Answer:
To determine the equation of the line that passes through the point (2, -10) and is
perpendicular to the line y= -2.
Let m be the slope of the required line,
we know that,
Product of the slope of the line and its perpendicular is -1
we get,
Slope of the line y=-2 is 0
Since for the y=mx+c its slope is m, for y=-2 the coefficient of x is 0, we get slope of y=-2 is 0
Product of the slope of the line and its perpendicular is -1
Substitute the values of the slopes we get,
[tex]m(0)=-1[/tex][tex]m=-\frac{1}{0}[/tex]
It is not defined, but consider slope as m=-1/0
We know that equation of the line with slope m and passes through the point (x1,y1) is,
[tex]y-y1=m(x-x1)[/tex]
The equation of the line with slope m=-1/0 and passes through the point (2,-10) is
[tex]y+10=-\frac{1}{0}(x-2)[/tex][tex]-1(x-2)=0(y+10)[/tex][tex]-x+2=0[/tex][tex]x=2[/tex]
Hence the required equation of the straight line is x=2
Answer is: x=2.
