SOLUTION: Find the equation of the line perpendicular to y=7x+1 and passing at a distance of √2 from (4,-2). Then graph (i need it ASAP i beg of you)
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Question 1027803: Find the equation of the line perpendicular to y=7x+1 and passing at a distance of √2 from (4,-2). Then graph (i need it ASAP i beg of you)
Found 2 solutions by rothauserc, Alan3354:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The line perpendicular to y = 7x +1 has a slope which is the negative reciprocal of the slope of the given line.
:
slope for given line is 7, therefore the perpendicular line's slope is -1/7
:
we have
:
y = -x/7 + b
:
x +7y -7b = 0 (this is the standard form for line, Ax +By +C = 0, note that C = -7b
:
The distance from a point to a line is
:
d = |Ax +By +C| / square root(A^2 +B^2)
:
We are given
:
square root(2) = | 4 + (7(-2)) -7b | / square root(1 + 49)
:
square root(2) = | 4 -14 -7b | / square root(50)
:
5 * 2 = | -10 -7b |
:
| -10 -7b | = 10
:
we have two cases to solve
:
1) -10 -7b = 10
b = -20/7
:
2) -(-10 -7b) = 10
10 +7b = 10
b = 0
:
***********************************
we have two lines
:
y = -x/7
y = -x/7 -20/7
************************************
:
The graph of these three lines is y = 7x+1 (red line), y = -x/7 (green line), y = -x/7 -20/7 (blue line)
:
:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the equation of the line perpendicular to y=7x+1 and passing at a distance of sqrt(2) from (4,-2). Then graph (i need it ASAP i beg of you)
===============
Another method:
All points a distance of sqrt(2) from (4,-2) is a circle,
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Find the 2 points on the circle with a slope of the tangent of -1/7.
----
m = -(x-h)/(y-k) for a circle with its center at (h,k)
(x-4)/(y+2) = 1/7
7x-28 = y+2
y = 7x-30
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Sub for y in and solve for x, then y.
Then you have 2 points and a slope of -1/7
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