Final answer:
To find the equation of the line perpendicular to 3x+8y-11=0 that passes through (0,-8), we first determined the original line's slope to be -3/8 and then used the negative reciprocal which is 8/3 as the slope for the perpendicular line. The final equation for the perpendicular line is y = (8/3)x - 8.
Step-by-step explanation:
The task at hand requires finding the equation of a line that is perpendicular to the given line 3x+8y-11=0 and that passes through the point (0,-8). We know from algebra that two lines are perpendicular if the slope of one line is the negative reciprocal of the slope of the other line. First, we need to identify the slope of the given line by rewriting it in slope-intercept form (y = mx + b).
By rearranging the given equation:
- 3x + 8y - 11 = 0
- 8y = -3x + 11
- y = (-3x/8) + 11/8
The slope (m) is -3/8. For the perpendicular line, the slope will be the negative reciprocal of this, which is 8/3. Now using the point-slope form y - y1 = m(x - x1) where m is the slope and (x1, y1) is the point the line passes through, we get:
- y - (-8) = (8/3)(x - 0)
- y + 8 = (8/3)x
- y = (8/3)x - 8
Therefore, the equation of the perpendicular line that passes through the point (0,-8) is y = (8/3)x - 8, where 8/3 is the slope and -8 is the y-intercept.