Math, asked by NIHARIKARATHOD, 10 months ago

find the value of x such that PQ is equal to QR where the coordinates of p 6, -1 Q 13 and r x 8 respectively ​

Answers

Answered by Anonymous
1

\huge\bf{Answer:-}

Given:

find the value of x such that PQ is equal to QR where the coordinates of p 6, -1 Q 13 and r x 8 respectively .

Hence :

\sf = (1 - 6) {}^{2}  + (3 + 1) {}^{2}  = (x - 1) {}^{2}  + (8 - 3) {}^{2}  \\ \sf = 21 + 4 = x {}^{2}  + 1 - 2 + 25 \\  \sf = x {}^{2}  - 2x + 1 - 2x + 25 \\ \sf = x {}^{2}  + x - 3x - 3 = 0 \\ \sf = x(x + 1) - 3(x + 1) = 0 \\ \sf = x + 1 = 0..x - 3 = 0 \\ \sf = x =  - 1 \:  \:  \:  \: or \:  \:  \:  \:  = 3

Answered by anishka56
5

\huge\boxed\boxed{Answer}

 {1 - 6}^{2}  +  {3 + 1}^{2}  =  {x - 1}^{2}  +  {8 - 3}^{2}

21 + 4 =  {x}^{2}  + 1 - 2 + 25

 {x}^{2}  - 2x + 1 - 2x + 25

 {x}^{2}  + x - 3x - 3 = 0

x \:  \:  x + 1 - 3 \:  \: x + 1 = 0

x + 1 = 0 \:  \:  \: x - 3 = 0

x =  - 1 \:  \:  \: or \:  \:  \:  = 3

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