Final answer:
To find c when given the system of perpendicular lines intersecting at (1,-5), we use the given point of intersection and the fact that the product of slopes for perpendicular lines is -1. By substituting the point into both line equations and setting a = b based on the slope relationship, we find that c = 13.
Step-by-step explanation:
To find the value of c in the system of perpendicular lines ax - 2y = c and 2x + by = -c that intersect at the point (1,-5), we can use the fact that these lines are perpendicular and the point of intersection is known.
For two lines to be perpendicular, the product of their slopes must be -1. From the first equation, the slope can be obtained by rewriting it in slope-intercept form, which gives us y = (a/2)x - c/2. Therefore, the slope is a/2 for the first line. For the second line, rewriting 2x + by = -c in slope-intercept form gives y = -2/b × x - c/b, and the slope is -2/b.
The product of the slopes a/2 and -2/b should be -1, which gives us (a/2) × (-2/b) = -1 or a/b = 1. Now, substituting the point of intersection (1,-5) into the equation ax - 2y = c gives us a × 1 - 2 × (-5) = c, leading to c = a + 10. Since a/b = 1, we can assume a = b for non-zero a and b, hence c = a + 10 = b + 10.
Substituting the same point (1,-5) into the second equation 2x + by = -c gives us 2× 1 + b × (-5) = -c, which further simplifies to 2 - 5b = -c. Assuming a = b for non-zero a and b from the previous discussion, we replace b with a to get 2 - 5a = -c which combined with c = a + 10 gives us a single equation: 2 - 5a = -(a + 10).
Solving this we get: 2 - 5a = -a - 10, which simplifies to -4a = -12, giving us a = 3. Subsequently, c = a + 10 = 3 + 10, resulting in c = 13.