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,
$\Rightarrow 6x-3y=-3$
$\Rightarrow 2x-y=-1$