Question 1027803
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
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we have
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y = -x/7 + b
:
x +7y -7b = 0  (this is the standard form for line, Ax +By +C = 0, note that C = -7b
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The distance from a point to a line is
:
d = |Ax +By +C| / square root(A^2 +B^2)
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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
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we have two cases to solve
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1) -10 -7b = 10
b = -20/7
:
2) -(-10 -7b) = 10
10 +7b = 10
b = 0
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we have two lines
:
y = -x/7
y = -x/7 -20/7
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:
The graph of these three lines is y = 7x+1 (red line), y = -x/7 (green line), y = -x/7 -20/7 (blue line)
:
{{{ graph( 300, 200, -10, 10, -5, 7, 7x+1, -x/7, -x/7-(20/7)) }}}
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