find the point on the line 6x+7y-5=0 closest to (2,2)
6x+7y-5=0
+5 +5
6x+7y=5
-6x -6x
7y=5-6x
/7 /7
y=(5-6x)/7
To find the point on the line closest to a given point, you can make use of the perpendicular distance formula. In this case, the given line is 6x + 7y - 5 = 0, and the given point is (2,2).
1. First, let's rearrange the given line equation to the standard form of a line (y = mx + b):
6x + 7y - 5 = 0
7y = -6x + 5
y = (-6/7)x + 5/7
2. The slope-intercept form of a line (y = mx + b) tells us that the slope of the line is -6/7.
3. Perpendicular lines have slopes that are negative reciprocals of each other. Thus, the slope of the line perpendicular to the given line is 7/6.
4. We can find the equation of the perpendicular line passing through the point (2,2) using the point-slope form (y - y1 = m(x - x1)):
y - 2 = (7/6)(x - 2)
6y - 12 = 7x - 14
7x - 6y + 2 = 0
5. Now, we have a system of two equations:
6x + 7y - 5 = 0 (equation of the given line)
7x - 6y + 2 = 0 (equation of the perpendicular line)
6. Solving the system of equations yields the coordinates of the point of intersection, which is the closest point on the given line to the given point (2,2).
By solving the system of equations, we find x = 16/61 and y = 22/61. Hence, the point on the line 6x + 7y - 5 = 0 closest to (2,2) is approximately (16/61, 22/61).