Question

Whats the equation for a line thats perpendicular to 6x-2y=8 when it passes through (6,-5)

( i hope i wrote that right sorry)

Answer 1

Equation of required line is:
`\mathbf{x+3y=-9 }`

Explanation:

We need to find equation for a line that's perpendicular to
`6x-2y=8` when it passes through (6,-5)

We will use the slope-intercept form:
`y=mx+b` where m is slope and b is y-intercept.

Finding Slope m:

When two lines are perpendicular there slopes are
`m=-(1)/(m)`

Finding slope of given equation first:
`6x-2y=8`

Writing in slope-intercept form:

`6x-2y=8\\-2y=-6x+8\\y=(-6x)/(-2)+(8)/(-2)\\y=3x-4`

Comparing it with slope intercept form:
`y=mx+b` , m= 3

So, slope of given equation is 3

Now, the slope of required equation will be
`m=-(1)/(m)`

So, Slope m=-1/3

Finding y-intercept

Using slope m=-1/3 and point (6,-5) we can find y-intercept

`y=mx+b\\-5=(-1)/(3)(6)+b\\-5=-2+b\\b=-5+2\\b=-3`

So, y-intercept is b=-3

Now, finding the equation of required line having Slope m=-1/3 and y-intercept b= -3

`y=mx+b\\y=-(1)/(3)x-3`

Writing in standard form:

`y=-(1)/(3)x-3\\y=(-x-9)/(3)\\3y=-x-9\\x+3y=-9`

So, equation of required line is:
`\mathbf{x+3y=-9 }`