The-point-A-has-coordinate-1-5-and-the-point-B-has-coordinates-7-1-The-perpendicular-bisector-of-AB-meets-the-x-axis-at-C-and-the-y-axis-at-D-Calculate-the-length-of-CD-

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Question Number 25200 by tawa tawa last updated on 06/Dec/17

$$\mathrm{The}\mathrm{point}\mathrm{A}\mathrm{has}\mathrm{coordinate}(-1,-5)\mathrm{and}\mathrm{the}\mathrm{point}\mathrm{B}\mathrm{has}\mathrm{coordinates}(7,1).$$$$\mathrm{The}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{AB}\mathrm{meets}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathrm{at}\mathrm{C}\mathrm{and}\mathrm{the}\mathrm{y}-\mathrm{axis}\mathrm{at}$$$$\mathrm{D}.\mathrm{Calculate}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{CD}$$

Answered by mrW1 last updated on 06/Dec/17

$$MiddleofAB=M(\frac{-1+7}{2},\frac{-5+1}{2})=M(3,-2)$$$$C(x_C,0)$$$$D(0,y_D)$$$$CM\mathrm{\perp }BA:$$$$\Rightarrow \frac{-2-0}{3-x_C}\times \frac{1-(-5)}{7-(-1)}=1$$$$\Rightarrow \frac{-2}{3-x_C}=\frac{4}{3}$$$$\Rightarrow 12-4x_C=-6$$$$\Rightarrow x_C=\frac{18}{4}=\frac{9}{2}$$$$DM\mathrm{\perp }BA:$$$$\Rightarrow \frac{-2-y_D}{3-0}\times \frac{1-(-5)}{7-(-1)}=1$$$$\Rightarrow \frac{-2-y_D}{3}=\frac{4}{3}$$$$\Rightarrow -6-3y_D=12$$$$\Rightarrow y_D=-\frac{18}{3}=-6$$$$CD=\sqrt{{\left(\frac{9}{2}\right)}^2+{(-6)}^2}=\frac{15}{2}$$

Commented by tawa tawa last updated on 06/Dec/17

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by tawa tawa last updated on 06/Dec/17

$$\mathrm{please}\mathrm{sir},\mathrm{help}\mathrm{me}\mathrm{with}\mathrm{the}\mathrm{other}\mathrm{ones}.$$

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