Find the equation of the line that passes through the point (4, -1) and is perpendicular to the 3x-5y=20
2007-06-05 02:42:40 · 4 answers · asked by SadToday22 3 in Science & Mathematics ➔ Mathematics
First, I will convert 3x - 5y = 20 into slope-intercept form: 5y = 3x - 20 ==> y = (3/5)x - 4. This line has a slope equal to 3/5, so a line perpendicular to it would have slope equal to the opposite reciprocal, -5/3. Using the slope-point form with the given point (4, -1), I get y + 1 = (-5/3)(x - 4), which you may convert to any other form you like.
2007-06-05 02:46:08 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
-5y = -3x + 20
y = 3/5x - 4
the slope of the given line is 3/5, so the slope of the line you are looking for is -5/3 (negative reciprocal)
y - y1 = m(x - x1)
y - -1 = (-5/3)(x - 4)
y + 1 = (-5/3)x + 20/3
y = (-5/3)x + 17/3
2007-06-05 09:49:57 · answer #2 · answered by hawkeye3772 4 · 0⤊ 0⤋
y= -5/3x+3/20
2007-06-05 09:58:49 · answer #3 · answered by Anonymous · 0⤊ 0⤋
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2007-06-05 09:51:19 · answer #4 · answered by Anonymous · 0⤊ 1⤋