NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.1 Question 9
NCERT Class X Chapter 7: Coordinate Geometry
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.
Given:
Points: Q(0, 1), P(5, -3), R(x, 6)
To Find:
1. The values of \( x \) such that Q is equidistant from P and R.
2. The distances QR and PR.
Formula:
The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
$$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$Solution:
Step 1: Let the coordinates be Q(0, 1), P(5, -3), and R(x, 6). Since Q is equidistant from P and R, we have:
$$ \text{QP} = \text{QR} $$Step 2: Find QP using the distance formula:
$$ \text{QP} = \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} $$Step 3: Find QR using the distance formula:
$$ \text{QR} = \sqrt{(x - 0)^2 + (6 - 1)^2} = \sqrt{x^2 + 25} $$Step 4: Set QP = QR and solve for \( x \):
$$ \sqrt{41} = \sqrt{x^2 + 25} $$ Squaring both sides: $$ 41 = x^2 + 25 $$ $$ x^2 = 16 $$ $$ x = 4 \quad \text{or} \quad x = -4 $$Step 5: For \( x = 4 \), R(4, 6):
QR:
$$ \text{QR} = \sqrt{(4 - 0)^2 + (6 - 1)^2} = \sqrt{16 + 25} = \sqrt{41} $$PR:
$$ \text{PR} = \sqrt{(4 - 5)^2 + (6 - (-3))^2} = \sqrt{(-1)^2 + (9)^2} = \sqrt{1 + 81} = \sqrt{82} $$Step 6: For \( x = -4 \), R(-4, 6):
QR:
$$ \text{QR} = \sqrt{(-4 - 0)^2 + (6 - 1)^2} = \sqrt{16 + 25} = \sqrt{41} $$PR:
$$ \text{PR} = \sqrt{(-4 - 5)^2 + (6 - (-3))^2} = \sqrt{(-9)^2 + (9)^2} = \sqrt{81 + 81} = \sqrt{162} $$ But \( \sqrt{162} = \sqrt{81 \times 2} = 9\sqrt{2} \). Alternatively, you may leave as \( \sqrt{162} \).Result:
The values of \( x \) are 4 and -4.
The distances are:
QR = \( \sqrt{41} \) units (for both values of \( x \)).
PR = \( \sqrt{82} \) units (if \( x = 4 \)),
PR = \( \sqrt{162} \) units (if \( x = -4 \)).
Result:
The distance between the points (2, 3) and (4, 1) is \(2\sqrt{2}\).
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