Find the distance between points P (-3, 4) and Q (1, 6).
\(Let\ A(x_A;\ y_A)\ and\ B(x_B;\ y_B)\ then\ the\ distance\ between\ AB\ is:\\\\|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\)
P(-3; 4); Q(1; 6)
subtitute
\(|PQ|=\sqrt{(1-(-3))^2+(6-4)^2}=\sqrt{4^2+2^2}=\sqrt{16+4}=\sqrt{20}\\\\=\sqrt{4\cdot5}=\sqrt4\cdot\sqrt5=2\sqrt5\)
Answer: The distance between P and Q is 2√5.
The distance between two points are \(\sqrt{20}\) unit.
The distance between two points \((x_{1},y_{1})\) and \((x_{2},y_{2})\) is computed as,
\(Distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }\)
Given points are, P (-3, 4) and Q (1, 6).
Substitute points in above formula.
\(Distance=\sqrt{(1+3)^{2}+(6-4)^{2} } \\\\Distance=\sqrt{16+4} =\sqrt{20}\)
The distance between two points are \(\sqrt{20}\) unit.
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