# Question-68212

Question Number 68212 by peter frank last updated on 07/Sep/19
Answered by $@ty@m123 last updated on 07/Sep/19 $${Let}\:{required}\:{equation}\:{of}\:{line}: \\$$$${y}={m}_{\mathrm{1}} {x}+{c}\:\:\:…..\left(\mathrm{1}\right) \\$$$${Given}\:{line}:\:\mathrm{4}{x}+\mathrm{3}{y}=\mathrm{21} \\$$$${Its}\:{slope}:\:{m}_{\mathrm{2}} =\frac{−\mathrm{4}}{\mathrm{3}} \\$$$${ATQ}, \\$$$${m}_{\mathrm{1}} .{m}_{\mathrm{2}} =−\mathrm{1} \\$$$$\Rightarrow{m}_{\mathrm{1}} ×\frac{−\mathrm{4}}{\mathrm{3}}=−\mathrm{1} \\$$$$\Rightarrow{m}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{4}} \\$$$${Substituting}\:{it}\:{in}\:\left(\mathrm{1}\right), \\$$$${y}=\frac{\mathrm{3}}{\mathrm{4}}{x}+{c} \\$$$${It}\:{passes}\:{through}\:\left(\mathrm{1},−\mathrm{3}\right) \\$$$$\Rightarrow\:−\mathrm{3}=\frac{\mathrm{3}}{\mathrm{4}}+{c} \\$$$$\Rightarrow{c}=\frac{−\mathrm{15}}{\mathrm{4}} \\$$$$\therefore{required}\:{equation}\:{of}\:{line}: \\$$$${y}=\frac{\mathrm{3}}{\mathrm{4}}{x}−\frac{\mathrm{15}}{\mathrm{4}} \\$$$$\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{15} \\$$ Commented by peter frank last updated on 07/Sep/19 $${thank}\:{you} \\$$ Answered by$@ty@m123 last updated on 07/Sep/19
$$\left(\mathrm{2}\right)\:{B}^{'} \subset{A}' \\$$$$\Rightarrow\xi−{B}\subset\xi−{A} \\$$$$\Rightarrow{A}\subset{B} \\$$
Answered by Rasheed.Sindhi last updated on 07/Sep/19
$$\mathrm{B}'\subset\mathrm{A}'\Rightarrow\mathrm{A}'\cap\mathrm{B}'=\mathrm{B}' \\$$$$\Rightarrow\left(\mathrm{A}'\cap\mathrm{B}'\right)'=\left(\mathrm{B}'\right)' \\$$$$\Rightarrow\left(\mathrm{A}'\right)'\cup\left(\mathrm{B}'\right)'=\mathrm{B} \\$$$$\Rightarrow\mathrm{A}\cup\mathrm{B}=\mathrm{B} \\$$$$\Rightarrow\mathrm{A}\subseteqq\mathrm{B} \\$$
Commented by peter frank last updated on 08/Sep/19
$${I}\:{did}\:{not}\:{real}\:{understood}\:{sir} \\$$
Commented by Rasheed.Sindhi last updated on 08/Sep/19
$$\mathrm{Used}\:\mathrm{some}\:\mathrm{laws}: \\$$$$\:^{\bullet} \mathrm{P}\subset\mathrm{Q}\Rightarrow\mathrm{P}\cap\mathrm{Q}=\mathrm{P} \\$$$$\:^{\bullet} \:\left(\mathrm{P}'\right)'=\mathrm{P} \\$$$$\:^{\bullet} \:\left(\mathrm{P}\cap\mathrm{Q}\right)^{'} =\mathrm{P}'\cup\mathrm{Q}' \\$$$$\:^{\bullet} \:\mathrm{P}\cup\mathrm{Q}=\mathrm{R}\Rightarrow\mathrm{P}\subseteq\mathrm{R}\:\&\:\mathrm{Q}\subseteq\mathrm{R} \\$$
Commented by peter frank last updated on 13/Sep/19
$${thank}\:{you} \\$$