SOLUTION: An equation of the line passing through (3,4) and perpendicular to y=7x+8 is
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Question 782057: An equation of the line passing through (3,4) and perpendicular to y=7x+8 is
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
Slope-intercept form is y = mx + b
m is the slope
b is the y-intercept
In the line y = 7x + 8
slope = 7
Perpendicular lines have slopes which are negative reciprocals.
Perpendicular slope is -1/7
On the new line,
slope = -1/7 and the given point is (3,4)
y - y1 = m(x - x1) {point-slope form}
y - 4 = (-1/7)(x - 3) {substituted into point-slope form}
y - 4 = (-1/7)x + 3/7 {used distributive property}
y = (-1/7)x + 31/7 {added 4 to each side}
For more help from me, visit: www.algebrahouse.com
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