Question

Example 5 : Find a point on the y-axis which
is equidistant from the points A6, 5) and
B(-4, 3).​

Answer 1

Answer :

• The coordinates of the point on y-axis which is equidistant from the points A( 6 , 5 ) and B ( -4 , 3 )

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• Let the coordinates of the point be C ( x , y )
• Given that the point C is in y-axis
• x coordinate of the point = 0
• AC = BC

$\underline{\boxed{\sf Distance \ formula = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}$

$:\implies \sf AC= BC$

$:\implies \sf \sqrt{(0-6)^2+(y-5)^2} = \sqrt{(0+4)^2+(y-3)^2}$

$:\implies \sf \sqrt{(-6)^2+y^2+5^2-2 \times y \times 5} = \sqrt{4^2+y^2+3^3- 2 \times y \times 3}$

$:\implies \sf \sqrt{36+y^2+25-10y} = \sqrt{16+y^2+9-6y}$

$:\implies \sf \sqrt{y^2-10y+61} = \sqrt{y^2-6y+25}$

$:\implies \sf y^2-10y+61 = y^2-6y+25$

$:\implies \sf -10y+6y = 25-61$

$:\implies \sf -4y = -36$

$:\implies\underline{\boxed{\pink{\mathfrak{y=9}}}}$

Answer 2

Answer:

$\: { \huge{ \underline{ \sf{ \red{Solution:}}}}}$

⇒ We know that a point on the y - axis is of the form (0, y)

⇒ So, let the point P(0, y) be equidistant from A and B

$\to{ \sf{PA = \sqrt{ {(6 - 0)}^{2} + {(5 - y)}^{2} } }}$

$\to{ \sf{PB = \sqrt{ {( - 4 - 0)}^{2} + {(3 - y)}^{2} } }}$

$\to{ \sf{PA² = PB²}}$

So,

${ \sf{ {(6 - 0)}^{2} + {(5 - y)}^{2} = {( - 4 - 0)}^{2} + {(3 - y)}^{2} }}$

${ \sf{36 + 25 + {y}^{2} - 10y = 16 + 9 + {y}^{2} - 6y}}$

${ \sf{4y = 36}}$

$\boxed{ \sf{y = 9}}$

So, the required point is (0, 9)

$\underline{ \underline{ \sf{Let's \: check \: our \: solution:}}}$

$\implies{ \sf{AP = \sqrt{ {(6 - 0)}^{2} + {(5 - 9)}^{2} } }}$

${ \sf{AP = \sqrt{36 + 16} = \sqrt{52} }}$

$\implies{ \sf{BP = \sqrt{ {( - 4 - 0)}^{2} + {(3 - 9)}^{2} } }}$

${ \sf{BP = \sqrt{16 + 36} = \sqrt{52} }}$

${ \therefore{ \sf{ \pink{(0, 9) \: is \: equidistant \: from \: (6, 5) \: and \: (4, 3)}}}}$