If the point A(2,-4)is equidistant from …
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Sia ? 6 years, 4 months ago
According to the question,we are given that,
PA = QA
⇒ PA2 = QA2
⇒ (3 – 2)2 + (8 + 4)2 = (–10 – 2)2 + (y + 4)2
⇒ 12 + 122 = (–12)2 + y2 + 16 + 8y
⇒ y2 + 8y + 16 – 1 = 0
⇒ y2 + 8y + 15 = 0
⇒ y2 + 5y + 3y + 15 = 0
⇒ y(y + 5) + 3(y + 5) = 0
⇒ (y + 5) (y + 3) = 0
⇒ y + 5 = 0 or y + 3 = 0
⇒ y = –5 or y = –3
So, the co–ordinates are P(3, 8), Q1(–10, –3), Q2(–10, –5).
Now, {tex}P Q _ { 1 } ^ { 2 }{/tex} = (3 + 10)2 + (8 + 3)2 = 132 + 112
⇒ {tex}P Q _ { 1 } ^ { 2 }{/tex} = 169 + 121
⇒ {tex}P Q _ { 1 } = \sqrt { 290 }{/tex} units
and {tex}P Q _ { 2 } ^ { 2 }{/tex} = (3 + 10)2 + (8 + 5)2 = 132 + 132
= 132[1 + 1]
⇒{tex}P Q _ { 2 } ^ { 2 }{/tex}= 132 × 2
⇒ {tex}P Q _ { 2 } = 13 \sqrt { 2 }{/tex} units
Hence, y = –3, –5 and PQ = {tex}\sqrt { 290 }{/tex} units and {tex}13 \sqrt { 2 }{/tex}units.
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