Question

Find an equation of the line perpendicular to the line 3x + 6y = 5 and passing through the point (1, 3). Write the equation in the standard form.

Answer 1

Hint: To solve the given question, we will first find out the slope of the line which is perpendicular to the given line with the help of the formula, \[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\] where \[\theta =\dfrac{\pi }{2}.\] After finding the slope of the line, we will write the equation of the new line in the slope intercept form: y = mx + c. To find the value of c, we will put the values of x, y, and m in the equation. After getting the slope intercept form, we will write the equation in the standard form: Px + Qy = R.

Complete step by step answer:
To start with, we will first find out the slope of the line which is perpendicular to the line 3x + 6y = 5. Let the slope of the given line be \[{{m}_{1}}\] and the slope of the new line be \[{{m}_{2}}.\] The formula of the angle between the two given lines whose slopes are \[\alpha \] and \[\beta \] is given by:
\[\tan \theta =\left| \dfrac{\alpha -\beta }{1+\alpha \beta } \right|\]
where \[\theta \] is the angle between the two lines. In our case, it is \[\dfrac{\pi }{2}.\] We have to find the value of \[{{m}_{2}}.\] With the help of the standard form of the line, we will first find out the value of \[{{m}_{1}}.\]
\[3x+6y=5\]
\[\Rightarrow 6y=5-3x\]
\[\Rightarrow 6y=-3x+5\]
\[\Rightarrow y=\dfrac{-3x}{6}+\dfrac{5}{6}\]
\[\Rightarrow y=\dfrac{-x}{2}+\dfrac{5}{6}\]
Now, y = mx + c is the slope intercept form. So, \[{{m}_{1}}=\dfrac{-1}{2}.\] Now, we will put the value \[{{m}_{1}}\] in \[\tan \theta .\] So, we can say that,
\[\tan \dfrac{\pi }{2}=\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]
\[\Rightarrow \dfrac{1}{0}=\dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}\]
\[\Rightarrow 1+{{m}_{1}}{{m}_{2}}=0\]
\[\Rightarrow {{m}_{1}}{{m}_{2}}=-1\]
\[\Rightarrow \left( \dfrac{-1}{2} \right){{m}_{2}}=-1\]
\[\Rightarrow {{m}_{2}}=2\]
Now, we have to find out the slope of the new line, so we will write the slope intercept form of this line. The slope intercept form of this line is,
\[y={{m}_{2}}x+c\]
Now, this line passes through the point (1, 3). So, we can say that,
\[\Rightarrow 3=2\left( 1 \right)+c\]
\[\Rightarrow c=1\]
Thus, the slope intercept form of the line is: \[y=2x+1\]
Now, we will write this equation in the standard form, i.e. of the form: Px + Qy = R where P is positive. Thus, we have,
\[y=2x+1\]
\[\Rightarrow 2x-y+1=0\]
\[\Rightarrow 2x-y=-1\]
Thus, the standard form of the line is 2x – y = – 1.

Note: The alternate method of solving the above question is shown: If we are given two equation of the line which are perpendicular to each and one of the lines has the equation, \[ax+by={{c}_{1}}\] then the other line will have the equation \[bx-ay={{c}_{2}}.\] In our case, a = 3, b = 6 and \[{{c}_{1}}=5.\] Thus the equation of the line perpendicular to the original line is: \[6x-3y={{c}_{2}}.\] Now, we know that this passes through (1, 3). So,
\[\Rightarrow 6\left( 1 \right)-3\left( 3 \right)={{c}_{2}}\]
\[\Rightarrow {{c}_{2}}=-3\]
Thus, we will get,
\[\Rightarrow 6x-3y=-3\]
\[\Rightarrow 2x-y=-1\]