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ABCD-is-a-square-AC-is-a-diagonal-If-the-coordinate-of-A-C-are-5-8-and-7-4-Find-the-coordinate-of-B-and-D-

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Question Number 58984 by Tawa1 last updated on 02/May/19
ABCD is a square, AC is a diagonal. If the coordinate of A, C are (− 5, 8) and (7, − 4) . Find the coordinate of B and D.
$$\mathrm{ABCD}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{square},\:\mathrm{AC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{diagonal}.\:\mathrm{If}\:\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\:\mathrm{A},\:\mathrm{C} \\ $$$$\mathrm{are}\:\:\left(−\:\mathrm{5},\:\mathrm{8}\right)\:\mathrm{and}\:\left(\mathrm{7},\:−\:\mathrm{4}\right)\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\:\mathrm{B}\:\mathrm{and}\:\mathrm{D}. \\ $$
Answered by tanmay last updated on 02/May/19
AC eqn (y−8)=((8+4)/(−5−7))(x+5) y−8=−x−5 x+y=3→ y=−x+3 slope m=(−1) the eqn of st line passing through(−5,8) y−8=m_1 (x+5) the angle between x+y=3 and (y−8)=m_1 (x+5) is 45^0 tan45^o =((m_1 −(−1))/(1+m_1 (−1))) m_1 +1=1−m_1 m_1 =0 so eqn is (y−8)=0×(x+5) y=8←eqn of AD the eqn ∥ to y=8 passing through (7,−4) is y=−4 →eqn of BC so eqn of AB x=−5 ( eqn of CD is x=7 so AB→x+5=0 CD is x−7=0 BC is y+4=0 DA is y−8=0 A(−5,8) B(−5,−4) C(7,−4) D(7,8)
$${AC}\:{eqn}\:\left({y}−\mathrm{8}\right)=\frac{\mathrm{8}+\mathrm{4}}{−\mathrm{5}−\mathrm{7}}\left({x}+\mathrm{5}\right) \\ $$$${y}−\mathrm{8}=−{x}−\mathrm{5} \\ $$$${x}+{y}=\mathrm{3}\rightarrow\:{y}=−{x}+\mathrm{3} \\ $$$${slope}\:{m}=\left(−\mathrm{1}\right) \\ $$$${the}\:{eqn}\:{of}\:{st}\:{line}\:{passing}\:{through}\left(−\mathrm{5},\mathrm{8}\right)\: \\ $$$${y}−\mathrm{8}={m}_{\mathrm{1}} \left({x}+\mathrm{5}\right) \\ $$$${the}\:{angle}\:{between}\:{x}+{y}=\mathrm{3}\:{and}\:\left({y}−\mathrm{8}\right)={m}_{\mathrm{1}} \left({x}+\mathrm{5}\right) \\ $$$${is}\:\mathrm{45}^{\mathrm{0}} \\ $$$${tan}\mathrm{45}^{{o}} =\frac{{m}_{\mathrm{1}} −\left(−\mathrm{1}\right)}{\mathrm{1}+{m}_{\mathrm{1}} \left(−\mathrm{1}\right)} \\ $$$$ \\ $$$${m}_{\mathrm{1}} +\mathrm{1}=\mathrm{1}−{m}_{\mathrm{1}} \:\:{m}_{\mathrm{1}} =\mathrm{0} \\ $$$${so}\:{eqn}\:{is}\:\left({y}−\mathrm{8}\right)=\mathrm{0}×\left({x}+\mathrm{5}\right) \\ $$$${y}=\mathrm{8}\leftarrow{eqn}\:{of}\:{AD} \\ $$$${the}\:{eqn}\:\parallel\:{to}\:{y}=\mathrm{8}\:{passing}\:{through}\:\left(\mathrm{7},−\mathrm{4}\right)\:{is} \\ $$$${y}=−\mathrm{4}\:\rightarrow{eqn}\:{of}\:{BC} \\ $$$${so}\:{eqn}\:{of}\:{AB}\:\:{x}=−\mathrm{5}\:\left(\right. \\ $$$${eqn}\:{of}\:{CD}\:{is}\:{x}=\mathrm{7} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{A}}{B}\rightarrow{x}+\mathrm{5}=\mathrm{0} \\ $$$$\:{CD}\:\:{is}\:{x}−\mathrm{7}=\mathrm{0} \\ $$$${BC}\:{is}\:{y}+\mathrm{4}=\mathrm{0} \\ $$$${DA}\:{is}\:{y}−\mathrm{8}=\mathrm{0} \\ $$$$ \\ $$$${A}\left(−\mathrm{5},\mathrm{8}\right)\:\:{B}\left(−\mathrm{5},−\mathrm{4}\right)\:\:\:{C}\left(\mathrm{7},−\mathrm{4}\right)\:\:{D}\left(\mathrm{7},\mathrm{8}\right) \\ $$
Commented by Tawa1 last updated on 02/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay last updated on 02/May/19
thank you...
$${thank}\:{you}… \\ $$

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