Question

Find a point on the y-axis, which is equidistant from the points A (6,5) and B (-4,3).​

Answer 1

Answer:

$\bigstar{\bold{Point\:on\:Y\:axis=(0,9)}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Point A = (6,5)
• Point B = (-4, 3)

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Point on the y axis which is equidistant from the point A and B

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Let the point be P.

→ Given that the point is on Y axis, hence the point would be (0 , y)

→ Also given points are equidistant

→ Hence,

  Distance of PA = Distance of PB

→ First find the distance between PA by using distance formula

  $\sf {PA=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2} } }$

 where x₁ = 0 , x₂ = 6 , y₁ = y, y₂ = 5

→ Substitute the datas,

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

 $\sf {PA=\sqrt{36+(5-y)^{2} }---(1) }$

→ Now find the distance of PB

  where x₁ = 0, x₂ = -4, y₁ = y, y₂ = 3

→ Substituting the datas,

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

  $\sf {PB=\sqrt{16+(3-y)^{2} }----(2) }$

→ It is given that the LHS of equation 1 and 2 are equal, hence RHS must also be equal.

 $\sf {\sqrt{36+(5-y)^{2} } =\sqrt{16+(3-y)^{2} }$

→ Squaring on both sides,

  36 + ( 5 - y )² = 16 + ( 3 - y )²

→ Expanding the brackets and simplifying,

  36 + 25 - 10y + y² = 16 + 9 - 6y + y²

→ Cancelling y² on both sides,

  36 + 25 - 10y = 16 + 9 - 6y

  61 - 10y = 25 - 6y

  61 - 25 = -6y + 10y

  36 = 4 y

    y = 36/4

    y = 9

→ Hence the point is ( 0, y ) = ( 0 , 9 )

$\boxed{\bold{Point\:on\:Y\:axis=(0,9)}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The distance between two points A and B is given by the formula,

  $\sf {AB=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2} } }$