Question

How do you write an equation of a line passing through (2, -2), perpendicular to $y=-x+3$ ?

Answer 1

We will use $y=mx+b$ first to find the value of the slope of the equation $y=-x+3$ . We will then use the formula ${{m}_{p}}=-\dfrac{1}{m}$ , which is the slope of a perpendicular line. We will 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 (2, -2), and an equation $y=-x+3$ . We know that $y=-x+3$ 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 $y=-x+3$ with $y=mx+b$ . We will get
$m=-1\text{ and }b=3$ .
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.
$\Rightarrow {{m}_{p}}=-\dfrac{1}{-1}=1$
We will now use a 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 (2, -2) . We will get
$\left( y-\left( -2 \right) \right)=1\left( x-2 \right)$
$\Rightarrow \left( y+2 \right)=x-2$
Let us take 2 from LHS to RHS. We will get
$y=x-2-2$
$\Rightarrow y=x-4$

Hence, the required equation of the line is $y=x-4$ .

Note: The parameters of slope intercept equation, $y=mx+b$ , may be misunderstood as b instead of m. You may make mistake by writing the point-slope formula as $\left( x-{{x}_{1}} \right)=m\left( y-{{y}_{1}} \right)$ instead of $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$ . Also, you may substitute for m in this equation as -1 instead of 1 (value of ${{m}_{p}}$ ).