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Question Number 121393 by faysal last updated on 07/Nov/20
a square′s one diagonal′s terminal points are (5,0), (9,4) what are the coordinate of other diagonal′s terminal points
$${a}\:{square}'{s}\:{one}\:{diagonal}'{s}\:{terminal}\:{points}\:{are} \\ $$$$\left(\mathrm{5},\mathrm{0}\right),\:\left(\mathrm{9},\mathrm{4}\right)\:{what}\:{are}\:{the}\:{coordinate}\:{of} \\ $$$${other}\:{diagonal}'{s}\:{terminal}\:{points} \\ $$
Answered by TANMAY PANACEA last updated on 07/Nov/20
Disgonal eqn ((y−0)/(0−4))=((x−5)/(5−9))→x−y−5=0 mid point of(5,0) and(9,4)=(7,2) another diagonal is ⊥(x−y−5=0)is x+y=k which passes through (7,2)→7+2=k Eqn anothdr diagonal x+y=9 the side of square make 45^o with diagonal x−y=5 snd which passes through (5,0)is y=m(x−5) →slope of x−y=5 is 1 tan45^0 =((1−m)/(1+m))→m=0 eqn of side is y=0 (x axis) solve y=0 and x+y=9 (9,0)is one terminal of another diagonal the side of square make 45^o with diagonal x−y=5 and passes through(9,4) is (y−4)=m_1 (x−9) tan45^o =((1−m_1 )/(1+m_1 ))→m_1 =0 eqn of anothdr side y=4 solve y=4 and x+y=9 another terminal(5,4) Rewuired answer is (9,0) and (5,4)
$${Disgonal}\:{eqn}\:\:\frac{{y}−\mathrm{0}}{\mathrm{0}−\mathrm{4}}=\frac{{x}−\mathrm{5}}{\mathrm{5}−\mathrm{9}}\rightarrow{x}−{y}−\mathrm{5}=\mathrm{0} \\ $$$${mid}\:{point}\:{of}\left(\mathrm{5},\mathrm{0}\right)\:{and}\left(\mathrm{9},\mathrm{4}\right)=\left(\mathrm{7},\mathrm{2}\right) \\ $$$${another}\:{diagonal}\:{is}\:\bot\left({x}−{y}−\mathrm{5}=\mathrm{0}\right){is} \\ $$$${x}+{y}={k}\:{which}\:{passes}\:{through}\:\left(\mathrm{7},\mathrm{2}\right)\rightarrow\mathrm{7}+\mathrm{2}={k} \\ $$$${Eqn}\:{anothdr}\:{diagonal}\:\:{x}+{y}=\mathrm{9} \\ $$$${the}\:{side}\:{of}\:{square}\:{make}\:\mathrm{45}^{{o}} \:{with}\:{diagonal} \\ $$$${x}−{y}=\mathrm{5}\:{snd}\:{which}\:{passes}\:{through}\:\left(\mathrm{5},\mathrm{0}\right){is} \\ $$$${y}={m}\left({x}−\mathrm{5}\right)\:\rightarrow{slope}\:{of}\:{x}−{y}=\mathrm{5}\:{is}\:\mathrm{1} \\ $$$${tan}\mathrm{45}^{\mathrm{0}} =\frac{\mathrm{1}−{m}}{\mathrm{1}+{m}}\rightarrow{m}=\mathrm{0}\: \\ $$$${eqn}\:{of}\:{side}\:{is}\:{y}=\mathrm{0}\:\:\left({x}\:{axis}\right) \\ $$$${solve}\:{y}=\mathrm{0}\:\:{and}\:{x}+{y}=\mathrm{9} \\ $$$$\left(\mathrm{9},\mathrm{0}\right){is}\:{one}\:{terminal}\:{of}\:{another}\:{diagonal} \\ $$$${the}\:{side}\:{of}\:{square}\:{make}\:\mathrm{45}^{{o}} \:{with}\:{diagonal}\:{x}−{y}=\mathrm{5} \\ $$$${and}\:{passes}\:{through}\left(\mathrm{9},\mathrm{4}\right)\:{is}\:\left({y}−\mathrm{4}\right)={m}_{\mathrm{1}} \left({x}−\mathrm{9}\right) \\ $$$${tan}\mathrm{45}^{{o}} =\frac{\mathrm{1}−{m}_{\mathrm{1}} }{\mathrm{1}+{m}_{\mathrm{1}} }\rightarrow{m}_{\mathrm{1}} =\mathrm{0}\:\:{eqn}\:{of}\:{anothdr}\:{side}\:{y}=\mathrm{4} \\ $$$${solve}\:{y}=\mathrm{4}\:\:{and}\:{x}+{y}=\mathrm{9}\:\: \\ $$$$\boldsymbol{{another}}\:\boldsymbol{{terminal}}\left(\mathrm{5},\mathrm{4}\right) \\ $$$$\boldsymbol{{R}}{ewuired}\:{answer}\:{is}\:\left(\mathrm{9},\mathrm{0}\right)\:{and}\:\left(\mathrm{5},\mathrm{4}\right) \\ $$$$ \\ $$
Answered by mr W last updated on 07/Nov/20
((5+9)/2)=7 ((0+4)/2)=2 ((4−0)/(9−5))=1 (√((9−5)^2 +(4−0)^2 ))=4(√2) (7,2)±((4(√2))/2)((1/( (√2))),−(1/( (√2))))=(7±2,2∓2) =(9,0) and (5,4)
$$\frac{\mathrm{5}+\mathrm{9}}{\mathrm{2}}=\mathrm{7} \\ $$$$\frac{\mathrm{0}+\mathrm{4}}{\mathrm{2}}=\mathrm{2} \\ $$$$\frac{\mathrm{4}−\mathrm{0}}{\mathrm{9}−\mathrm{5}}=\mathrm{1} \\ $$$$\sqrt{\left(\mathrm{9}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{0}\right)^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{7},\mathrm{2}\right)\pm\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\left(\mathrm{7}\pm\mathrm{2},\mathrm{2}\mp\mathrm{2}\right) \\ $$$$=\left(\mathrm{9},\mathrm{0}\right)\:{and}\:\left(\mathrm{5},\mathrm{4}\right) \\ $$

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