Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 64311 by aseer imad last updated on 16/Jul/19

Commented by Tony Lin last updated on 17/Jul/19

$${letA}^{\prime }=A-(1,3)=(0,0),C^{\prime }=C-(1,3)=(4,5)$$$$D^{\prime }=\left[\begin{array}{c}\frac{\sqrt{2}}{2}0\\ 0\frac{\sqrt{2}}{2}\end{array}\right]\left[\begin{array}{c}cos\frac{\pi }{4}-sin\frac{\pi }{4}\\ sin\frac{\pi }{4}cos\frac{\pi }{4}\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]$$$$=\left[\begin{array}{c}\frac{1}{2}-\frac{1}{2}\\ \frac{1}{2}\frac{1}{2}\end{array}\right]\left[\begin{array}{c}4\\ 5\end{array}\right]$$$$=\left[\begin{array}{c}-\frac{1}{2}\\ \frac{9}{2}\end{array}\right]$$$$\Rightarrow D=D^{\prime }+(1,3)=(\frac{1}{2},\frac{15}{2})$$$$\because A\stackrel{\rightharpoonup }{B}=D\stackrel{\rightharpoonup }{C}$$$$\therefore B=(1,3)+(\frac{9}{2},\frac{1}{2})=(\frac{11}{2},\frac{7}{2})$$

Answered by Tanmay chaudhury last updated on 16/Jul/19

$$AC(y-3)=\frac{8-3}{5-1}(x-1)$$$$y-3=\frac{5}{4}(x-1)$$$$4y-5x-12+5=04y-5x-7=0iseanAC$$$$leteqnAB(y-3)=m(x-1)$$$$anglebetweeABandACis\frac{\pi }{4}$$$$tan\frac{\pi }{4}=\frac{m-\frac{5}{4}}{1+m\times \frac{5}{4}}$$$$1+\frac{5m}{4}=m-\frac{5}{4}$$$$\frac{5m}{4}-m=\frac{-5}{4}-1$$$$\frac{m}{4}=\frac{-9}{4}m=-9$$$$AB(y-3)=\frac{-9}{1}(x-1)$$$$y-3+9x-9=0$$$$y+9x-12=0\leftarrow eqnAB$$$$DC(y-8)=\frac{-9}{1}(x-5)$$$$y-8+9x-45=0$$$$y+9x-53=0\leftarrow eqnDC$$$$againtan\frac{\pi }{4}=\frac{\frac{5}{4}-m}{1+\frac{5m}{4}}$$$$1+\frac{5m}{4}=\frac{5}{4}-m$$$$\frac{9m}{4}=\frac{1}{4}m=\frac{1}{9}$$$$eqnAD(y-3)=\frac{1}{9}(x-1)$$$$9y-27-x+1=0$$$$9y-x-26=0\leftarrow eqnAD$$$$BC(y-8)=\frac{1}{9}(x-5)$$$$9y-72=x-5$$$$9y-x-67=0\leftarrow eqnBC$$$$solveeqnDCy+9x-53=0andAD9y-x-26=0$$$$tofindDy+9(9y-26)-53=0$$$$82y-234-53=0$$$$y=\frac{287}{82}=\frac{7}{2}sox=9\times \frac{7}{2}-26$$$$D(\frac{63-52}{2},\frac{7}{2})so\mathit{D}(\frac{11}{2},\frac{7}{2})$$$$solveABy+9x-12=0BC9y-x-67=0$$$$y+9(9y-67)-12=0$$$$82y=603+12y=\frac{615}{82}=\frac{15}{2}$$$$x=9\times \frac{15}{2}-67$$$$x=\frac{135-134}{2}=\frac{1}{2}$$$$B(\frac{1}{2},\frac{15}{2})so$$$$A(1,3)B(\frac{1}{2},\frac{15}{2}),C(5,8)D(\frac{11}{2},\frac{7}{2})plscheck$$$$Tanmay16.07.19$$$$$$$$$$$$$$

Commented by aseer imad last updated on 16/Jul/19

Thank you very much sir,it is right. Could've avoid the complexity of finding all equations. Instead finding one vertex and use mid point to find the other...

Commented by Tanmay chaudhury last updated on 17/Jul/19

$$iamgoingtorechekthesteps..$$

Answered by mr W last updated on 16/Jul/19

$$M=midpointofAC=midpointofBD$$$$M(\frac{1+5}{2},\frac{3+8}{2})=M(3,\frac{11}{2})$$$$\mathrm{\Delta }{x}_{MB}=\mathrm{\Delta }{y}_{AM}=\frac{11}{2}-3=\frac{5}{2}$$$$\Rightarrow x_B=3+\frac{5}{2}=\frac{11}{2}$$$$\mathrm{\Delta }{y}_{MB}=\mathrm{\Delta }{x}_{AM}=3-1=2$$$$\Rightarrow y_B=\frac{11}{2}-2=\frac{7}{2}$$$$\frac{x_D+\frac{11}{2}}{2}=3$$$$\Rightarrow x_D=6-\frac{11}{2}=\frac{1}{2}$$$$\frac{y_D+\frac{7}{2}}{2}=\frac{11}{2}$$$$\Rightarrow y_D=11-\frac{7}{2}=\frac{15}{2}$$$$\Rightarrow B(\frac{11}{2},\frac{7}{2}),D(\frac{1}{2},\frac{15}{2})$$

Commented by mr W last updated on 17/Jul/19

Answered by ajfour last updated on 17/Jul/19

Commented by ajfour last updated on 17/Jul/19

$$B(1+4+\frac{1}{2},3+\frac{1}{2})\Rightarrow B(\frac{11}{2},\frac{7}{2})$$$$D(1-\frac{1}{2},3+\frac{1}{2}+4)\Rightarrow D(\frac{1}{2},\frac{15}{2}).$$

Commented by aseer imad last updated on 22/Jul/19

Cool! thanks

Terms of Service

Privacy Policy

Contact: info@tinkutara.com