Find the point on the line −4x+2y+4=0 which is closest to the point (1,−5).
To find the point on the line −4x+2y+4=0 that is closest to the point (1, -5), we need to find the point on the line that intersects with the line perpendicular to it and passing through (1, -5).
First, we need to find the slope of the line −4x+2y+4=0. We can rearrange the equation into slope-intercept form (y = mx + b):
2y = 4x - 4
y = 2x - 2
The slope of this line is 2 (the coefficient of x).
A line perpendicular to this line will have a slope that is the negative reciprocal of 2, which is -1/2.
So the equation of the perpendicular line passing through (1, -5) is:
y - (-5) = (-1/2)(x - 1)
y + 5 = (-1/2)x + 1/2
y = (-1/2)x - 9/2
Now, we can find the intersection point of the two lines by setting the two equations equal to each other:
2x - 2 = (-1/2)x - 9/2
To solve for x, we can multiply both sides of the equation by 2 to eliminate the fraction:
4x - 4 = -x - 9
Now, move the x term to one side and the constant term to the other side:
4x + x = -9 + 4
5x = -5
x = -1
Substituting this value of x back into one of the equations, we can find the y-coordinate:
y = 2(-1) - 2
y = -4
So, the point (-1, -4) on the line −4x+2y+4=0 is closest to the point (1, -5).
To find the point on the line −4x+2y+4=0 which is closest to the point (1,−5), we need to compute the perpendicular distance between the line and the given point. This can be done using the formula for the distance between a point and a line.
The equation of the given line is −4x+2y+4=0. We can rewrite this equation in slope-intercept form (y = mx + b) by isolating y:
2y = 4x - 4
y = 2x - 2
From this, we can see that the slope (m) of the line is 2.
Now, let's compute the perpendicular distance between the line and the point (1,−5). We know that the line with slope 2 is perpendicular to the line connecting the point (1,−5) to the line itself.
The slope of the line connecting the point (1,−5) to the line with slope 2 is the negative reciprocal of 2. Therefore, the slope of the connecting line is -1/2.
Now, we can find the equation of the line connecting the point (1,−5) to the line with slope 2. Using point-slope form (y - y1 = m(x - x1)), we have:
y - (-5) = -1/2(x - 1)
y + 5 = -1/2x + 1/2
2y + 10 = -x + 1
x + 2y = -9
Now, we have a system of equations:
−4x+2y+4 = 0
x + 2y = -9
By solving this system, we can find the intersection point, which is the point on the line −4x+2y+4=0 closest to the point (1,−5).
If (a,b) is a point not on Ax + By + C = 0
then the shortest distance D from (a,b) to the line is
D = |aA + bB + C|/√(A^2 + B^2)
First change your equation into general form:
4x - 2y - 4 = 0
D = |1(4) + (-5)(-2) - 4|/√(16+4)
= 10/√20
= 10/2√5
= 5/√5
= √5