已知菱形ABCD的对角线及点D、M坐标，求点A的坐标

Let's work through this problem step by step, using core properties of rhombuses and coordinate geometry:

Step 1: Find coordinates of Point B (since M is the midpoint of BD) #

Rhombuses have diagonals that bisect each other, so M is the midpoint of both AC and BD. Using the midpoint formula:

Midpoint formula: If ( M(x_m, y_m) ) is the midpoint of ( B(x_b, y_b) ) and ( D(x_d, y_d) ), then ( x_m = \frac{x_b + x_d}{2} ) and ( y_m = \frac{y_b + y_d}{2} )

We know ( D(1,1) ) and ( M(2,-1) ), so rearrange to solve for B's coordinates:

• ( x_b = 2x_m - x_d = 2*2 - 1 = 3 )
• ( y_b = 2y_m - y_d = 2*(-1) - 1 = -3 )

Point B is (3, -3).

Step 2: Use rhombus properties (perpendicular diagonals, BD=2AC) #

Two key properties we'll use here:

• Diagonals of a rhombus are perpendicular to each other
• Given ( BD = 2AC )

First, calculate the slope of BD to find the slope of AC (perpendicular lines have slopes that are negative reciprocals):

• Slope of BD: ( k_{BD} = \frac{1 - (-3)}{1 - 3} = \frac{4}{-2} = -2 )
• Slope of AC: ( k_{AC} = \frac{1}{2} ) (negative reciprocal of -2)

Since AC passes through M(2,-1), the line equation for AC is:
( y + 1 = \frac{1}{2}(x - 2) ) → simplified to ( y = \frac{1}{2}x - 2 )
Any valid point A must lie on this line.

Step 3: Calculate lengths to find Point A's exact coordinates #

First, find the length of BD using the distance formula:
( BD = \sqrt{(3-1)^2 + (-3-1)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{20} = 2\sqrt{5} )

Given ( BD = 2AC ), so ( AC = \sqrt{5} ). Since M is the midpoint, the length of AM is half of AC: ( AM = \frac{\sqrt{5}}{2} )

Let ( A(x, y) ). Using the distance formula for AM:

Distance formula: ( AM = \sqrt{(x - 2)^2 + (y + 1)^2} = \frac{\sqrt{5}}{2} )

Substitute ( y = \frac{1}{2}x - 2 ) into the equation and solve:

1. Square both sides to eliminate square roots:
   ( (x - 2)^2 + \left( \frac{1}{2}x - 2 + 1 \right)^2 = \left( \frac{\sqrt{5}}{2} \right)^2 )
2. Simplify terms:
   ( (x-2)^2 + \left( \frac{1}{2}x - 1 \right)^2 = \frac{5}{4} )
3. Expand and combine like terms:
   ( x^2 -4x +4 + \frac{1}{4}x^2 -x +1 = \frac{5}{4} )
   ( \frac{5}{4}x^2 -5x +5 = \frac{5}{4} )
4. Multiply through by 4 to clear fractions:
   ( 5x^2 -20x +20 = 5 ) → ( 5x^2 -20x +15 =0 ) → ( x^2 -4x +3=0 )
5. Factor the quadratic:
   ( (x-1)(x-3)=0 ) → ( x=1 ) or ( x=3 )

Find corresponding y-values with ( y = \frac{1}{2}x -2 ):

• If ( x=1 ): ( y = \frac{1}{2}(1) -2 = -\frac{3}{2} ) → A(1, -3/2)
• If ( x=3 ): ( y = \frac{1}{2}(3) -2 = -\frac{1}{2} ) → A(3, -1/2)

Step 4: Verify both solutions #

Both points satisfy all rhombus conditions:

• For ( A(1, -3/2) ), point C is (4-1, -2 - (-3/2))=(3, -1/2). AC length is ( \sqrt{(3-1)^2 + (-1/2 +3/2)^2} = \sqrt{5} ), BD length is ( 2\sqrt{5} ), and AC is perpendicular to BD (their slopes multiply to -1).
• For ( A(3, -1/2) ), point C is (4-3, -2 - (-1/2))=(1, -3/2). Same length and perpendicularity checks pass.

Both are valid coordinates for point A.

内容的提问来源于stack exchange，提问作者mampu