Respuesta :
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
[tex]y=x-3[/tex]Recall that the equation of a line in slope-intercept form is given by
[tex]y=mx+b[/tex]Where m is the slope and b is the y-intercept.
Comparing the given equation with the above standard form we see that,
[tex]slope=m=1[/tex]Since we are given that the lines are perpendicular so the slope of the other line must be negative reciprocal of the given line.
[tex]m_2=-\frac{1}{m_1}=-\frac{1}{1}=-1[/tex]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
[tex]y-y_1=m\mleft(x-x_1\mright)[/tex]Let us substitute the value of slope and the given point into the above equation.
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-(-2)_{}=-1(x-2_{}) \\ y+2_{}=-1(x-2_{}) \end{gathered}[/tex]Solving the equation for y.
[tex]\begin{gathered} y+2_{}=-1(x-2_{}) \\ y+2_{}=-x+2_{} \\ y=-x+2-2 \\ y=-x \end{gathered}[/tex]Therefore, the required equation of the line is
[tex]y=-x[/tex]