Question

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Answer 1

Answer:

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Answer 2

❏ SolutioN :

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$\blacksquare\:\:\footnotesize{\underline{\underline{Given}}}$

$\footnotesize{\text{point A(0,2) is equidistant from point B(3,p) and (p,5)}}$

$\blacksquare\:\:\footnotesize{\underline{\underline{To\: Find}}}$

$\footnotesize{\text{value of p = ? and AB = ?}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\footnotesize{\text{so , the distance AC and AB will be equal }}$

$\therefore\:\:\footnotesize{AB=AC}$

$\implies\footnotesize{\sqrt{(3-0)^2+(p-2)^2}=\sqrt{(p-0)^2+(5-2)^2}}$

$\implies\footnotesize{\sqrt{(3)^2+(p^2-4p+4)}=\sqrt{(p)^2+(3)^2}}$

$\implies\footnotesize{\sqrt{9+p^2-4p+4}=\sqrt{p^2+9}}$

$\implies\footnotesize{9+p^2-4p+4=p^2+9}$

$\implies\footnotesize{-4p+4=0}$

$\implies\footnotesize{-4p=-4}$

$\implies\footnotesize{4p=4}$

$\implies\footnotesize{p=\dfrac{4}{4}}$

$\implies\footnotesize{\red{\boxed{p=1}}}$

$\therefore\:\:\footnotesize{\text{point A(0,2) , point B(3,1) , point C(1,5)}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\therefore\:\:\footnotesize{AB=\sqrt{(3-0)^2+(1-2)^2}}$

$\implies:\footnotesize{AB=\sqrt{(3)^2+(-1)^2}}$

$\implies:\footnotesize{AB=\sqrt{9+1}}$

$\implies:\footnotesize{\boxed{AB=\sqrt{10}\: units}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.08mm}\multiput(0,0)(1,0){41}{\line(0,1){40}}\multiput(0,0)(0,1){41}{\line(1,0){40}}\linethickness{0.4 mm}\put(20,0){\line(0,1){40}}\put(0,20){\line(1,0){40}}\multiput(23,19)(3,0){6}{\line(0,1){2}}\multiput(17,19)(-3,0){6}{\line(0,1){2}}\multiput(19,23)(0,3){6}{\line(1,0){2}}\multiput(19,17)(0,-3){6}{\line(1,0){2}}\put(29,23){\circle*{0.5}}\put(23,35){\circle*{0.5}}\put(20,26){\circle*{0.5}}\put(29,23){\circle{2}}\put(23,35){\circle{2}}\put(20,26){\circle{2}}\put(16,26){ A }\put(31,23){ B }\put(25,35){ C }\put(20,26){\line(1,3){3}}\put(20,26){\line(3,-1){9}}\end{picture}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$