Question

20.) P = (6.-2)= (1.3), R = (x, 8), such
PQ = QR, then the value of x is
​

Answer 1

Answer:

x is both -4 and 6

Step-by-step explanation:

P = (6.-2), Q= (1.3), R = (x, 8)

PQ=QR

$\sqrt{(6-1)^{2}+(3-(-2))^{2} } =\sqrt{(x-1)^2 +{(8-3)^{2} }}$

$\sqrt{(5)^{2}+(5)^{2} } =\sqrt{(x-1)^2 +{(5)^{2} }}$

$5\sqrt{2 } =\sqrt{x^2-2x+26 }}$

50=x²-2x+26

x²-2x-24=0

x²-6x+4x-24=0

x(x-6)+4(x-6)=0

(x+4)(x-6)=0

x=-4 ,x=6

x can either be -4 or 6