Question

find the value of x for which the distance between the points p(4 -5) and (12, x) is 10 units

Answer 1

Answer:

p(4,-5), q(12,x)

By using distance formula

distance PQ= √(x1-y1)²+(x2-y2)²

10=√[4-(12)]²+(-5-x)²

10=√[-8]²+(-5-x)²

10=√64+25+x²+10x

10=√x²+10x+89

Squaring on both sides

(10)²=(√x²+10x+89)²

100=x²+10x+89

0=x²-10x+89-100

x²-10x-11=0

x²-(11-1)x-11=0

x²-11x+x-11=0

x(x-11)+1(x-11)=0

(x+1)(x-11)=0

x+1=0. , x-11=0

x=-1. , x=11

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Answer 2

CHAPTER -Coordinate geometry

SOLUTION:-

$\color{aqua} \sf PQ=10$

$\sf \implies PQ²=(10)²=100$

$\sf \implies(12-4)²+(x+5)²=100$

$\sf \implies 8²+(x+5)²=100$

$\sf \implies (x+5)²=(100-64)=36=6²$

$\sf \implies x+5= \pm6$

$\sf \implies x+5=6 \: or \: x+5=-6$

$\implies \sf x=1 \: or \: x=-11$

$\sf \red{ \: Hence, \: x=1 \: or \: x=-1 \: is \: the \: required \: solution }$