Question

На оси х найдите точку равноудаленную от точек а(1;3) и в(4;2)

Answer 1

Я думаю что точка 2,75

Answer 2

I can help you with your math question.

To find the point on the x-axis that is equidistant from the points A(1;3) and B(4;2), you need to use the distance formula. The distance between a point $(x_1;y_1)$ and a point $(x_2;y_2)$ is given by:

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

Since the point we are looking for is on the x-axis, its y-coordinate is zero. Let's call it C(x;0). Then the distance between A and C is:

$$d_{AC}=\sqrt{(x-1)^2+(0-3)^2}=\sqrt{x^2-2x+10}$$

And the distance between B and C is:

$$d_{BC}=\sqrt{(x-4)^2+(0-2)^2}=\sqrt{x^2-8x+20}$$

Since we want these distances to be equal, we can set them equal to each other and solve for x:

$$\sqrt{x^2-2x+10}=\sqrt{x^2-8x+20}$$

Squaring both sides, we get:

$$x^2-2x+10=x^2-8x+20$$

Simplifying, we get:

$$6x=10$$

So, x is:

$$x=\frac{10}{6}=\frac{5}{3}$$

Therefore, the point C is $(\frac{5}{3};0)$. This is the point on the x-axis that is equidistant from A and B. You can check this by plugging in the values of x and y into the distance formula and verifying that they are equal. I hope this helps you understand the solution.Ответь+подробно.+На+оси+х+найдите+точку+равноудаленную+от+точек+а(1;3)+и+в(4;2)