Question

Find the equation for the line that passes through the point (2-2), and that is perpendicular to the line with the equation - x + y = -3.

Answer 1

We are asked to find the equation of a line that is perpendicular to the line -x + y = -3 and passes through the point (2, -2)

Re-writing the given equation in slope-intercept form

`y=x-3`

Recall that the equation of a line in slope-intercept form is given by

`y=mx+b`

Where m is the slope and b is the y-intercept.

Comparing the given equation with the above standard form we see that,

`slope=m=1`

Since we are given that the lines are perpendicular so the slope of the other line must be negative reciprocal of the given line.

`m_2=-(1)/(m_1)=-(1)/(1)=-1`

So the slope of the required equation is -1

Since we are also given that the line passes through the point (2, -2)

The point-slope form of the equation of a line is given by

`y-y_1=m\mleft(x-x_1\mright)`

Let us substitute the value of slope and the given point into the above equation.

`\begin{gathered} y-y_1=m(x-x_1) \\ y-(-2)_{}=-1(x-2_{}) \\ y+2_{}=-1(x-2_{}) \end{gathered}`

Solving the equation for y.

`\begin{gathered} y+2_{}=-1(x-2_{}) \\ y+2_{}=-x+2_{} \\ y=-x+2-2 \\ y=-x \end{gathered}`

Therefore, the required equation of the line is

`y=-x`