Distance and Section Formula

We can easily measure the distance between two points using a meter scale. However, imagine you were not provided a scale and were asked to measure the distance between two points. In such cases, it is best to use the distance and section formula.

The distance formula is used to find the distance between two defined points on a graph (in the absence of a scale).

The distance between two points $A=(x_1,y_1)$ and $B=(x_2,y_2)$ is given by the formula

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

Plot $A$ and $B$ on the graph. Draw $AC$ and $BD$ perpendicular to the $x$-axis. Also draw $AE$ perpendicular to $BD$. Now, we observe that triangle $ABE$ forms a right-angled triangle. Then by the Pythagorean theorem,

$\begin{aligned} {AB}^2 & = {AE}^2 + {BE}^2\\ AB & = \sqrt{{AE}^2 + {BE}^2}\\ &= \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}.\ _\square \end{aligned}$

From this proof we can derive the following corollary:

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

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

Find the distance between the points $A=(2,3)$ and $B=(5,7)$.

ANSWER

We have

$\begin{aligned} AB & = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}\\ & = \sqrt{{(5-2)}^2 + {(7-3)}^2}\\ & = \sqrt{3^2 + 4^2}\\ & = \sqrt{9+16}\\ & = \sqrt{25}\\ & = 5.\ _\square \end{aligned}$

Find the distance of the point $P=(7,1)$ from the origin.

ANSWER

We have

$\begin{aligned} OP & = \sqrt{x^2 + y^2}\\ & = \sqrt{7^2 + 1^2}\\ & = \sqrt{49 + 1}\\ & = \sqrt{50}\\ & = 5\sqrt{2}.\ _\square \end{aligned}$

Find the coordinates of points on the $x$-axis which are at a distance of $5$ units from the point $(6,-3)$.

ANSWER

Let the co-ordinates of the point on the $x$-axis be $(x,0)$. Since, distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},$ we have

$\begin{aligned} 5 &= \sqrt{(x - 6)^2 + (0 - 3)^2} \\ 25 &= x^2 - 12x + 36 + 9 \\ x^2 - 12x + 20 &= 0 \\ (x - 2)(x - 10) &= 0 \\ x &= 2 \text{ or }10. \end{aligned}$

Therefore, the required points on the $x$-axis are $(2,0)$ and $(10,0)$. $_\square$

Find the point on the $y$-axis which is equidistant from the points $(12,3)$ and $(-5,10)$.

ANSWER

Let the required point on the $y$-axis be $(0,y).$ Then the distance between $(12,3)$ and $(0,y)$ is equal to the distance between $(0,5)$ and $(-5,10):$

$\begin{aligned} \sqrt{(12 - 0)^2 + (3 - y)^2} &= \sqrt{(-5 - 0)^2 + (10 - y)^2} \\ 144 + 9 + y^2 - 6y &= 25 + 100 + y^2 - 20y \\ y &= -2. \end{aligned}$

Therefore, the required point is $(0,-2)$. $_\square$

Condition for three points to be collinear

Three points $A,B,C$ are said to be collinear if and only if the sum of distances between any two points is equal to distance between that point and third point. That means,

$\begin{aligned} AB + BC & = AC \quad \text{or} \\ AB + AC & = BC \quad \text{or} \\ AC + BC & = AB. \end{aligned}$

Verify whether the points $A = (1,-1), B = (6,4),$ and $C = (4,2)$ are collinear or not.

ANSWER

We have

$\begin{aligned} AB & = \sqrt{(6 - 1)^2 + (4 + 1)^2} = \sqrt{50} = 5\sqrt{2} \\ BC & = \sqrt{(4 - 6)^2 + (2 - 4)^2} = \sqrt{8} = 2\sqrt{2} \\ AC & = \sqrt{(4 - 1)^2 + (2 + 1)^2} = \sqrt{18} = 3\sqrt{2}. \end{aligned}$

As, ${\color{#3D99F6}BC + AC} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} = {\color{#3D99F6}AB}$, the given points are collinear. $_\square$

Sometimes we are given four points and asked to comment on the nature of the quadrilateral which is formed by joining them. For this, we have to recall the following:

A quadrilateral is a

• rectangle, if its opposite sides are equal and diagonals are equal;
• square, if all its sides are equal and diagonals are equal;
• parallelogram, if its opposite sides are equal;
• rhombus, if its sides are equal.

Show that the points $A=(-3,0), B=(1,-3), C=(4,1)$ are the vertices of an isosceles right-angled triangle. Also, find the area of the triangle.

ANSWER

We have

$\begin{aligned} AB & = \sqrt{{(1-(-3))}^2 + {(-3-0)}^2}\\ & = \sqrt{4^2 + {(-3)}^2} = \sqrt{16 + 9}\\ & = \sqrt{25} =5 \\ \\ BC & = \sqrt{{(4-1)}^2 + {(1-(-3)/)}^2}\\ & = \sqrt{3^2 + {4}^2} = \sqrt{9 + 16}\\ & = \sqrt{25} = 5 \\ \\ CA & = \sqrt{{(4-(-3))}^2 + {(1-0)}^2}\\ & = \sqrt{7^2 + {1}^2} = \sqrt{49 + 1}\\ & = \sqrt{50} = 5\sqrt{2}. \end{aligned}$

Since ${AB}^2 +{BC}^2= 5^2 + 5^2= 50= {CA}^2,$ triangle $ABC$ is a right-angled triangle.
Now, since $AB = BC,$ triangle $ABC$ is an isosceles triangle.
Hence, triangle $ABC$ is a right-angled isosceles triangle. $_\square$

The area of triangle $ABC$ is

$\begin{aligned} \dfrac{1}{2} × AB × BC = \dfrac{1}{2} × 5 ×5 = 12.5.\ _\square \end{aligned}$

Show that the points $A=(2,-2), B=(8,4), C=(5,7), D=(-1,1)$ are the vertices of a rectangle. Also, find the area of the rectangle.

ANSWER

We have

$\begin{aligned} AB = \sqrt{{(8-2)}^2 + {(4+2)}^2}& = \sqrt{72}\\& = 6\sqrt{2}\\\\ BC = \sqrt{{(5-8)}^2 + {(7-4)}^2}& = \sqrt{18}\\& = 3\sqrt{2}\\\\ CD = \sqrt{{(-1-5)}^2 + {(1-7)}^2}& = \sqrt{72}\\& = 6\sqrt{2}\\\\ DA = \sqrt{{(2+1)}^2 + {(-2-1)}^2}& = \sqrt{18}\\& = 3\sqrt{2}, \end{aligned}$

which implies $AB = CD$ and $BC = DA.$ Now,

$\begin{aligned} AC = \sqrt{{(5-2)}^2 + {(7+2)}^2}& = \sqrt{90}\\& = 3\sqrt{10}\\\\ BD = \sqrt{{(-1-8)}^2 + {(1-4)}^2}& = \sqrt{90}\\& = 3\sqrt{10}, \end{aligned}$

which implies $AC = BD.$

Thus, $ABCD$ is a quadrilateral whose opposite sides are equal and the diagonals are equal, which implies that $ABCD$ is a rectangle.

The area of quadrilateral $ABCD$ is

$\begin{aligned} (\text{Area of } ABCD) & = AB × BC\\ & = 6\sqrt{2} × 3\sqrt{2}\\ & = 36.\ _\square \end{aligned}$

Find the area of a circle which has its center at $(5,-3)$ and passes through the point $(-7,2)$.

ANSWER

The radius of the circle $r$ is equal to the distance between the points $(5,-3)$ and $(-7,2):$

$r = \sqrt{(-7 - 5)^2 + (2 + 3)^2} = \sqrt{144 + 69} = \sqrt{169} = 13.$

Therefore, the area of the circle is $\pi (13)^2 = 530.66.\ _\square$

Main article : Section Formula

The section formula gives the coordinates of a point which divides the line joining two points in a ratio, internally or externally.

If a point $P = (x , y)$ divides the line joining two points $A = (a , b)$ and $B = (c , d)$ in the ratio $m : n$ internally, then the coordinates of $P$ are given by

$P~(x , y) = \left (\dfrac{c \cdot m + a \cdot n}{m + n} , \dfrac{d \cdot m + b \cdot n}{m + n} \right).$

If $P$ divides $AB$ externally in the ration $m : n$, then

$P~(x , y) = \left(\dfrac{c \cdot m - a \cdot n}{m - n} , \dfrac{d \cdot m - b \cdot n}{m - n} \right).$

Find the coordinates of point $P$ which divides the line joining $A = (4 , -5)$ and $B = (6 , 3)$ in the ratio $2 : 5$.

ANSWER

Let the coordinates of $P$ be $(x , y).$ Then

$\begin{aligned} P~(x , y) & = \left(\dfrac{2 \times 6 + 5 \times 4}{2 + 5} , \dfrac{2 \times 3 + 5 \times -5}{2 + 5} \right) \\ & = \left(\dfrac{12 + 20}{7} , \dfrac{6 - 25}{7} \right) \\ & = \left(\dfrac{32}{7} , -\dfrac{19}{7} \right).\ _\square \end{aligned}$