Question

Find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units.​

Answer 1

Answer:

p (2,-3)is second quadrant

10,is first quadrant

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Answer 2

$\frak{Given}\begin{cases}\sf{\;\;Coordinates\;of\;P={\textsf{\textbf{(2,\;-3)}}}}\\\sf{\;\;Coordinates\;of\;Q={\textsf{\textbf{(10,\;y)}}}}\\\sf{\;\;Distance\;b/w\;the\;points={\textsf{\textbf{10\;units}}}}\end{cases}$

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Need to find: The value of y.

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$\qquad\qquad\footnotesize{\underline{\bf{\dag}\frak{\;By\;using\;Distance\;formula:}}}$

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$\star\;\boxed{\pink{\frak{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=PQ}}}$

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Here,

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• x₁ = 2
• x₂ = 10
• y₁ = -3
• y₂ = y
• PQ = 10 units

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Therefore,

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$:\implies\sf{\sqrt{(10-2)^2+(y-(-3))^2}=10}\\\\\\\\:\implies\sf{\sqrt{(8)^2+(y+3)^2}=10}$

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$\qquad\qquad\footnotesize{\underline{\bf{\dag\;}\frak{Applying\;identity,\;(a+b)=a^2+2ab+b^2:}}}$

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$:\implies\sf{64+y^2+2(y)(3)+(3)^2=(10)^2}\\\\\\\\:\implies\sf{64+y^2+6y+9=100}\\\\\\\\:\implies\sf{y^2+6y+64+9-100=0}\\\\\\\\:\implies\sf{y^2+6y+73-100=0}\\\\\\\\:\implies\sf{y^2+6y-27=0}\\\\\\\\:\implies\sf{y^2+9y-3y-27=0}\\\\\\\\:\implies\sf{y(y+9)-3(y+9)=0}\\\\\\\\:\implies\sf{(y+9)(y-3)=0}\\\\\\\\:\implies\sf{y+9=0\;\;or\;\;y-3=0}\\\\\\\\:\implies\underline{\boxed{\purple{\frak{y=-9\;\;or\;\;y=3}}}}{\;\bigstar}$

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$\therefore\;{\underline{\sf{Hence,\;the\;value\;of\;y\;is\;{\textsf{\textbf{-9}}}\;or\;{\textsf{\textbf{3}}}}.}}$⠀⠀