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Given vector OA=(3,-4), vector OB=(6,-3), vector OC=(5-x,-3-y), if A, B, C can form a triangle, find the condition that x, y satisfies.
A, B, C can form a triangle, then the three points are no longer a straight line, when AB=OB-OA=(6,-3)-(3,-4)=(3,1) BC=OC-OB=(-x-1,-y) AB=rBC (3,1)=(-rx-1r,-yr)3=-rx-1r=-(x+1) r1=yr divide:3=-(x+1)/yy=-(1/3) x-(1/3) when y=-(1/3) x-(1,...
Vector OA=(3,-4), OB=(6,-3), OC=(5-x,-3-y).(1) If A, B, C can form a triangle, find the condition that x, y should satisfy.(2) If isosceles triangle ABC,∠B is a right angle, find the value of x, y.
(1) AB=(3,1).|AB|=√10.
BC=(-1-x,-y),|BC|=√[(1+x)2+y2]
CA=(-2+x,-1+y),|CA|=√[(2-x)2+(1-y)2]
If A, B, C can form a triangle, x, y should satisfy the following conditions ①②③
1√10.[(1+X)2+y2]>√[(2-x)2+(1-y)2]
2√10<√[(1+X)2+y2][(2-x)2+(1-y)2]
③√[(2-X)2+(1-y)2]10>√[(1+x)2+y2]
(2)10=(1+X)2+y2
10+(1+X)2+y2=(2-x)2+(1-y)2
Solve to (x, y)=(0,3) or (-2,-3).
Given the vector OA=(3,-4), OB=(6,-3), OC=(5-x,-3-y)(where O is the coordinate origin)(1) If A, B, C are collinear... Given the vector OA=(3,-4), OB=(6,-3), OC=(5-x,-3-y)(where O is the coordinate origin)(1) If A, B, C are collinear, find the expression of y about x (2) If △ABC is an isosceles right angle triangle with angle B.
1.
Vector AB=OB-OA=(3,1), vector CB=OB-OC=(x+1, y),
If A, B and C are collinear, then AB vector and CB vector are collinear;
Therefore:3y-(x+1)=0,
Get: y=(x+1)/3
2.
Isosceles triangle at a right angle to B,
Then: AB⊥CB, AB2=CB2
AB⊥CB:3(x+1)+y=0,1
AB2= CB2:10=(x+1)2+ y2,2
1 2 Two-form coupling column, and the solution is: x=0, y=-3; or x=-2, y=3;
So: x=0, y=-3 or x=-2, y=3;
If I don't understand, please, Hi, I,
If the vector a=(x,3), x€R, then x=4 is |a|=5? A Sufficient Unnecessary B Necessary Inadequate C Necessary D Inadequate and Inadequate
If |a|=√3^2+4^2=5, then |a|=5=>√x^2+3^2=5=> x=±4, then it is sufficient and unnecessary condition A
X=4 then |a|=√3^2+4^2=5 when |a|=5=>√x^2+3^2=5=> x=±4, so it is sufficient and unnecessary condition A
If 2(vector a + vector x)=3(vector b - vector x), then vector x =
0
Given the point A (-2,0), B (3,0), the moving point P (x, y) satisfies the square of the vector PA point multiplication vector PB=y, then what is the trajectory of the point P
The trajectory is a parabola. I just learned the trajectory of the equation this year.\x0d Steps:\x0dPA vector=(-2-x,-y)\x0dPB vector=(3-x,-y)\x0dPA point PB=x^2-x-6y^2=y^2\x0d\x0d Reduction: x^2-x-6=0\x0d Recall that the quadratic function in junior high school is a parabola,\x0d so the image is parabola.
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