Question

Write the equation of the line that is perpendicular to the line y = x + 4 and passes through the point (−6, 3)

Answer 1

We can tell from the perpendicular line that its slope is 1, so we need to have a sloe of its negative reciprocal for our line, or -1. we can use this as m for the point slope form:
\[y - y1 = m (x - x1)\]
where y1 is 3
x1 is -6

Answer 2

the slope of a perpendicular line follows the format \[\large m = -\frac1m\] where m is the slope of the original line and -1/m is the slope of the perpendicular line. we would take the slope given to us from our equation y = x+4, where the slope is m=1, and plug it into this equation. \[m= -\frac11=-1\]so the slope of our perpendicular line is -1.
Now we're going to use the point (-6,3) and plug it into the point-slope form to find the equation of the tangent line. \[\large y-y{1}=m(x-x_{1}) \] plug in -6 and 3 into the x1 and y1 values. \[\large y-3=-1(x-(-6))\]simplify \[\large y-3 = -1(x+6)\]\[\large y-3 = -x-6\] Can you finish it off? :)

Answer 3

y=-2/3-1?

Answer 4

Line is y=mx+b
We need values for m and b
m = slope, = negative reciprocal of line perpendicular to it
From y = x + 4
Slope of perpendicular line is 1
m for our line= -1
y=(-1)x+b
Passes through (−6, 3)
3=(-1)(-6)+b
3=6+b
3-6=b
-3=b
Equation is y=(-1)x-3
y=-x-3
x+y=-3