Find the point on the line 2x+7y-6=0 which is closest to the point (-5,5.
The distance from (m,n) to Ax+By+C=0 is
|Am+Bn+C|/sqrt(A^2+B^2)
All these values appear above. Use 'em.
sqrt((x-m)^2 + (y-n)^2) = distance above, x and y fit the line equation. Minimize that distance.
To find the point on the line 2x+7y-6=0 which is closest to the point (-5,5), we can use the concept of perpendicular distance.
1. Begin by solving the given equation for y to obtain it in slope-intercept form:
2x + 7y - 6 = 0
7y = -2x + 6
y = (-2x + 6)/7
The slope-intercept form of the line is y = (-2/7)x + 6/7.
2. Recall that the line perpendicular to another line has a negative reciprocal slope. Therefore, the slope of the line perpendicular to 2x + 7y - 6 = 0 is 7/2.
3. Use the given point (-5,5) and the perpendicular line's slope to find the equation of the perpendicular line. Using the point-slope form, we have:
y - y1 = m(x - x1)
y - 5 = (7/2)(x - (-5))
y - 5 = (7/2)(x + 5)
y - 5 = (7/2)x + 35/2
y = (7/2)x + 45/2
The perpendicular line's equation is y = (7/2)x + 45/2.
4. Now, we have two lines: 2x + 7y - 6 = 0 (original line) and y = (7/2)x + 45/2 (perpendicular line). We can solve these two linear equations simultaneously to find their point of intersection.
Substituting y = (7/2)x + 45/2 into the original line's equation:
2x + 7((7/2)x + 45/2) - 6 = 0
2x + (49/2)x + 315/2 - 6 = 0
(53/2)x + 309/2 = 0
(53/2)x = -309/2
x = (-309/2) * (2/53)
x = -309/53
Plugging this value of x back into the equation y = (7/2)x + 45/2:
y = (7/2)(-309/53) + 45/2
y = -2163/106 + 45/2
y = (-2163 + 2385)/106
y = 222/106
y = 111/53
Hence, the point on the line 2x + 7y - 6 = 0 closest to the point (-5,5) is (-309/53, 111/53).