Question

Find a if point Q(0,1) is equidistant from the points
P(5,-3)and are R(x,6)​

Answer 1

Step-by-step explanation:

Given:-

a point Q(0,1) is equidistant from the points

P(5,-3)and are R(x,6)

To find:-

Find the value of x ?

Solution:-

Given points are P(5,-3) , Q(0,1) and R(x,6)

Given that

Point Q is equidistant from P and R

P__________Q____________R

=> PQ = QR

Distance between P and Q = Distance between Q and R

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units

Distance between P and Q:-

Let (x1, y1)=P(5,-3) => x1 = 5 and y1 = -3

Let (x2, y2)=Q(0,1)=> x2=0 and y2=1

Distance between P and Q = PQ

=>√[(0-5)^2+(1-(-3))^2]

=>√[(-5)^2+(1+3)^2]

=>√(25+4^2)

=>√(25+16)

PQ = √41 units

Distance between Q and R:-

Let (x1, y1)=Q(0,1)=> x1=0 and y1=1

Let (x2, y2)=R(x,6)=> x2=x and y2=6

Distance between Q and R=

=>√[(x-0)^2+(6-1)^2]

=>√(x^2+5^2)

QR =√(x^2+25) units

We have

PQ=QR

=>√41 = √(x^2+25)

On squaring both sides then

=>(√41)^2 = [ √(x^2+25) ] ^2

=>41 = x^2+25

=>x^2 = 41-25

=>x^2= 16

=>x=±√16

=>x = ±4

The values of x are 4 or -4

Answer:-

The values of x for the given problem are 4 and -4

Used formula:-

The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)^2+(y2-y1)^2] units