Math, asked by sofia2647, 4 months ago

guys please answer this​

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Answered by OfficialPk
74

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\mathsf\red{Question \: 12}

\mathsf{Let \: p(0,y) \: be \: on \: y-axis}

\mathsf{Given \: Points \: are \: equidistant}

\mathsf{AP \: = \: BP}

\mathsf{\sqrt{{(0-6)}^{2}+{(y-5)}^{2}} \: = \: \sqrt{{(0+4)}^{2}+{(y-3)}^{2}}}

\mathsf{Squaring \: on \: both \: sides}

\mathsf{{6}^{2}+{(y-5)}^{2} \: = \: {4}^{2} + {(y-3)}^{2}}

\mathsf{36+{y}^{2}-10y+25 \: = \: 16+{y}^{2}-6y+9}

\mathsf{4y \: = \: 36}

\mathsf{y \: = \: \frac{36}{4}}

\mathsf{y \: = \: 9}

\mathsf\red{\therefore \: The \: point \: on \: y-axis \: is \: (0,9)}

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\mathsf\red{Question \: 13}

\mathsf{From \: the \: question \: A(4,-3) ; B(8,5); \: l =3 ; \: m=1}

\mathsf{(\frac{lx_2 + mx_1}{l+m} \: , \: \frac{ly_2 + my_1}{l+m})}

\mathsf{(\frac{(3×8)+(1×4)}{3+1} \: , \frac{(3×5)+(1×(-3))}{3+1})}

\mathsf{(\frac{24+4}{4} \: , \frac{15-3}{4})}

\mathsf{(\frac{28}{4} \: , \frac{12}{4})}

\mathsf{(7,3)}

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Answered by pradoshG
1

Answer:

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