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Given that point P is a point in the plane of triangle ABC and satisfies 3PA + 5pb + 2pc = 0, let the area of ABC be s, then the area of triangle PAC is
3 (PA vector + Pb vector) + 2 (Pb vector + PC vector) = 0 vector, take AB midpoint m, BC midpoint n, take PA vector + Pb vector = 2pm vector, Pb vector + PC vector = 2pn vector, Pb vector + PC vector = 2pn vector, x0d6pm vector + 4pn vector = 0 vector, x0d3pm vector + 2pn vector = 0 vector, Mn is the center line of triangle ABC, P is on Mn
In triangle ABC, vector am = 1 / 3, vector AB, vector an = 1 / 3, vector AC, BN and cm intersect at point P
Prove that Mn is parallel to BC by vector method
If the coordinates of a, B and C are (0,0) (3,0) (2,4), the coordinates of point P are obtained
BC=AC-AB
MN=AN-AM=1/3AC-1/3AB=1/3(AC-AB)
BC/MN=3
So Mn is parallel to BC
In a triangle, the point P is a point on AB, and the vector CP = 2 / 3, the vector Ca + 1 / 3, the vector CB, q is the midpoint of BC
CP vector = CB vector + BP vector = 2 / 3CA vector + 1 / 3CB vectorization simple get BP vector = 2 / 3ba vector, then p is ab's trisection point, below you have not finished
In the plane rectangular coordinate system, the vertex a (0,0) B (5,0) C (8,4) d (3,4) of the quadrilateral ABCD can find the diagonal AC and BD of the quadrilateral ABCD, which are perpendicular to each other
First draw the rectangular coordinates
Make de ⊥ AB in E
∴AE=8-5=3
BE=8-4=4
CD = radical (8-3) & sup2; - (4-4) & sup2; = 5
∴AB=BC=CD=AD
The quadrilateral ABCD is a diamond (all four sides of the diamond are equal)
⊥ AC ⊥ BD (the diagonals of the diamond are perpendicular to each other)
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