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Find the equation of a line that is perpendicular to y-4=0 and passes through (-1,6)...
Find the equation of a line that is perpendicular to y-4=0 and passes through (-1,6) and line is parallel to x+3=0 and passes through (-6,-7). Two perpendicular lines have the same x-intercept. An equation of one line is y=3x+1. Find the equation of the other line.
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y - 4 = 0 is the equation of a horizontal line. The line perpendicular to it is a vertical line. As is passes through (-1, 6), the equation of the line is x = -1.
The line x + 3 = 0 is a vertical line. A line parallel to it and passing through (-6, -7) has the equation x = -6.
The slope of the line y = 3x + 1 is 3, the slope of a line perpendicular to it is `-1/3` . The x-intercept of the line is `(-1/3, 0)` . The equation of a line with slope `-1/3` and passing through `(-1/3, 0)` is `y/(x + 1/3) = -1/3`
=> `3y + x + 1/3 = 0`
=> 3x + 9y + 1 = 0
Posted by justaguide on May 5, 2013 at 3:52 PM (Answer #1)
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