Question

PQ is a straight line of 13 units. If P has the coordinates (5, -3)and Q has the co ordinates (-7, y); find the values of y?​

Answer 1

Answer:

PQ=9

√(X2-X1)²+(Y2-Y1)²

√(-7-5)²+(y+3)²

=√(49+25+70)+y²-6y+9

=√

Step-by-step explanation:

√144+y²-6y+9

=153+

Answer 2

Answer:

$\red { Value \: of \: y } \green { \:2 \: Or \: -8 }$

Step-by-step explanation:

$Let \: P(5,-3) = (x_{1} , y_{1} ) , \:and \\</p><p>Q(-7,y) = (x_{2} , y_{2} )$

$PQ = 13 \: units \: (Given)$

$\implies PQ^{2} = 13^{2}$

$\implies (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} = 169$

$\implies ( -7-5)^{2} + (y+3)^{2} = 169$

$\implies (-12)^{2} + (y+3)^{2} = 169$

$\implies 144 + (y+3)^{2} = 169$

$\implies (y+3)^{2} = 169 - 144$

$\implies (y+3)^{2} = 25$

$\implies y+3 = ± \sqrt{25}$

$\implies y+3 = ±5$

$\implies y = ±5 - 3$

$\implies y = 5 - 3 \:Or \: y = -5 - 3$

$\implies y = 2 \: Or \: y = -8$

Therefore.,

$\red { Value \: of \: y } \green { \:2 \: Or \: -8 }$

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