Length of a Line Segment

Length of a Line Segment

Secondary 3

Length of a Line Segment

• A line segment is part of a line with two end-points.

• For any two points \(A\left( {{x}_{1}},{{y}_{1}} \right)\) and \(B\left( {{x}_{2}},{{y}_{2}} \right)\), the line segment joining \(A\) and \(B\) can be found by applying Pythagoras Theorem.

• The length of the line segment joining \(A\left( {{x}_{1}},{{y}_{1}} \right)\) and \(B\left( {{x}_{2}},{{y}_{2}} \right)\), is

\(AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\)

Since length is a positive quantity, the negative \(\sqrt{\fbox{}}\) is not considered.

Tip:

\(AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\)

QuantityFormulaHow it works
Distance \(AB\)

The horizontal and vertical differences are the legs of a right-angled triangle, so the distance is the positive hypotenuse. Reversing both subtraction orders gives the same length because the differences are squared.

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

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