Question

find the value of x such that PQ=QR where the coordinates of P,Q and R are (6, -1), (1,3) and (x,8) respectively.

Answer 1

PQ=QR
root((6-1)^2+(-1-3)^2) =root((1-x)^2 +(3-8)^2)
squaring on both sides
(6-1)^2 +(-4)^2 =(1-x) ^2 +(-5)^2
(5)^2+(4)^2=x^2-2x+1 +25
25+16 =x^2-2x+26
41=x^2-2x+26
x^2-2x+26-41=0
x^2-2x-15=0
x^2-5x+3x-15=0
x(x-5)+3(x-5)=0
(x-5)(x+3)=0
x-5=0
x=5
x+3=0
x=-3
x=5,-3
x=5 is considered

Answer 2

Your Answer :-

Using Distance Formula ,

PQ = $=> \sqrt{(6-1)^{2}+(-1-3)^{2}}$

PQ = $=>\sqrt{25+16}$

PQ = $\sqrt{41}$

then ,

As Given Pq = Qr .

so ,

41 = $(x-1)^{2} + (8-3)^{2}$

$41 = x^{2} +1-2x+25$

$41=x^{2}+1-2x+25$

$16=x^{2}+1-2x$

$x^{2}+1-2x-16=0$

$x^{2}-2x-15=0$

$x^{2}-5x+3x-15=0$

x(x-5) + 3(x-5) = 0

(x+3)(x-5) = 0

x = 5 units

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