Question

Write the equation of a line that is perpendicular to y=-0.3x+6 and that passes through the point (3,-8)

Answer 1

To find the equation of a line that is perpendicular to y=-0.3x+6 and passes through the point (3,-8), we need to find the negative reciprocal of the slope of the given line. The equation of the perpendicular line is y = 3x - 17.

Step-by-step explanation:

To find the equation of a line that is perpendicular to y=-0.3x+6 and passes through the point (3,-8), we need to find the negative reciprocal of the slope of the given line. The given line has a slope of -0.3, so the negative reciprocal is 3. Therefore, the slope of the perpendicular line is 3.

Now we can use the point-slope form of a linear equation to write the equation of the perpendicular line. The equation will be:

y - y1 = m(x - x1)

Plugging in the values x1 = 3, y1 = -8, and m = 3, we get:

y - (-8) = 3(x - 3)

Simplifying, we have:

y + 8 = 3x - 9

And finally, rearranging the equation to the standard form, we get:

y = 3x - 17

Answer 2

we know that

if two lines are perpendicular

then

the product of their slopes is equal to minus one

so

`m1*m2=-1\\m2=-1/m1`

in this problem the slope of the given line is

`m1=-0.3=-(3)/(10)`

the value of m2 is

`m2=-1/(-3/10)=(10)/(3)`

with the point
`(3,-8)` and the slope m2 find the equation of the line

the equation of the line in the form point-slope is

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

substitute the values

`y+8=(10)/(3)*(x-3)`

`y=(10)/(3)x-10-8`

`y=(10)/(3)x-18`

therefore

the answer is

`y=(10)/(3)x-18`