Answer: Therefore, the equation of the line that is perpendicular to the given line and passes through the point (5, -4) is y = (6/5)x - 10.
Explanation:
To find the equation of a line that is parallel to the given line 5x + 6y = 7 and passes through the point (5, -4), we can use the slope-intercept form of the equation, which is y = mx + b, where m is the slope of the line and b is the y-intercept.
Since the given line has a slope of -5/6, the slope of the parallel line will also be -5/6. To find the y-intercept of the parallel line, we can substitute the coordinates of the point (5, -4) into the equation y = mx + b and solve for b. This gives us the equation -4 = (-5/6)(5) + b, which simplifies to -4 = -25/6 + b. Solving for b, we find that b = 1/6.
Therefore, the equation of the line that is parallel to the given line and passes through the point (5, -4) is y = (-5/6)x + (1/6).
To find the equation of a line that is perpendicular to the given line 5x + 6y = 7 and passes through the point (5, -4), we need to find the slope of the perpendicular line. The slope of a line perpendicular to another line is the negative reciprocal of the original line's slope. Therefore, the slope of the perpendicular line will be -6/-5 = 6/5.
To find the y-intercept of the perpendicular line, we can substitute the coordinates of the point (5, -4) into the equation y = mx + b and solve for b. This gives us the equation -4 = (6/5)(5) + b, which simplifies to -4 = 6 + b. Solving for b, we find that b = -10.
Therefore, the equation of the line that is perpendicular to the given line and passes through the point (5, -4) is y = (6/5)x - 10.