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Given the vector OA=(2,2), OB=(4,1), there is a point P on the X axis, so that AP×BP has the minimum value, then P point coordinates
P (x,0)
A (2,2) B (4,1)
AP*BP
=(X-2,-2)*(x-4,-1)
= X^2-6x+8+2
=(X-3)^2+1
AP*BP=1 when x=3
The coordinates of p are p (3,0)
Given that P is a point in a triangle ABC, and 3-vector AP+4-vector BP+5-vector CP=vector O, extend AP to BC at point D, if vector AB=vector a, vector AC=vector b, vector a, b represents vector AP, AD Given that P is a point in the triangle ABC, and 3-vector AP+4-vector BP+5-vector CP=vector O, extending AP to BC at point D, if vector AB=vector a, vector AC=vector b, vector a, b represents vector AP, AD
3AP+4BP+5CP=0, i.e.:3AP+4(AP-AB)+5(AP-AC)=12AP-4AB-5AC=0, i.e.: AP=4AB/12+5AC/12=(4a+5b)/12, order: AP=kPD, order: PD=(4AB+5AC)/(12k), order: AD=xAB+yAC, order: x+y=1, order: PD=AD-AP=(x-4/12) AB+(y-5/12) AC, order: x-1/3=1/(3k), y-5/12=5...
Given A (-4,0), B, C; two points move on the Y-axis and X-axis, the moving point P satisfies vector BC=vector CP, vector AB*vector BP=0 (1) Find the trajectory equation of the moving point P (2) Let the straight line passing through point A intersect with the trajectory of point P at two points E and F. Let D (4,0) find the value of KDE+KDF.
The point P (x, y), B (0, b), C (c,0) can be set by the title meaning. Then the vector BC=(c,-b), CP=(x-c, y), AB=(4, b), BP=(x, y-b). From the title meaning, the vector BC=CP, AB·BP=0.===>(c,-b)=(x-c, y),(4, b)·(x, y-b)=0.==>-b=y,4x+b (y-b)=0. The trajectory equation of the point P can be obtained by b: y2=2x.(...
Given points A (2,-1), B (-3,-2), if vector A B =1/3 vector BP, find the coordinates of point P
Vector AB=(-3-2,-2+1)=(-5,-1)
Vector BP=3 Vector AB=(-15,-3)
Coordinates of point P = vector BP + coordinates of point B
=(-15,-3)+(-3,-2)=(-18,-5)
Given the point O (0,0), A (1,2), B (4,5), and vector OP=vector OA+tvector AB. Given the point O (0,0), A (1,2), B (4,5), and vector OP=vector OA+t vector AB. When t changes, does point P move on a fixed line?
OA=(1,2), AB=(3,3), let P (x, y) then x=1+2t, y=2+3t, the linear equation is 3x-2y+1=0, i.e. no matter how t changes, point P is on the fixed line.
OA=(1,2), AB=(3,3), Let P (x, y) then x=1+2t, y=2+3t. The linear equation is 3x-2y+1=0, i.e. no matter how t changes, the point P is on the straight line.
Given O (0,0), A (2,-1), B (1,3), vector OP=vector OA+t vector AB 1. Find the value of t, and P is on the x-axis. P on the y-axis? Point P is in the fourth quadrant? 2. Whether O, A, B, P can be the four vertices of parallelogram. Given O (0,0), A (2,-1), B (1,3), vector OP=vector OA+t vector AB 1. Find the value of t, P on the x-axis? P on the y-axis? Point P is in the fourth quadrant? 2. Whether O, A, B, P can be the four vertices of parallelogram.
Solution 1) vector AB=(1-2,3+1)=(-1,4)
Vector OP=vector OA+t vector AB
=(2-T,4t-1)
Then P (2-t,4t-1)
If P is on the x-axis,4t-1=0, i.e. t=1/4
If P is on the y-axis,2-t=0, i.e. t=2
If P is in the fourth quadrant,2-t >0, and 4t-1
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