What is the equation, in slope-intercept form, of the line that is perpendicular to the liney - 4 = -2(x - 6) and passes through the point (-2,-2)?O y=-x - 10O y= {x + 10O y=zx-1O y=2x +1Mark this and returnSave and ExitNext

Accepted Solution

Answer:[tex]y=\frac{1}{2} x-1[/tex]Step-by-step explanation:The initial line is given in "point-slope form" ([tex]y-y_0=m(x-x_0)[/tex] therefore we can easily extract from it the slope, and obtain from it what should be the slope of a line perpendicular to it:Given the expression: [tex]y-4=-2(x-6)[/tex]we understand that the slope is "-2", and the point through which the line passes is the point (6,4) on the plane. We are concerned about the slope, since a line perpendicular to the given line must have a slope that corresponds to the "opposite of the reciprocal" of the original slope.The opposite of "-2" is "2", and the reciprocal of this is "1/2".Therefore the slope of the perpendicular line must be "1/2". Now we can build the equation of the line perpendicular line through the point (-2,-2) by using again the "point-slope form":[tex]y-y_0=m(x-x_0)\\y-(-2)=\frac{1}{2} (x-(-2))\\y+2=\frac{1}{2} (x+2)[/tex]Solving for y in this last expression we express the line in slope_y-intercept form:[tex]y+2=\frac{1}{2} (x+2)\\y+2=\frac{1}{2} x+1\\y=\frac{1}{2} x+1-2\\y=\frac{1}{2} x-1[/tex]