Question

Coo:
162
9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the
th
distances QR and PR.
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Answer 1

Given:-

Q(0, 1) is equidistant from P(5, -3) and R(x, 6)

To find:-

If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution:-

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

Given that Q is the equidistant from P and R

P______________Q______________R

=> Distance between P and Q = Distance between Q and R

=> PQ = QR

I)Length of PQ :-

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

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

We know that

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

=> PQ = √[(0-5)^2+(1+3)^2]

=> PQ =√[(-5)^2+(4)^2]

=> PQ = √(25+16)

=> PQ =√41 units -----------(1)

ii)Length of QR:-

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

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

We know that

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

=> QR =√[(x-0)^2+(6-1)^2]

=> QR =√[x^2+(5)^2]

=> QR =√(x^2+25) units ----------(2)

From (1)&(2)

(1) = (2)

=> √41= √(x^2+25)

On squaring both sides then

=> [√41]^2 = [√(x^2+25)]^2

=> 41 = x^2+25

=> x^+25 = 41

=>x^2 = 41-25

=>x^2 = 16

=> x=±√16

=>x = ±4

The value of x is 4 or -4

Then the point R=(4,6) or (-4,6)

QR = √(x^2+25)

=>√(4^2+25)

=>√(16+25) units

=> √41 units

Taking x = 4

Distance between P and R :-

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

Let (x2, y2)=(4,6)=> x2=4 and y2=6

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

=> √[(4-5)^2+(6+3)^2]

=> √[(-1)^2+(9)^2]

=>√(1+81)

=>√82 units

(or)

Taking x = -4

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

Let (x2, y2)=(-4,6)=> x2=-4 and y2=6

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

=> √[(-4-5)^2+(6+3)^2]

=> √[(-9)^2+(9)^2]

=>√(81+81)

=>√162 units

=> √2×81

=> 9√2 units

Answer:-

The value of x = 4or -4

The distance between Q and R = √41 units

The distance between P and R = 9√2 units

or √82 units

Used formulae:-

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