Question

What is the equation of the line that is perpendicular to -x+y=7 and passes through (-1,-1)?

a.-x+y=2
b.-x-y=2
c.x-y=2
d.-x-y=0
e. x+y=0

Answer 1

Answer:

$-x - y = 2$.

Step-by-step explanation:

If the slope of a line in a plane is $m$ and the $y$-intercept of that line is $b$, the slope-intercept equation of that line would be $y = m\, x + b$.

Rearrange the equation of the given line $-x + y = 7$ in the slope-intercept form to find the slope of this line:

$-x + y = 7$.

$x -x + y = x + 7$.

$y = x + 7$.

Notice that in the slope-intercept equation of this given line, the coefficient of $x$ is $1$. Thus, the slope of the given line would be $m_{1} = 1$.

Two lines in a plane are perpendicular to one another if the product of their slopes is $(-1)$. In other words, if $m_{1}$ and $m_{2}$ are the slopes of two lines perpendicular to each other, then $m_{1}\, m_{2} = (-1)$.

Since $m_{1} = 1$ for the given line, the slope of the line perpendicular to this given line would be:

$m_{2} = (-1) / m_{1} = (-1) / 1 = -1$.

If the slope of a line in a plane is $m$, and that line goes through the point $(x_{0},\, y_{0})$, the equation of that line in point-slope form would be:

$y - y_{0} = m\, (x - x_{0})$.

The slope of the line in question is $m = (-1)$. It is given that this line goes through the point $(-1,\, -1)$, where $x_{0} = (-1)$ and $y_{0} = (-1)$. Thus, the equation of this line in point-slope form would be:

$y - y_{0} = m\, (x - x_{0})$.

$y - (-1) = (-1)\, (x - (-1))$.

$y + 1 = -(x + 1)$.

Rearrange this equation to match the format of the choices:

$y + 1 = -x - 1$.

$x + y = -2$.

$-x - y = 2$.