Question

Write the equation of a perpendicular and parallel line to the following: (in picture)

Answer 1

anIf we have an equation of the form

`y=mx+b`

then the equation of the perpendicular will be

`y=-(1)/(m)x+c`

where c is any arbitrary constant determined by the point the line must pass through,

Now in our case, we have

`y=-10x+20`

Therefore, the equation of the perpendicular line will be

`y=(1)/(10)x+c`

We must choose c such that the above line passes through (-2,2).

Putting in y = 2 and x = 2 from (-2, 2) gives

`2=(1)/(10)(-2)+c`
`2=-(1)/(5)+c`
`\therefore c=(11)/(5).`

Hence, the equation of a line perpendicular y = -10x + 20 and passing through (-2, 2) is

`y=(1)/(10)+(11)/(5)`

Now we find the equation of a line parallel to y = -10x+20 and passing through (-2,2).

Now, the slope of the parallel line is the same as that of the original equation; therefore the equation for the parallel line we have is

`y=-10x+k`

We find k by using the point (-2,2) and substituting x = -2 and y = 2 into the above equation to get

`2=-10(-2)+k`
`2=20+k`
`\therefore k=-18.`

Hence, the equation of the parallel line is

`y=-10x-18`