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In the plane rectangular coordinate system, given o (0,0) a (1,0) B (1,1) C (0,2), if any three of the four oabc points are selected, the probability of forming a right triangle is 0
There are four ways to choose any three of the four points in oabc
OA⊥OB,AB⊥OA,BC⊥OB
It can form three right triangles
If any three of the four oabc points are selected, the probability of forming a right triangle is 3 / 4
In the plane rectangular coordinate system, what's the probability that six points and three points can form a three angle
In the plane rectangular coordinate system, from any three of the six points a (0,0), B (2,0), C (1,1), D (0,2), e (2,2) f (3,3), what is the probability that these three points can form a triangle?
Please write down the solution in detail,
But I'm a bit stupid. How can I know that DCB and ACEF are in a straight line
(C63-C33-C43)/C63
=3/4
C63 is a kind of arbitrary three points
C33, because DCB is in a straight line, subtract this
C43, ACEF four point line, take three points to subtract
You should understand
In fact, the upstairs answer is very good, just a little mistake, give it to him, to encourage new people
In the plane rectangular coordinate system, a, B and C are not on the same line and form a triangle with area of 3
A (- 1,3) B (4, - 1) P (C, 0), find X
If you want 200 dollars, you'll get it. A(-1,3)B(4,-1)C(X,0)
I don't know high school stuff
Let the analytic formula of AB intersection X-axis on P line AB be y = KX + B substituting into a (- 1,3), B (4, - 1) - K + B = 34k + B = - 1K = - 4 / 5, B = 11 / 5Y = - 4x / 5 + 11 / 5 substituting into y = 0, - 4x / 5 + 11 / 5 = 0, x = 11 / 4, so the P coordinate is (11 / 4,0) △ ACP is based on CP, and the distance from a to X axis is high; △ BCP is based on CP, and the distance from B to X axis is high
In the plane rectangular coordinate system, three points from six points constitute a triangle
In the plane rectangular coordinate system, take three points from six points a (1,1) B (1,2) C (1,3) d (2,1) e (2,2) f (3,1) to form a triangle
A1/40 B39/40 C17/20 D3/20
Total number of choices: 6c3 = 20, draw a graph, collinear ABC can't, ADF can't, CEF can't, the rest can be, 17 kinds, choose C
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