Find the point on the line 7x+5y−2=0 which is closest to the point (−6,−2).
The idea is to find the family of lines perpendicular to L:7x+5y-2=0.
Out of that, we find the particular line L1 passing through P(-6,-2).
The intersection of L and L1 gives the required point (closest to P).
Family of lines perpendicular to L passing through P0(x0,y0):
L : 7x+5y-2=0
L1 : 5(x-x0)-7(y-y0)=0
in the above, we switch the coefficients of x and y, and change the sign of one of them. To make the line pass through P0, we subtract the coordinates of P0 from x and y respectively.
For P0(-6,-2),
L1 : 5(x-(-6))-7(y-(-2))=0
=> 5x-7y+16=0
Finally, the required point is the intersection point of L and L1, namely, the solution of the system of equations
7x+5y-2=0 and
5x-7y+16=0
Solving by elimination or by Cramer's rule, we obtain: (-33/37, 61/37)
Substitute values to verify that the solution is correct:
7x+5y=7(-33/37)+5(61/37)=2
5x-7y=5(-33/37)-7(61/37)=-16
ok
L1 :
To find the point on the line 7x + 5y - 2 = 0 that is closest to the point (-6,-2), we need to find the point on the line that has the minimum distance to the given point.
The distance between two points (x1, y1) and (x2, y2) can be calculated using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, the given point is (-6, -2), and we need to find a point (x, y) on the line 7x + 5y - 2 = 0 that minimizes the distance.
To find this point, we'll use the concept of orthogonal projection. The point on the line that is closest to the given point is the orthogonal projection of the given point onto the line.
The slope of the line 7x + 5y - 2 = 0 can be found by rearranging the equation in the form y = mx + c, where m is the slope:
7x + 5y - 2 = 0
5y = -7x + 2
y = (-7/5)x + 2/5
The slope of the line is -7/5.
The slope of a line perpendicular to the given line is the negative reciprocal of its slope. Therefore, the slope of the line perpendicular to the given line is 5/7.
Now, let's find the equation of the line perpendicular to the given line, passing through the given point (-6, -2). We'll use the point-slope form of a line:
y - y1 = m(x - x1)
Substituting the values, we have:
y - (-2) = (5/7)(x - (-6))
y + 2 = (5/7)(x + 6)
y + 2 = (5/7)x + 30/7
y = (5/7)x + 16/7
Now, we have two equations representing two lines:
1) The equation of the given line: y = (-7/5)x + 2/5
2) The equation of the perpendicular line passing through (-6, -2): y = (5/7)x + 16/7
Next, we'll find the intersection point of these two lines, which will give us the point on the given line that is closest to the given point.
To find the intersection point, we'll set the equations of the two lines equal to each other:
(-7/5)x + 2/5 = (5/7)x + 16/7
Multiplying both sides by 35 (the least common multiple of 5 and 7) to eliminate fractions, we get:
-49x + 14 = 25x + 80
Combining like terms:
-74x = 66
Dividing both sides by -74:
x = -66/74 = -33/37
Substituting this value of x into either of the equations, we can find the value of y:
y = (-7/5)(-33/37) + 2/5
y = 231/185
So, the intersection point of the two lines is (-33/37, 231/185).
Therefore, the point on the line 7x + 5y - 2 = 0 that is closest to the point (-6, -2) is (-33/37, 231/185).