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It is known that two points a (- 1, - 2) B (3,6) find a point P on the straight line L: 3x-3y-10 = 0, (1) minimize the distance between point P and a, B, (2) maximize the difference between point P and a, B, (3) minimize the sum of squares of the distance between point P and a, B
Make a symmetric point a 'about the line L: 3x-3y-10 = 0, and connect a'B with L, that is p
Point P coordinate solution: let AA 'intersect L and C, AA' and l be perpendicular and pass through a, and find that AA 'is y = - x-3, so C is (1 / 6, - 19 / 6)
C is the midpoint of AA ', so a' is (4 / 3, - 13 / 3), so the equation of a'B is y = 31 / 5 (x-3) + 6
P can be obtained by combining a'B and L
It is known that the intersection point of the straight line x-3y + 10 = 0 and 3x + 8y-4 = 0 is p
(emergency!) the intersection point of x-3y + 10 = 0 and 3x + 8y-4 = 0 is known as P (1). Find the coordinate of P (2). Find the equation of the line L passing through P and perpendicular to the line L: 4x-3y-6 = 0!
1. Simultaneous equations
x-3y+10=0 (1)
3x+8y-4=0 (2)
(2)-(1)*3
3x+8y-4-3(x-3y+10)=0
17y=34
y=2,x=-4
P(-4,2)
2. The slope of 4x-3y-6 = 0 is 4 / 3
The slope of L is - 3 / 4
The point oblique equation of L is Y-2 = (- 3 / 4) (x + 4)
Given that line L passes through the intersection of two lines 3x-y-10 = 0 and X + Y-2 = 0, and the distance between line L and point a (1,3) and point B (5,2) is equal, the equation of line L is obtained
From 3x − y − 10 = 0x + y − 2 = 0 & nbsp; When the line L is parallel to AB, the slope is KAB = 3 − 21 − 5 = - 14, so the equation of the line L is y + 1 = - 14 (x-3), that is, x + 4Y + 1 = 0. When the line L passes through the midpoint n (3,52) of AB, the equation of the line L is y + 1 = - 14 (x-3), because the line L passes through M and N, and Mn is perpendicular to the X axis In conclusion, the equation of line L is x + 4Y + 1 = 0 or x = 3
RELATED INFORMATIONS
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