Question

The distance between the point P(3a,4a+1) and Q(a, a+1) is 52. Find the value(s) of a if a ε Z^+.

Answer 1

Answer:

a = 4√13

Step-by-step explanation:

We have,

P = ((3a), (4a + 1))

Q = (a, (a + 1))

Let ((3a), (4a + 1)) = (x1, y1)

And (a, (a + 1)) = (x2, y2)

We know that,

By Distance formula,

Distance between two Points

= √[(x1 - x2)² + (y1 - y2)²]

But we are given,

Distance between P and Q = 52 units

So, substituting the values,

52 = √[(3a - a)² + (4a + 1 - (a + 1))²]

52 = √[(2a)² + (4a + 1 - a - 1)²]

52 = √[4a² + (3a)²]

52 = √[4a² + 9a²]

52 = √[13a²]

Squaring both sides, we get,

52² = (√(13a²))²

2704 = 13a²

a² = 2704/13

a² = 208

a = √208

a = √(2 × 2 × 2 × 2 × 13)

a = √(2² × 2² × 13)

a = 2 × 2 × √13

a = 4√13

Hence,

a = 4√13

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