A line which is passing through \[\left( 1,\text{ }2 \right)\]
To Find: The equation of a straight line.
By using the formula,
The equation of line is \[[y\text{ }-\text{ }{{y}_{1}}~=\text{ }m(x\text{ }-\text{ }{{x}_{1}})]\]
Here, \[sin\text{ }\theta \text{ }=\text{ }3/5\]
We know, \[sin\text{ }\theta \text{ }=\text{ }perpendicular/hypotenuse\]
\[=\text{ }3/5\]
So, according to Pythagoras theorem,
\[{{\left( Hypotenuse \right)}^{2}}~=\text{ }{{\left( Base \right)}^{2}}~+\text{ }{{\left( Perpendicular \right)}^{2}}\]
\[{{\left( 5 \right)}^{2}}~=\text{ }{{\left( Base \right)}^{2}}~+\text{ }{{\left( 3 \right)}^{2}}\]
Or,
\[\left( Base \right)\text{ }=\surd (25-9)\]
\[{{\left( Base \right)}^{2}}~=\surd 16\]
Or,
\[Base\text{ }=\text{ }4\]
Hence, \[tan\text{ }\theta \text{ }=\text{ }perpendicular/base\]
\[=\text{ }3/4\]
The slope of the line, \[m\text{ }=\text{ }tan\text{ }\theta \]
\[=\text{ }3/4\]
The line passing through \[({{x}_{1}},{{y}_{1}})\text{ }=\text{ }\left( 1,2 \right)\]
The required equation of line is \[y\text{ }-\text{ }{{y}_{1}}~=\text{ }m(x\text{ }-\text{ }{{x}_{1}})\]
Now, substitute the values, we get
\[y\text{ }-\text{ }2\text{ }=~\left( {\scriptscriptstyle 3\!/\!{ }_4} \right)~\left( x\text{ }-\text{ }1 \right)\]
\[4y\text{ }-\text{ }8\text{ }=\text{ }3x\text{ }-\text{ }3\]
So,
\[3x\text{ }-\text{ }4y\text{ }+\text{ }5\text{ }=\text{ }0\]
∴ The equation of line is \[3x-4y\text{ }+\text{ }5\text{ }=\text{ }0\]