Question

Consider the line y=-2/7x-7

and the point is (-7,-4) what would the perpendicular line be and the parallel be

Answer 1

a) The equation of line perpendicular to above line and passes through point (-7,-4) is
`\mathbf{y=(7)/(2)x+(41)/(2)}`

b) The equation of line parallel to above line and passes through point (-7,-4) is
`\mathbf{y=-(2)/(7)x-6}`

Explanation:

We are given the line:
`y=-(2)/(7)x-7`

The slope of the above equation is:
`m=-(2)/(7)` (By comparing with general form
`y=mx+b` where m is slope)

a) Find equation of line perpendicular to above line and passes through point (-7,-4)

If two lines are perpendicular there slopes are opposite reciprocal i.e
`m_1=-(1)/(m_2)`

The slope of new line will be:
`m = (7)/(2)`

Now finding y-intercept using slope and point (-7,-4)

`y=mx+b\\-4=(7)/(2)(-7)+b \\-4=(-49)/(2)+b \\b=-4+(49)/(2)\\b=(-8+49)/(2)\\b=(41)/(2)`

So, the equation of line perpendicular to above line and passes through point (-7,-4) will be:

`y=mx+b\\y=(7)/(2)x+(41)/(2)`

So, the equation of line perpendicular to above line and passes through point (-7,-4) is
`\mathbf{y=(7)/(2)x+(41)/(2)}`

b) Find equation of line parallel to above line and passes through point (-7,-4)

If two lines are parallel there slopes are same i.e
`m_1=m_2`

The slope of new line will be:
`m = -(2)/(7)`

Now finding y-intercept using slope and point (-7,-4)

`y=mx+b\\-4=-(2)/(7)(-7)+b \\-4=-2(-1)+b\\-4=+2+b \\b=-4-2\\b=-6`

So, the equation of line parallel to above line and passes through point (-7,-4) will be:

`y=mx+b\\y=-(2)/(7)x-6`

So, the equation of line parallel to above line and passes through point (-7,-4) is
`\mathbf{y=-(2)/(7)x-6}`