣}3+12−5\mathrm{∣}}{\sqrt{5}}=\frac{10}{\sqrt{5}}\)

Square both sides to find \({d}^{2}\): \({d}^{2}=\frac{100}{5}=20\)