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Given the line 2x-3=9 and the point (4,-1) find the lines through the point that are (a) parallel to the given line and (b) perpendicular to it

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I must assume that you meant 2x - 3y = 9.

A line parallel to this line has exactly the same format, but a different constant on the right: 2x - 3y = c. Substituting the given values of x and y, we get:

2(4) - 3(-1) = 8+3 = 11. Thus, c = 11, and the equation of the new parallel line is

11 - 2x

2x - 3y = 11. Solving for y, -3y = 11 - 2x, or y = -------------

-3

so the slope of the parallel lilnes is 2/3.

A line perpendicular to the given line has a slope which is the negative reciprocal of 2/3: -3/2.

This perpendicular line has the equation

y + 1 = (-3/2)(x-4) (point-slope form), or

2y + 2 = -3(x-4) = -3x + 12. Then 2y = -3x + 12, and y = (-3/2)x + 6 (the perpendicular line's equation)

User Timclutton
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