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
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,
\[\Rightarrow 6x-3y=-3\]
\[\Rightarrow 2x-y=-1\]
A perpendicular line is a straight line through a point. It makes an angle of 90 degrees with a particular point through which the line passes. Coordinates and line equation is the prerequisite to finding out the perpendicular line.
Consider the equation of the line is ax + by + c = 0 and coordinates are (x1, y1), the slope should be − a/b. If one line is perpendicular to this line, the product of slopes should be -1. Let m1 and m2 be the slopes of two lines, and if they are perpendicular to each other, then their product will be -1.
\[\large Perpendicular\;Lines;\;m_{1}\times m_{2}=-1\]
\[\large Slope\;m=\frac{-a}{b}\]
\[\large Perpendicular\;Line \; equation:\; (y-y_{1})=m(x-x_{1})\]
Solved Example
Question: Check whether 2x + 3y + 5 = 0 and 3x – 2y + 1 = 0 are perpendicular or not.
Solution:
The given equations of lines are:
2x + 3y + 5 = 0 and 3x – 2y + 1 = 0
To check whether they are perpendicular to each other, find out the slopes of both lines. If the product of their slopes is -1, these lines are perpendicular to each other.
Slope formula is; m =
\(\begin{array}{l}\frac{-a}{b}\end{array} \)
Slope for first line,
\(\begin{array}{l}m_{1}\end{array} \)
=
\(\begin{array}{l}\frac{-a}{b}\end{array} \)
=\(\begin{array}{l}\frac{-2}{3}\end{array} \)
Slope for second line,
\(\begin{array}{l}m_{2}\end{array} \)
=\(\begin{array}{l}\frac{-a}{b}\end{array} \)
=\(\begin{array}{l}\frac{-3}{-2}\end{array} \)
\(\begin{array}{l}m_{1}\times m_{2}\end{array} \)
=\(\begin{array}{l}\frac{-2}{3}\end{array} \)
\(\begin{array}{l}\times\end{array} \)
\(\begin{array}{l}\frac{-3}{-2}\end{array} \)
= -1Since the product of slope is -1, the given lines are perpendicular to each other.