Question

Determine the equation of the line that passes through the point (2, -10) and isperpendicular to the line y= -2.

Answer 1

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,

`m(0)=-1`
`m=-(1)/(0)`

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,

`y-y1=m(x-x1)`

The equation of the line with slope m=-1/0 and passes through the point (2,-10) is

`y+10=-(1)/(0)(x-2)`
`-1(x-2)=0(y+10)`
`-x+2=0`
`x=2`

Hence the required equation of the straight line is x=2

Answer is: x=2.