求 x 和 y 之间的关系，使得点 (x, y) 与 (6, 5) 和 (-4, 3) 等距

数学是一门与数字和计算相关的学科。并且，根据计算类型，数学分为不同的分支，如代数、几何、算术等。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。

坐标几何

坐标几何就像是几何和代数之间通过图形的联系。使用坐标几何可以获得两点之间的距离，三角形在笛卡尔平面上的面积等。在坐标几何中，点表示在笛卡尔平面上。笛卡尔平面是由两条垂直线形成的平面，这两条线是 x 轴（水平轴）和 y 轴（垂直轴）。

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 (6, 5) 和 (-4, 3) 等距。

解决方案：

Equidistant means having equal distance. (x, y) is equidistant from points (6, 5) and (-4, 3), which means a distance of the point (x, y) from both the point is equal. Let’s suppose A and B be the name of points (6, 5) and (-4, 3) respectively and C be the point having coordinated (x, y). 

According to the problem statement, d₁ ( CB)  is equal to d₂ (AC).

Use distance formula to find d₁ and d₂ values,

• d₂ (Distance between (x, y) and (6, 5)) 

= $\sqrt{(x-6)^2+(y-5)^2}$

= $\sqrt{(x^2-12x+36)+(y^2-10y+25)}$

=√(x² – 12x + y² – 10y + 61)

• d₁ (Distance between (x, y) and (-4, 3))

=$\sqrt{(x - (-4))^2 + (y - 3)^2}$

= $\sqrt{(x^2 + 8x + 16) + (y^2 - 6y + 9)}$

=√(x² + 8x + y² – 6y + 25)

According to the question, d₁ = d₂

After substituting the values,

√(x² – 12x + y² – 10y + 61) = √(x² + 8x + y² – 6y + 25)

Squaring both sides,

(x² – 12x + y² – 10y + 61) = (x² + 8x + y² – 6y + 25)

x² – x² + y² – y² -10y + 6y – 12x – 8x = 25 – 61

-4y – 20x = – 36 

4y + 20x = 36

y + 5x = 9 

y = 9 – 5x

Hence, the relation between x and y such that the point (x, y) is equidistant from (6, 5) and (-4, 3) is y = 9 -5x

类似问题

问题 1：求 x 和 y 之间的关系，使得点 (x, y) 与 (0, 0) 和 (6, 6) 等距。

解决方案：

(x, y) is equidistant from point (0, 0) and (6, 6), which means distance of point (x, y) from both the point is equal.

• d₁(Distance between (x, y) and (0, 0)) 

= $\sqrt{(x-0)^2+(y-0)^2}$

= $\sqrt{(x^2)+(y^2)}$

• d₂(Distance between (x, y) and (6, 6))

= $\sqrt{(x - 6)^2 + (y - 6)^2}$

= $\sqrt{(x^2-12x+36)+(y^2-12y+36)}$

= √(x² – 12x + y² – 12y + 72)

According to the question, d₁ = d₂

After substituting the values:

$\sqrt{(x^2)+(y^2)}$= √(x² – 12x + y² – 12y + 72)

Squaring both sides,

x² + y² = x² – 12x + y² – 12y + 72

x² – x² + y² – y² = -12x – 12y + 72

-72 = -12x – 12y

72 = 12x + 12y

6 = x + y

x = 6 – y

Hence, the relation between x and y such that the point (x, y) is equidistant from (0, 0) and (6, 6) is x = 6 – y

问题 2：求 x 和 y 之间的关系，使得点 (x, y) 与 (5, 0) 和 (0, 5) 等距。

解决方案：

(x, y) is equidistant from point (5,0) and (0,5), which means distance of point (x,y) from both the point is equal.

• d₁ (Distance between (x, y) and (5, 0))

= $\sqrt{(x-5)^2+(y-0)^2}$

= $\sqrt{(x^2-10x+25)+(y^2)}$

= √{x² – 10x +25 + y²}

• d₂ (Distance between (x, y) and (0, 5))

= $\sqrt{(x-0)^2+(y-5)^2}$

= $\sqrt{(x^2)+(y^2-10y+25)}$

=√(x² + y² – 10y + 25)

According to the question, d₁ = d₂

After substituting the values:

√(x² – 10x + 25 + y²) = √(x² + y² – 10y + 25)

Squaring both sides,

x² – 10x + 25 + y² = x² + y² – 10y + 25 

x² – x² + y² – y² = -10y +10x + 25 – 25

10y = 10x

y = x

Hence, the relation between x and y such that the point (x, y) is equidistant from (0, 5) and (5, 0) is x = y