求 x 和 y 之间的关系，使得点 (x, y) 与点 (7, 1) 和 (3, 5) 等距

数学与数字和计算有关。数学根据运算或计算的类型（如代数、几何、算术等）分为不同的分支。几何是处理形状及其属性的数学分支。处理涉及坐标的点、线和平面的几何称为坐标几何。

坐标

平面上任意一点的位置可以表示为(x,y)，这些对称为该点的坐标，x是平面上一点的水平值。这个值也可以称为x坐标或横坐标， y是平面上一点的垂直值。该值可以称为 y 坐标或纵坐标。在坐标几何中，点表示在笛卡尔平面上。

笛卡尔平面

它是由两条垂直线组成的平面，即 x 轴（水平轴）和 y 轴（垂直轴）。笛卡尔平面中点的位置可以使用有序对 (x, y) 来表示。

距离公式

该公式用于计算平面上两点之间的距离。换句话说，它给出了连接两点后可以形成的线段的长度。设两点为 A 和 B，坐标分别为 (x ₁ , y ₁ ) 和 (x ₂ , y ₂ )。然后，两点之间的距离为，

Distance (d)= $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1 )^2}$

求 x 和 y 之间的关系，使得点 (x, y) 与 (7, 1) 和 (3, 5) 等距。

解决方案：

Equidistant means having equal distance. (x, y) is equidistant from point (7, 1) and (3, 5), which means distance of point (x, y) from both the point is equal.

Lets suppose A and B be the name of point (7, 1) and (3, 5) respectively and C be the point having coordinated (x, y). 

So, according to the question, d₁ and d₁ are equal.

CB = AC

Use distance formula to find d₁ and d₂ values.

• AC (Distance between (x, y) and (7, 1))

= $\sqrt{(x - 7)^2 + (y- 1 )^2}$

= $\sqrt{(x^2 -14x +49) + (y^2-2y+1) = \sqrt{(x^2 + 14x + y^2 -2y +50 )}$

• CB (Distance between (x, y) and (3, 5))

= $\sqrt{(x-3)^2+(y-5)^2}$

= $\sqrt{(x^2-6x+9)+(y^2-10y+25)}$

= $\sqrt{(x^2-6x+y^2-10y+34)}$

As, CB = AC

$\sqrt{(x^2-14x+y^2-2y+50)} = \sqrt{(x^2-6x+y^2-10y+34)}$

Squaring both sides,

(x² – 14x + y² – 2y + 50) = (x² – 6x + y² – 10y + 34)

x² – x² + y² – y² – 14x + 6x – 2y + 10y = 34 – 50 

-8x + 8y = -16

8x – 8y = 16

x – y = 2

x = 2 + y

Hence, relation between x and y such that the point(x, y) is equidistant from (7, 1) and (3 , 5) is x = 2 + y

示例问题

问题 1：求 x 和 y 之间的关系，使得点 (x, y) 与 (-5, -5) 和 (5, 5) 等距。

解决方案：

Equidistant means having equal distance. (x, y) is equidistant from point (-5, -5) and (5, 5), which means distance of point (x, y) from both the point is equal.

Lets suppose A and B be the name of point (-5, -5) and (5, 5) respectively and C be the point having coordinated (x, y). 

So, according to the question, d₁ and d₂ are equal.

CB = AC

Use distance formula to find d₁ and d₂ values.

• AC (Distance between (x , y) and (-5, -5))

$\sqrt{(x^2-14x+y^2-2y+50)} = \sqrt{(x^2-6x+y^2-10y+34)}$

• CB (Distance between (x , y) and (5 , 5))

$= \sqrt{(x-5)^2+(y-5)^2} = \sqrt{(x^2-10x+25)+(y^2-10y+25)} = \sqrt{(x^2-10x+y^2-10y+50)}$

As, CB = AC

$\sqrt{(x^2-10x+y^2-10y+50)} = \sqrt{(x^2+10x+y^2+10y+50)}$

Squaring both sides,

(x² – 10x + y² – 10y + 50) = (x² + 10x + y² + 10y + 50)

x² – x² + y² – y² = 10x + 10x + 10y + 10y + 50 – 50

0 = 20x + 20y 

x = -y 

Hence, relation between x and y such that the point (x, y) is equidistant from (-5, -5) and (5, 5) is x = -y

问题 2：求 x 和 y 之间的关系，使得点 (x, y) 与 (-2, 0) 和 (2, 0) 等距。

解决方案：

(x, y) is equidistant from point (-2, 0) and (2, 0), which means distance of point (x, y) from both the point is equal. Let’s suppose A and B be the name of point (-2, 0) and (2, 0) respectively and C be the point having coordinated (x, y). So, according to the question, CB and AC are equal.

CB = AC

Use distance formula to find the distances CB and AC.

• AC (Distance between (x, y) and (-2, -0))

$= \sqrt{(x-(-2)^2+(y-0)^2} = \sqrt{(x + 2)^2 + (y )^2} = \sqrt{(x^2+ 4x + 4)+(y^2 )}$

• CB (Distance between (x, y) and (2, 0))

$= \sqrt{(x-2)^2+(y-0)^2} = \sqrt{(x^2-4x+4)+(y^2)}$

As, CB = AC

\sqrt(x^2-4x+4)+(y^2)  = \sqrt(x^2+ 4x + 4)+(y^2 )

Squaring both sides,

(x² – 4x + 4) + (y²) = (x² + 4x + 4) + (y²)

x² – x² + y² – y² = 4x + 4x  + 4 – 4

0 = 8x

x = 0

x = 0  means y-axis.

Therefore, all the points that lie on y-axis is equidistant from point (-2, 0) and (2, 0).

Hence, relation between x and y such that the point (x, y) is equidistant from (-2, 0) and (2, 0) is x = 0