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

\[AB = \sqrt{{(x_2 - x_1)}^2 + {(y_2 - y_1)}^2}.\ _\square\]

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,

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

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 \[OP = \sqrt{{(x-0)}^2 + {(y-0)}^2} = \sqrt{x^2 + y^2}.\ _\square\]

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

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

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

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We have \[\begin{align} OP & = \sqrt{x^2 + y^2}\\ & = \sqrt{7^2 + 1^2} = \sqrt{49 + 1}\\ & = \sqrt{50} = 5\sqrt{2}.\ _\square \end{align}\]

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.

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We have

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

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

\[\begin{align} \dfrac{1}{2} × AB × BC = \dfrac{1}{2} × 5 ×5 = 12.5.\ _\square \end{align}\]

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.

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We have

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

which implies \(AB = CD\) and \(BC = DA.\) Now,

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

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

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

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