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Consider the line y=9x-5.
Find the equation of the line that is parallel to this line and passes through the point
(5, -4).

Find the equation of the line that is perpendicular to this line and passes through the point (5, -4).

Respuesta :

texaschic101
y = 9x - 5...the slope here is 9...a parallel line will have the same slope.

y = mx + b
slope(m) = 9
(5,-4)...x = 5 and y = -4
now we sub and solve for b, the y int
-4 = 9(5) + b
-4 = 45 + b
-4 - 45 = b
-49 = b
so ur parallel line is : y = 9x - 49

y = 9x - 5...slope is 9. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So the slope we need is -1/9.

y = mx + b
slope(m) = -1/9
(5,-4)...x = 5 and y = -4
now we sub and solve for b, the y int
-4 = -1/9(5) + b
-4 = -5/9 + b
-4 + 5/9 = b
-36/9 + 5/9 = b
- 31/9 = b

so ur perpendicular line is : y = -1/9x - 31/9

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