Question

PLSSSSS HELP!!!!!

Write the equation of a line that is perpendicular to the given line and that passes through the given point.
y- 4 = (x+3); (-7,8)

O A. y-8 = -2/5(x-7)
OB. y-8 = -2/5(–7)
O C. y-8 = -2/5(x + 7)
O D.y-8 = -2/5(x +7)

Answer 1

`\displaystyle y - 8 = -(2)/(5)\, (x + 7)` is perpendicular to
`\displaystyle y - 4 = (5)/(2) \, (x + 3)` and goes through the point
`(-7,\, 8)`.

Explanation:

Consider a line that has a slope of
`m` and goes through the point
`(x_(0),\, y_(0))`. The point-slope equation of this line would be:

`y - y_(0) = m\, (x - x_(0))`.

The equation of "the given line" in this question is in the point-slope form. Compare the equation of this given line to
`y - y_(0) = m\, (x - x_(0))`.
`m = (5/2)`.

The coefficient of
`x` would be
`m = (5 / 2)`. In other words, the slope of this given line would be
`(5/2)`.

If two lines are perpendicular to one another, the product of their slopes would be
`(-1)`.

Since the slope of the given line is
`(5 / 2)`, the slope of a line perpendicular to this line would be:

`\displaystyle (-1)/(5 / 2) = -(2)/(5)`.

The question requested that this line should go through the point
`(-7,\, 8)`. Since the slope of that line is found to be
`(-2/5)`, the point-slope equation of that line would be:

`\displaystyle y - 8 = -(2)/(5)\, (x - (-7))`.

Simplify this equation to get:

`\displaystyle y - 8 = -(2)/(5)\, (x + 7)`.