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