Question

Distance between the points (1,2) and (1,6)​

Answer 1

Coordinate Geometry

Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane, then the distance between A and B is given by,

$\boxed{\bf{\;AB = \sqrt{{(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2}}}\:}$

The given points are A(1, 2) and B(1, 6). Then,

$\;\;x_1 = 1, \: y_1 = 2 \; \: \& \: \; x_2 = 1, \: y_2 = 6$

Now, on substituting the given values in the formula, we get the following results:

$\implies AB = \sqrt{{(1 - 1) }^{2} + {(6 - 2}^{2}} \\\\\\ \implies AB = (1)^2 + (4)^2 \\\\\\ \implies AB = 1 + 16 \\\\\\ \implies \red{\boxed{\bf{\gray{AB = 17}}}}$

Hence, the distance between the given points is 17 units.

$\rule{300}{2}$

Extra Information

1. Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane, then the distance between A and B is given by,

$\boxed{\tt{\;AB = \sqrt{ {(x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2}\;}}}$

2. The distance of the point P(x, y) from the origin O(0, 0) is given by,

$\boxed{\tt{\;OP = \sqrt{x^2 + y^2\;}}}$

3. The area of a ∆ABC with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by,

$\boxed{\tt{\;Area = \dfrac{1}{2}\bigg|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\bigg|\;}}$

4. Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear only when,

$\boxed{\tt{\;x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) = 0\;}}$

5. Let A(x₁, y₁) and B(x₂, y₂) be two points in the coordinate plane and C(x, y) be the point which divides AB internally in the ratio m₁ : m₂. Then, the coordinates of C will be:

$\boxed{\tt{\:C = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\bigg)\;}}$

6. Let us consider a line segment joining the points A and B and Let C (x, y) be the midpoint of AB, then coordinates of C(x, y) is given by,

$\boxed{\tt{\;(x, y) = \bigg(\dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2}\bigg)\;}}$

7. Any point on the x-axis is of the form (x, 0).

8. Any point on the y-axis is of the form (0, y).