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Category: Coordinate Geometry

Question-106029

Question Number 106029 by ajfour last updated on 02/Aug/20 Commented by ajfour last updated on 02/Aug/20 $${Find}\:{coordinates}\:{of}\:{A}\left({x}_{{A}} ,\:{y}_{{A}} \right)\:\:{if} \\ $$$${all}\:{four}\:{quadrants}\:{receive}\:{equal} \\ $$$${area}\:{portions}\:{of}\:\bigtriangleup{ABC}.\:\:\left({x}_{{A}} \geqslant{y}_{{A}} \right)…

Question-171512

Question Number 171512 by infinityaction last updated on 16/Jun/22 Answered by som(math1967) last updated on 17/Jun/22 $$\overset{\frown} {{DL}}=\overset{\frown} {{LM}}=\overset{\frown} {{MB}} \\ $$$$\therefore\:\angle{DAL}=\angle{LAM}=\angle{MAB}=\mathrm{30} \\ $$$${ar}.\bigtriangleup{ALM}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}{sin}\mathrm{30}×\mathrm{8}=\mathrm{16}{cm}^{\mathrm{2}} \\…

Find-the-equation-of-a-circle-which-touches-the-line-x-3y-13-0-and-passes-through-the-points-6-3-and-4-1-

Question Number 170890 by nadovic last updated on 02/Jun/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{which} \\ $$$$\mathrm{touches}\:\mathrm{the}\:\mathrm{line}\:{x}−\mathrm{3}{y}+\mathrm{13}\:=\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{points}\:\left(\mathrm{6},\:\mathrm{3}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{4},\:−\mathrm{1}\right). \\ $$ Answered by aleks041103 last updated on 02/Jun/22…

Question-105001

Question Number 105001 by ajfour last updated on 25/Jul/20 Commented by ajfour last updated on 25/Jul/20 $${If}\:{both}\:{circles}\:{have}\:{unit}\:{radius},\:{and} \\ $$$${regions}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\:{have}\:{equal}\:{areas}, \\ $$$${find}\:{eq}.\:{of}\:{both}\:{circles}. \\ $$ Answered by…

Question-170323

Question Number 170323 by cortano1 last updated on 20/May/22 Answered by mr W last updated on 20/May/22 $$\frac{{AC}}{\mathrm{sin}\:\mathrm{18}°}=\frac{{AD}}{\mathrm{sin}\:\mathrm{12}°}\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{{DB}}=\frac{\mathrm{sin}\:\alpha}{{AD}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\mathrm{12}°} \\…