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Find k so that the line through (4, -3) and (k.1) is

a. parallel to 3x + 5y = 10,
b. perpendicular to 4x - 3y = - 1
a. k=

User Schaemelhout
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2.4k points

1 Answer

20 votes
20 votes

Answer:

a. k = -8/3 = -2 2/3

b. k = 32/5 = 6.4

Explanation:

You want to find the values of k that place the point (k, 1) on the line through the point (4, -3) when that line is (a) parallel to 3x +5y = 10, and (b) perpendicular to 3x +5y = 10.

a. Parallel

The equation of the parallel line will have the same x- and y-coefficients, but will have a constant that make the equation true at the point (4, -3).

3x +5y = 3(4) +5(-3) = 12 -15 = -3

The equation of the parallel line is

3x +5y = -3

When y=1, the value of k is ...

3k +5(1) = -3

3k = -8

k = -8/3 = -2 2/3 . . . . . . on line parallel to 3x+5y=10

b. Perpendicular

The equation of the perpendicular line will have swapped x- and y-coefficients, with one of them negated. The constant will be chosen to make the equation true at the point (4, -3).

5x -3y = 5(4) -3(-3) = 20 +9 = 29

The equation of the perpendicular line is

5x -3y = 29

When y=1, the value of k is ...

5k -3(1) = 29

5k = 32

k = 32/5 = 6.4 . . . . . . on th eline perpendicular to 3x+5y=10

Find k so that the line through (4, -3) and (k.1) is a. parallel to 3x + 5y = 10, b-example-1
User Wesley Franks
by
3.4k points