Question

How do you write an equation of a line passing through (5, -3), perpendicular to $y=6x+9$ ?

Answer 1

Hint: From the given equation, we can see that it is in the slope-intercept form. We have to use $y=mx+b$ and compare it with $y=6x+9$ to get the value of m and b. We will use the formula for slope of a perpendicular line, that is, ${{m}_{p}}=-\dfrac{1}{m}$ . We then substitute the values in the point-slope formula, $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ to get the required equation.

Complete step by step answer:
We are given a point (5, -3) and an equation $y=6x+9$ . We know that $y=6x+9$ is of the form $y=mx+b$ , where m is the slope of the line b is the y-intercept.
Now, let us compare the equation $y=6x+9$ with $y=mx+b$ . We will get
$m=6\text{ and }b=9$ .
Let us denote the slope of the perpendicular as ${{m}_{p}}$ .
We know that the slope of a perpendicular line is given by
${{m}_{p}}=-\dfrac{1}{m}$ .
Let us now substitute the values.
$\Rightarrow {{m}_{p}}=-\dfrac{1}{6}$
We will now use point-slope formula to find the equation of the required line. We can write the point-slope formula as
$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ , where m is the slope and $\left( {{x}_{1}},{{y}_{1}} \right)$ denotes the point through which the line passes.
Now, we have to substitute the values in the above formula. Here, the slope is ${{m}_{p}}$ and the point is (5, -3) . We will get
$\left( y-\left( -3 \right) \right)=-\dfrac{1}{6}\left( x-5 \right)$
$\Rightarrow \left( y+3 \right)=-\dfrac{1}{6}x+\dfrac{5}{6}$
Let us take 3 from LHS to RHS. We will get
$y=-\dfrac{1}{6}x+\dfrac{5}{6}-3$
Now, we have to take the LCM.
$y=-\dfrac{1}{6}x-\dfrac{13}{6}$

Hence, the answer is $y=-\dfrac{1}{6}x-\dfrac{13}{6}$ .

Note: Slope intercept equation is the backbone of these types of problems. You may misunderstand the slope in $y=mx+b$ as b instead of m. Also, the value of m in the point-slope formula is the slope of perpendicular, that is, ${{m}_{p}}$ . You may make a mistake by substituting the value of m instead of ${{m}_{p}}$ .