For what value of 'a', the line 3x-ay+9=0 is perpendicular to the line 27x+ay=-10.
To determine the value of 'a' for which the two lines are perpendicular, we need to compare their slopes. The slope of a line can be determined by expressing it in slope-intercept form (y = mx + b), where 'm' is the slope.
The given line 3x - ay + 9 = 0 can be rearranged to slope-intercept form as follows:
ay = 3x + 9
y = (3/a)x + 9/a
So, the slope of this line is 3/a.
Similarly, the line 27x + ay = -10 can be rearranged to slope-intercept form:
ay = -27x - 10
y = (-27/a)x - 10/a
The slope of this line is -27/a.
If two lines are perpendicular, their slopes are negative reciprocals of each other. In other words, the product of their slopes is -1.
So, we can set up the equation:
(3/a) * (-27/a) = -1
Simplifying this equation:
-81/a^2 = -1
Cross-multiplying:
81 = a^2
Taking the square root of both sides:
|a| = 9
Therefore, the value of 'a' for which the line 3x - ay + 9 = 0 is perpendicular to the line 27x + ay = -10 is either 9 or -9.