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It is known that the three vertices of △ ABC are a (0,3), B (1,5), C (3, - 5). (I) find the linear equation of edge AB; (II) find the linear equation of middle line ad
The linear equation of AB is y − 35 − 3 = x − 01 − 0. The order is: 2x-y + 3 = 0; (II) from B (1,5), C (3, - 5), 1 + 32 = 2, 5 + (− 5) 2 = 0, and the midpoint of BC is d (2,0)
The vertices of the triangle are a (- 5,0), B (3, - 3), C (0,2),
To find the coordinates of the center of gravity; to find the equations of the height and center line of each side; to find the bisector equation of the angle ABC; to find the area of the triangle
AB edge midpoint (- 1, - 3 / 2), AB edge midline equation (two-point formula): (y + 3 / 2) / (x + 1) = (Y-2) / (x-0),
It is concluded that y = 7x / 2 + 2 （1）
BC edge midpoint (3 / 2, - 1 / 2), BC edge midline equation (two-point formula): (y + 1 / 2) / (x-3 / 2) = (y-0) / (x + 5),
The result is: y = - X / 13-5 / 13 （2）
(1) (2) the solutions of the simultaneous equations are: x = - 2 / 3, y = - 1 / 3; the coordinates of the center of gravity are (- 2 / 3, - 1 / 3)
AC edge midpoint (- 5 / 2,1), AC edge midline equation (two-point formula): (Y-1) / (x + 5 / 2) = (y + 3) / (x-3)
The result is: y = - 8x / 11-9 / 11
AB edge equation (two-point formula): (y-0) / (x + 5) = (y + 3) / (x-3), sorted out: y = - 3x / 8-15 / 8, slope = - 3 / 8,
The slope of AB contour line is 8 / 3, and the equation of AB contour line is Y-2 = 8 (x-0) / 3 (3)
BC boundary equation (two-point formula): (y + 3) / (x-3) = (Y-2) / (x-0), sorted out: y = - 5x / 2 + 3, slope = - 5 / 2,
The slope of BC contour line is 2 / 5, the equation of BC contour line is y-0 = 2 (x + 5) / 5, and it is sorted out that y = 2x / 5 + 1 (4)
AC edge equation (two-point formula): (y-0) / (x + 5) = (Y-2) / (x-0), sorted out: y = 2x / 5 + 2, slope = 2 / 5
AC contour line slope = - 5 / 2, AC contour line equation (point slope): y + 3 = - 5 (x-3) / 2, sorted out: y = - 5x / 2-9 / 2
(3) (4) the solution of simultaneous equations: x = - 15 / 34, y = 14 / 17
The equation of ABC bisector (two-point formula): (y-14 / 17) / (x + 15 / 34) = (y + 3) / (x-3), sorted out: y = - 10x / 9 + 1 / 3
|AC|=√(5²+2²)=√29
The simultaneous equations of y = 2x / 5 + 2 and y = - 5x / 2-9 / 2 are solved as follows: x = 65 / 29, y = 32 / 29,
That is, the intersection D (65 / 29,32 / 29) of AC contour line and AC
|AD|=√[(65/29+5)²+(32/29)²=√(1556/29)=2√(389/29)
S△ABC=1/2*|AC|*|AD|=1/2*√29*2√(389/29)=√389
The coordinates of the three vertices a, B and C of the triangle ABC are a (6, - 4), B (4, - 6) and C (7, - 6.5). Find the area of the triangle
Draw a picture, draw the height, the bottom is BC, the absolute value of BC = 7 - (- 4) = 3, make a vertical line to the right of point D, the height is DC, the absolute value of DC = - 6.5 - (- 4) = 2.5, the area is BC × DC × 0.5 = 3.75
The three vertices of a triangle are a (4,0), B (6,7), C (0,3). (1) the equation for finding the line with the height on the BC side; (2) the equation for finding the line with the middle line on the BC side; (3) the equation for finding the vertical bisector on the BC side
(1) The slope of the line on the BC side is k = 7-36-0 = 23, because the height on the BC side is perpendicular to BC, so the slope of the line on the BC side is - 32. The height on the BC side passes through point a (4,0), so the equation of the line on the BC side is y-0 = - 32 (x-4), that is, 3x + 2y-12 = 0. (2) it is known that the coordinates of the midpoint E on the BC side are (3,5) The equation of the middle line AE is y-0x-4 = 5-03-4, that is, 5x + y-20 = 0. (3) from (1), the slope of the straight line where BC side is located is k = 23, so the slope of the vertical bisector of BC side is - 32. From (2), the coordinate of the middle point e of BC side is (3,5), so the equation of the vertical bisector of BC side is Y-5 = - 32 (x-3), that is 3x + 2y-19 = 0
Seeking help to solve a geometry problem
In the parallelogram ABCD, BC = 2Ab, CE ⊥ AB, f is the midpoint of AD,
The value of ∠ AEF = 50 ° is calculated as ∠ B =?
A D
B C this is the arrangement of parallelogram ABCD
G is the midpoint of BC, connecting FG and eg, because ∠ AEF = 50 °, FG ‖ AB, Ge = GF, so ∠ FEG = ∠ EFG = 50 °, and ∠ CEF = 40 °, so ∠ GEC = 10 °, because eg = GC, ∠ BCE = 10 °, so ∠ B = 80 °
BAC = 90 degrees, e is the midpoint of AC, ad is perpendicular to be and f AB = AC
Verification: angle AEB = angle CED
Idea: AG is divided by a, BAC is handed over to be by G
It can be concluded that △ ABG ≌ △ CAD (ASA) → Ag = CD
It can be concluded that △ AEG ≌ △ CED (SAS) → ∠ AEB = ∠ CED
As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, ab = DC. Points e, F and G are on the edges AB, BC and CD respectively, AE = GF = GC. (1) prove that the quadrilateral aefg is a parallelogram; (2) prove that the quadrilateral aefg is a rectangle when ∠ FGC = 2 ∠ EFB
It is proved that: (1) in the trapezoidal ABCD, ab = DC, {B =} C. ∵ GF = GC, {C =} GFC, {} B =} GFC ∥ ab ∥ GF, i.e., AE} GF. ∵ AE = GF, {quadrilateral aefg is a parallelogram. (2) in the trapezoidal ABCD, the trapezoidal aefg is a parallelogram. (2) the trapezoidal aefg is a parallelogram The quadrilateral aefg is a rectangle
A geometry problem,
The bottom square p-abcd, PC = PD = CD = 2, the plane PCD, the vertical plane ABCD, the dihedral angle b-pd-c, the distance from a to the plane PBC?
Take PD midpoint E and connect EC and EB
Because PC = CD, CE is perpendicular to PD
Plane PCD is perpendicular to plane ABCD and CB is perpendicular to CD, so BC is perpendicular to PC
Because PC = CD, Pb = BD, be is perpendicular to PD
The angle BEC is the plane angle of the dihedral angle b-pd-c in the right triangle BCE CE = √ 3 CB = 2
tanBEC=2/√3 BEC=arctan2/√3
Take the midpoint F of PC to connect DF. Because CDP is an equilateral triangle, DF is perpendicular to CP and BC is perpendicular to plane CDP
Then BC is perpendicular to DF, so DF is perpendicular to PBC
So DF is the distance from a to plane PBC
In the equilateral triangle CDP, DF = √ 3 is obtained
It is known that: as shown in the figure, rectangle ABCD. (1) draw the symmetrical point C 'of point C with respect to the straight line where BD is located. (2) connect C' B and C 'D. if the overlapping area of △ C' BD and △ abd is equal to 23% of the area of △ abd, calculate the degree of ∠ CBD
(1) As shown in the figure; (2) the overlapping area of ∵ C ′ BD and ∵ abd is equal to 23% of the area of ∵ abd, and the two triangles are of equal height ∵ ed = 2ae. ∵∵ EBD = ∵ DBC and ∵ ad ∥ BC,
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