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Find equations of the lines through the given point parallel to the given line and perpendicular to the given line.

Point (7/8, 3/4)



Line 5x + 3y = 0

Verified Answer :

nobillionaireNobley

Equation of a line passing through point (x1, y1) with slope m is given by y - y1 = m(x - x1)
For perpendicular lines, the slope of the required line is given by -1 divided by the slope of the given line.
5x + 3y = 0
3y = -5x
y = -5/3x
slope of the given line is -5/3.
Thus, the slope of the required line = -1/(-5/3) = 3/5

y - 3/4 = 3/5(x - 7/8)
5(y - 3/4) = 3(x - 7/8)
5y - 15/4 = 3x - 21/8
40y - 30 = 24x - 21
24x - 40y = -30 + 21 = -9
24x - 40y = 9.

syed514

The given line -s 5x - 3y = 0, or y = 5x / 3, and so it has gradient 5 / 3. 
(a) A line parallel to it will thus also have a gradient of 5 / 3. You want the specific line through (3/4, 7/8). 
Since the line through (a, b) with gradient m is y - b = m(x - a), we have: 
y - 7/8 = 5/3 (x - 3/4) 
24y - 21 = 40(x - 3/4) = 40x - 30 
40x - 24y - 9 = 0 
(b) A line perpendicular to the given line will have a gradient m such that m 5 / 3 = -1. That is, m = -3 / 5 You want the specific line through (3/4, 7/8). 
Since the line through (a, b) with gradient m is y - b = m(x - a), we have: 
y - 7/8 = -3/5 * (x - 3/4) 
40y - 35 = -24(x - 3/4) = -24x + 18 
24x + 40y - 53 = 0

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