Final answer:
Equations representing a line perpendicular to the line 5x – 2y = -6 and passing through (5, -4) can be derived using the negative reciprocal slope of -2/5 and the point-slope form, yielding options (b) 2x + 5y = -10 and (d) y + 4 = -2/5(x - 5). Option (c) is not a valid equation as it does not represent the perpendicular line passing through (5, -4).
Step-by-step explanation:
The question asks for equations that represent a line perpendicular to the line 5x – 2y = -6 and passes through the point (5, -4). First, we find the slope of the given line by rewriting the equation in slope-intercept form: y = mx + b. For 5x – 2y = -6, dividing every term by -2 gives us y = (5/2)x + 3. Thus, the slope is 5/2. A line perpendicular to this would have a negative reciprocal slope which is -2/5. Now, using the point (5, -4) and the slope -2/5, we apply the point-slope form of the equation, y - y1 = m(x - x1), which gives us y + 4 = -2/5(x - 5), matching option (d).
To find a standard form equation like option (b) and (c), we can rearrange (d) and multiply by 5 to eliminate fractions, giving us 5y + 20 = -2x + 10, leading to 2x + 5y = -10, which corresponds to option (b). Note that option (c) is not equivalent as its rearrangement does not pass through the point (5, -4). Lastly, option (a) differs in slope and thus is not perpendicular, while option (e) contains a typographical error with 'D' instead of 'y', disqualifying it as well.