Question

Find the distance between each of the following pairs of points.
(1) A(2, 3), B(4, 1)

(2) P(–5, 7), Q(–1, 3)

(3) $\mathrm{R}\left(0,–3\right),\mathrm{S}\left(0,\frac{5}{2}\right)$

(4) L(5, –8), M(–7, –3)

(5) T(–3, 6), R(9, –10)

(6) $\mathrm{W}\left(\frac{–7}{2},4\right),\mathrm{X}\left(11,4\right)$

Answer 1

(1) A(2, 3), B(4, 1)
$$\begin{gathered} \mathrm{AB}=\sqrt{{\left(2-4\right)}^2+{\left(3-1\right)}^2}=\sqrt{{\left(-2\right)}^2+2^2} \\ =\sqrt{4+4} \\ =2\sqrt{2} \end{gathered}$$

(2) P(–5, 7), Q(–1, 3)
$$\begin{gathered} \mathrm{PQ}=\sqrt{{\left(-5-\left(-1\right)\right)}^2+{\left(7-3\right)}^2}=\sqrt{{\left(-4\right)}^2+4^2} \\ =\sqrt{16+16} \\ =4\sqrt{2} \end{gathered}$$
(3) $\mathrm{R}\left(0,–3\right),\mathrm{S}\left(0,\frac{5}{2}\right)$
$$\begin{gathered} \mathrm{RS}=\sqrt{{\left(0-0\right)}^2+{\left(-3-\frac{5}{2}\right)}^2}=\sqrt{{\left(\frac{-11}{2}\right)}^2} \\ =\sqrt{\frac{121}{4}} \\ =\frac{11}{2} \end{gathered}$$

(4) L(5, –8), M(–7, –3)
$$\begin{gathered} \mathrm{LM}=\sqrt{{\left(5-\left(-7\right)\right)}^2+{\left(-8-\left(-3\right)\right)}^2}=\sqrt{{12}^2+{\left(-5\right)}^2} \\ =\sqrt{144+25}=\sqrt{169} \\ =13 \end{gathered}$$

(5) T(–3, 6), R(9, –10)
$$\begin{gathered} \mathrm{TR}=\sqrt{{\left(-3-9\right)}^2+{\left(6-\left(-10\right)\right)}^2}=\sqrt{{\left(-12\right)}^2+{\left(16\right)}^2} \\ =\sqrt{144+256}=\sqrt{400} \\ =20 \end{gathered}$$

(6) $\mathrm{W}\left(\frac{–7}{2},4\right),\mathrm{X}\left(11,4\right)$
$$\begin{gathered} \mathrm{WX}=\sqrt{{\left(\frac{-7}{2}-11\right)}^2+{\left(4-4\right)}^2}=\sqrt{{\left(\frac{-29}{2}\right)}^2} \\ =\sqrt{\frac{841}{4}} \\ =\frac{29}{2} \end{gathered}$$