Question

A point p is at a distance of root 10 from the point(2,3).Find the coordinates of the p if its y-coordinate is twice of the x coordinate

Answer 1

Answer:

Step-by-step explanation:

y = 2x

√(x - 2)^2+ (y - 3)^2 = √10

x^2 + 4 - 4x + 4x^2 + 9 -12x = 10

5x^2 + 13 - 16x =10

5x^2 - 16x = - 3

5x^2 - 16x + 3 = 0

5x^2 - 15x - x + 3 =0

5x(x - 3) - (x - 3) = 0

(5x - 1) (x - 3) = 0

x = 1/5 and 3

if x = 1/5 then y = 2/5

if x = 3 then y = 6

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Answer 2

Given, the co-ordinates of point P whose ordinate is twice of its abscissa is (a, 2a)

Again, given, the point P is at a distance of root of 10 from the points (4,3)

=> √{(2 - a)2 + (3 - 2a)2 } = √10

=> (2 - a)2 + (3 - 2a)2  = 10

=> (4 + a2 - 4a) + (9 + 4a2 - 12a) = 10

=> 4 + a2 - 4a + 9 + 4a2 - 12a = 10

=> 13 + 5a2 - 16a = 10

=> 5a2 - 16a + 13 - 10 = 0

=> 5a2 - 16a + 3 = 0

=> 5a2 - 15a - a + 3 = 0

=>5a(a - 3) - 1(a - 3) = 0

=> (a - 3)*(5a - 1) = 0

=> a = 3, 1/5

So, 2a = 6, 3/5

Hence, the point is either (3, 6) or (1/5, 6/5)

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