Question

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​

Answer 1

Answer:

the values of. x is 4 or -4

and the distances QR and PR​ is QR=$\sqrt{41}$ and Hence PR=$\sqrt{82}$ OR 9$\sqrt{2}$.

Step-by-step explanation:

Since Q is a equidistant from P and R.

QP=QR

Find QP

x1=0,y1=1

x2=5,y2=-3

Therefore QP=$\sqrt{(x2-x1)^{2}+(y2-y1)^{2} }$

QP=$\sqrt{(5-0)^{2}+(-3-1)^{2} }$

                       =$\sqrt{41}$.

Find QR

x1=0,y1=1

x2=x,y2=6

Therefore QR=$\sqrt{(x2-x1)^{2}+(y2-y1)^{2} }$

QR=$\sqrt{(x-0)^{2}+(6-1)^{2} }$

     =$\sqrt{x^{2} +25}$.

Since QP=QR

$\sqrt{41}$=$\sqrt{x^{2} +25}$

Squaring both sides

41=$x^{2}$+25

or,$x^{2}$=16

oe,x=±4

so x=4 or x=-4.

Therefore the point R(x,6) is (4,6) or(-4,6)

Now we find the the distances QR and PR​

Finding QR

Taking x=4

QR=$\sqrt{x^{2} +25}$.

QR=$\sqrt{4^{2} +25}$

     =$\sqrt41}$.

Taking x=-4

QR=$\sqrt{x^{2} +25}$.

QR=$\sqrt{(-4)^{2} +25}$

    = $\sqrt{41}$

Hence QR=$\sqrt{41}$

Finding PR

x1=5,y1=-3

x2=x,y2=6

Therefore PR=$\sqrt{(x2-x1)^{2}+(y2-y1)^{2} }$

PR=$\sqrt{(x-5)^{2}+(6-3)^{2} }$

 PR=$\sqrt{(x-5)^{2} +(9)^{2} }$.

Taking x=4

PR=$\sqrt{(4-5)^{2} +(9)^{2} }$.

    =$\sqrt{82}$

Taking x=-4

PR=$\sqrt{(-4-5)^{2} +(9)^{2} }$  

   =9$\sqrt{2}$

Hence PR=$\sqrt{82}$ OR 9$\sqrt{2}$.

Answer 2

Value of x is -4.

We know that the distance between the two points is given by the Distance Formula = √[( x₂ - x₁ )2 + (y₂ - y₁)2]

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

So, PQ = QR

Hence by applying the distance formula we get,

√(5 - 0)² + (-3 - 1)² = √(0 - x)² + (1 - 6)²

√(5)² + (- 4)² = √(- x)² + (- 5)²

By squaring both the sides,

25 + 16 = x2 + 25

16 = x2

x = ± 4

Therefore, point R is (4, 6) or (- 4, 6).

Case (1): When point R is (4, 6),

Distance between P (5, - 3) and R (4, 6) can be calculated using the Distance Formula as,

PR = √(5 - 4)² + (- 3 - 6)²

= √1² + (- 9)²

= √1 + 81

= √82

Distance between Q (0, 1) and R (4, 6) can be calculated using the distance formula as,

QR = √(0 - 4)² + (1 - 6)²

= √(- 4)² + (- 5)²

= √16 + 25

= √41

Case (2): When point R is (- 4, 6)

Distance between P (5, - 3) and R (- 4, 6) can be calculated using the distance formula as,

PR = √(5 - (- 4))² + (- 3 - 6)²

= √(9)² + (- 9)²

= √81 + 81

= 9√2

Distance between Q (0, 1) and R (- 4, 6) can be calculated using the distance formula as,

QR = √(0 - (- 4))² + (1 - 6)²

= √(4)² + (- 5)²

= √16 + 25

= √41

Thus, we see that using R (- 4, 6) we get PR = QR. Thus, the point is R (- 4, 6). Hence, x = - 4.

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