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As shown in the figure, in the plane rectangular coordinate system, point O is the origin, the diagonal ob of diamond oabc is on the X axis, and the vertex A is in the inverse scale function y = 2 On the image of X, the area of the diamond is______ .
AC
∵ the quadrilateral oabc is a diamond,
∴AC⊥OB．
∵ point a is in the inverse proportional function y = 2
In the image of X,
The area of △ AOD = 1
2×2=1，
The area of lozenge oabc = 4 ×△ AOD = 4
In the plane rectangular coordinate system xoy, point B and point a (- 1,1) are symmetric about the origin o, P is a moving point, and the product of slope of straight line AP and BP is equal to − 1 2． (I) find the trajectory equation of the moving point P; (II) let the straight line AP and BP intersect with the straight line x = 3 at points m and N respectively, and ask whether there is a point P, so that an ∥ BM; if there is, calculate the coordinates of point p; if not, please explain the reasons
(1) Since point B and a (- 1,1) are symmetric about origin o, the coordinates of point B are (1, - 1). If the coordinates of point P are (x, y), then ∵ the product of slope of straight line AP and BP is ? the product of slope of straight line AP and BP is ? 12, ? the product of slope of line AP and BP is ? 12, which is simplified to x2 + 2Y2 = 3 (x ≠ ± 1)
In the plane rectangular coordinate system, point B and point a (- 1,1) are symmetric about origin o, P is a moving point, and the product of slope of straight line AP and BP is equal to − 1 3, then the trajectory equation of the moving point P______ .
Let the coordinates of point p be (x, y), ∵ the product of the slope of the straight line AP and BP is - 13, ᙽ y − 1x + 1 · y + 1 x − 1 = - 13, (x ≠± 1). X 2 + 3y2 = 4 (x ≠± 1) is obtained
In the plane rectangular coordinate system xoy, point B and point a (- 1.1) are symmetric with respect to origin o, and P is a moving point, and the product of slope of straight line AP and BP is equal to - 1 / 3. (1) find the trajectory equation of moving point P. (2) let the straight line AP and BP intersect with the straight line x = 3 respectively at points m, N. question: is there a point P that makes the area of triangle PAB and triangle PMN equal? If so, find the coordinates of point p; if not, explain the reason
This can be obtained from the knowledge of similar triangles. You only need to pass through point m to make a straight line parallel to the X axis, and then pass through points a and P respectively to sit at the intersection of the vertical lines of this line. The straight line is D, e, and the triangle mad is similar to the triangle MPE. Therefore, AP / Pb is equal to de / em, that is, (x0 + 1) / (3-x0). The right side of the equation is the same
In the rectangular coordinate system xoy, point B and point a (- 1,1) are symmetric about the origin o, P is a moving point, and the product of slope of straight line AP and BP is equal to - 1 / 3 (1) Seeking the trajectory equation of moving point P (2) Let the straight line AP and BP intersect with the straight line x = 3 respectively at points m, N. ask: is there a point P that makes the area of triangle PAB and triangle PMN equal? If so, find the coordinates of point p; if not, find the coordinates of point p; if not, explain the reasons First, I figured it out,
(1) If the coordinates of point P are (x, y) Y-1 / x + 1 * y + 1 / X-1 = 1 / 3, then x2 + 3y2 = 4 (x ≠± 1). (II) if there is a point P such that the areas of △ PAB and △ PMN are equal, and if the coordinates of point P are (x0, Y0), then 1 / 2p
It is known that point B and point a (- 1,1) are symmetric about origin o, and the product of slope of straight line AP and BP is - 1 / 3 1. Find the trajectory equation x ^ 2 + 3Y ^ 2 = 4 of the moving point P 2. Let the straight lines AP and BP intersect with the straight line x = 3 at points m and N respectively. Question: is there a point P that makes the area of △ PAB and △ PMN equal? If so, find the coordinates of point p; if not, explain the reasons Ask for the second question
(1) Because point B and a (- 1,1) are symmetric about origin o, the coordinates of point B are (1, - 1)
Let the coordinates of point p be (x, y)
y-1/x+1 * y+1/x-1 =1/3
After simplification, x 2 + 3Y 2 = 4 (x ≠ ± 1)
(2) If there is a point P such that the area of △ PAB and △ PMN is equal, let the coordinates of point p be (x0, Y0)
Then 1 / 2PA * pbsinapb = 1 / 2pm * pnsinmpn
Drawing and finding that APB and MPN are complementary
sinAPB=sinMPN
PA/PM=PN/PB
(x0+1)/(3-x0)=(3-x0)/(x0-1)
In other words, (3-x0) 2 = |x02-1|, the solution x0 = 5 / 3
x0^2+3y0^2=4
Y0 = positive and negative root sign 33 / 9
There is p (5 / 3, positive and negative root sign 33 / 9)
As shown in the figure, in the plane rectangular coordinate system, the image of the quadratic function y = x2 + BX + C intersects with the X axis at two points a and B, point a is on the left of the origin, and the coordinates of point B are The coordinates of point B are (3,0) and intersect with the y-axis at point C (0, - 3), (1) Find the relation of this quadratic function and the coordinates of point a; (what I did is y = x2-2x-3, a (- 1,0) (2) If point P is an understanding point on the parabola below the straight line BC, when the point P moves to what position, the area of △ BPC is the largest? Find out the coordinates of point P and the maximum area of △ BPC. (what I did is p (1, - 4), the maximum area is 3.5 (3) If point q is a moving point on a parabola, when point Q moves to what position, △ ACQ is a right triangle with AC as a right angle side Don't let pictures be inserted I have been working hard for a long time
(1) Substituting the point a (- 1,0) into the analytic formula, 1-B + C = 0,
From the parabola symmetry axis X = 1, we can get - B / 2 = 1
The solution is b = - 2, C = - 3, so the analytic formula of this quadratic function is y = x2-2x-3
(2) When y = 0, x = - 1 or 3, so the coordinates of point B are (3,0), and because y = (x-1) ^ 2-4,
Therefore, the coordinates of the vertex of the parabola are C (1, - 4), and the axis of symmetry ch of the parabola intersects the x-axis of the parabola, and the crossing point D is DM ⊥ X-axis to M. because eh / / DM, eh / DM = BH / BM, that is, eh / 12 = 2 / 6, eh = 4,
Therefore, EC and ab are perpendicular to each other, so the quadrilateral bcae is diamond. ① if the quadrilateral BCEF is a parallelogram, BF = EC = 8, and BF / / CE, then point F is (3,8); ② if the quadrilateral becf is a parallelogram, the coordinates of point F are (3, - 8); ③ if the quadrilateral BCFE is a parallelogram, f coincides with point a, so the coordinates of point F are (- 1, In addition, it should be emphasized that the three cases of parallelogram discussed just now are the three possible cases of taking be, BC and EC as diagonals
(3) If (2) we can get be / / AC, so the area of BD / / AC and △ pad is equal to the area of trapezoidal PACB, because △ pad is equal to trapezoidal PACB (because BD / / AC), if they are equal, then 1 / 2 * (Pb + AC) * H = 1 / 2 * de * h, so Pb + AC = De, so let Pb = a, 6 √ 5-a = 2 √ 5 + a,
So a = 2 √ 5, so point P coincides with point a, and the coordinates of point P are (1,4)
It doesn't seem like this
As shown in the figure, in the plane rectangular coordinate system, O is the coordinate origin. The image of the quadratic function y = - x2 + BX + 3 passes through point a (- 1,0), and the vertex is B (1) Find the analytic expression of the quadratic function and write the coordinates of vertex B; (2) If the coordinate of point C is (4,0), AE ⊥ BC, the perpendicular foot is point E, point D is on the straight line AE, de = 1, find the coordinates of point D
(1) ∵ the image of the quadratic function y = - x2 + BX + 3 passes through the point a (- 1,0), 0 = - 1-B + 3, and the solution is: B = 2, the analytic formula of the quadratic function is y = - x2 + 2x + 3, then the coordinates of the vertex B of the quadratic function image are (1,4); (2) the axis BF ⊥ X through the point B is the point F, in RT △ BCF, BF = 4, CF
As shown in the figure, in the plane rectangular coordinate system, O is the coordinate origin. The image of the quadratic function y = - x2 + BX + 3 passes through point a (- 1,0), and the vertex is B (1) Find the analytic expression of the quadratic function and write the coordinates of vertex B; (2) If the coordinate of point C is (4,0), AE ⊥ BC, the perpendicular foot is point E, point D is on the straight line AE, de = 1, find the coordinates of point D
(1) ∵ the image of the quadratic function y = - x2 + BX + 3 passes through the point a (- 1,0), 0 = - 1-B + 3, and the solution is: B = 2, the analytic formula of the quadratic function is y = - x2 + 2x + 3, then the coordinates of the vertex B of the quadratic function image are (1,4); (2) the axis BF ⊥ X through the point B is the point F, in RT △ BCF, BF = 4, CF
As shown in the figure, in the plane rectangular coordinate system, O is the coordinate origin. The image of the quadratic function y = - x2 + BX + 3 passes through point a (- 1,0), and the vertex is B (1) Find the analytic expression of the quadratic function and write the coordinates of vertex B; (2) If the coordinate of point C is (4,0), AE ⊥ BC, the perpendicular foot is point E, point D is on the straight line AE, de = 1, find the coordinates of point D
(1) ∵ the image of the quadratic function y = - x2 + BX + 3 passes through point a (- 1,0),
∴0=-1-b+3，
The solution is: B = 2,
The analytic expression of the quadratic function is y = - x2 + 2x + 3,
Then the coordinates of vertex B of the quadratic function image are (1,4);
(2) Passing through point B as BF ⊥ X axis, perpendicular foot as point F,
In RT △ BCF, BF = 4, CF = 3, BC = 5,
∴sin∠BCF=4
5，
In RT △ ace, sin ∠ ace = AE
AC，
And ∵ AC = 5, AE can be obtained
5＝4
5，
∴AE=4，
Point D is the axis of DH ⊥ x, and the foot perpendicular is point H,
It is easy to prove △ ADH ∽ ace,
∴AH
AE=DH
CE=AD
AC，
CE = 3, AE = 4,
If the coordinates of point D are (x, y), then ah = x + 1, DH = y,
① If point D is on the extension line of AE, then ad = 5,
X + 1
4＝y
3＝5
5，
∴x=3，y=3，
So the coordinates of point D are (3, 3);
② If point D is on line AE, then ad = 3
X + 1
4＝y
3＝3
5，
∴x＝7
5，y＝9
So the coordinates of point D are (7
5，9
5）．
To sum up, the coordinates of point D are (3,3) or (7)
5，9
5）．
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