The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is
The distance of the point \((-1,1)\) from the line \(12(x+6)=5(y-2)\) is
2
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5
Solution
The given line is \(12(x+6)=5(y-2)\)
\(\Rightarrow 12 x+72=5 y-10\)
or \(12 x-5 y+72+10=0 \Rightarrow 12 x-5 y+82=0\)
The perpendicular distance from $(x_{1}, y_{1})$ to the line $ax+by+c=0$ is $\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}$.
The point $(x_{1}, y_{1})$ is $(-1,1)$, therefore, the perpendicular distance from $(-1,1)$ to the line $12x-5y+82=0$ is $\frac{|-12-5+82|}{\sqrt{12^{2}+(-5)^{2}}}=\frac{65}{\sqrt{144+25}}=\frac{65}{\sqrt{169}}=5$.