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Given the vector OA=(3,4) OB=(6.-3), OC=(5-m,-3-m). If ABC can form a triangle, then m should satisfy? Answer m is not equal to -7/10
As long as vector AB is not parallel to vector AC.
Vector AB=(3,-7), vector AC=(2-m,-7-m).
Vector AB parallel to AC requires (2-m)/3=(-7-m)/(-7),
The result is m=-7/10.
So long as m is not equal to -7/10.
I'm sorry, the calculation was wrong.
As long as vector AB is not parallel to vector AC.
Vector AB=(3,-7), vector AC=(2-m,-7-m).
Vector AB parallel to AC requires (2-m)/3=(-7-m)/(-7),
The result is m=-7/10.
So long as m is not equal to -7/10.
I'm sorry, I was wrong.
Given the vector OA=(3,-4), OB=(6,-3) OC,=(5-m,-(3+m)). If the point A. B. C can form a triangle, I already know the standard answer, but why is there only three points in the brick answer that are not collinear without the condition that the sides of the triangle are larger than the third?
In high school, the math is almost returned to the teacher. But as long as the three points on the same straight line can form a triangle, not on the line this point itself can be deduced from both sides and greater than the third side, why add a condition to verify it? Is the answer M not equal to 1? I've done some calculations, I hope I did n' t lose my high school lessons.
In high school, most of the math was returned to the teacher. However, as long as the three points were not in the same straight line, they could form a triangle. The point that was off-line could already deduce that the two sides were larger than the three sides. Why add a condition to verify it? Was the answer M not equal to 1? I actually calculated it. I hope I didn't lose my high school lessons.
In high school, the math is almost all returned to the teacher. But as long as not in the same straight line on the 3 points can form a triangle, not on the line on this point itself has been able to deduce both sides and greater than the third side, why add a conditional verification? Is the answer M not equal to 1? I've done some calculations, I hope I have n' t lost my high school lessons.
Given vectors OA=(1,1), OB=(2,30, OC=(m+1, m-10), if points A, B, C form a triangle, the range of the number m Given vectors OA=(1,1), OB=(2,30, OC=(m+1, m-10) If points A, B, C can form a triangle, the range of the number m is
The title should be "Given vectors OA=(1,1), OB=(2,3), OC=(m+1, m-1) If points A, B, C can form a triangle, then the range of the number m"
Draw the coordinate axis, O is the coordinate origin. Draw vectors OA and OB respectively, and connect AB
It can be seen that as long as A, B, C three different lines, can form a triangle
I.e. kAC=kAB
(M-1-1)/(m+1-1)=(3-1)/(2-1)
M=-2
Given λ1 2=1 and λ1 vector OA 2 vector OB=vector OC, it is proved that A, B and C are collinear.
λ1OA 2OB=λ1OA+(1-λ1) OB=λ1OA-λ1OB+OB=OC
λ1BA=BC
So A, B, C are collinear
λ1OA 2OB=λ1OA+(1-λ1) OB=λ1OA-λ1OB+OB=OC
λ1BA=BC
So A, B, C are collinear.
Let o be a point outside the line of points a and b, and prove that the condition of the three-point collinear of abc is vector o c = x times vector o a + y times vector ob, where x + Y =1. Let o be a point connected with a point and b is a point outside of the line, and prove that the condition of abc three-point collinear is the vector o c = x times vector o a + y times vector ob x+Y=1. Let o be a point outside the line of points a and b, and prove that the condition of abc three-point collinear is the vector o c = x times vector o a + y times vector ob where x + Y =1.
First change y in the equation into 1-x, then bring it into the equation, then shift the terms on both sides, combine them, and finally the two sides of the equation are:
BC vector = xBA vector, which means that A, B and C are on the same line.
Of course, you can try the law of rebuttal. If you don't understand, please continue to ask.
First transform y in the equation into 1-x, then bring it into the equation, then shift the terms on both sides, combine them, and finally the two sides of the equation are:
BC vector = xBA vector, which means that A, B and C are on the same line.
Of course, you can try the law of rebuttal. If you don't understand, please continue to ask.
The point O is a point on the plane of △ABC and satisfies the vector OA×vector OB=vector OB×vector OC=vector OC×vector OA. Then the point O is △ABC (). A Center of gravity B Center of gravity C Center of gravity D Center of gravity. Why? Which one?
B
OA*OB=OB*OC=OA*OC
Then OA*(OB-OC)=OB*(OC-OA)=OC (OA-OB)=0
OA*CB=OB*AC=OC*BA=0
OA⊥BC, OB⊥AC, OC⊥AB
O is the vertical center of △ABC
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