Question

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Answer 1

QUESTION:

Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2 , - 3) and B is the point (1 , 4).

ANSWER:

• The cordinates of a point A = (3 , -10)

GIVEN:

• AB is diameter of a circle.

• centre is (2 , - 3).

• Point B is (1 , 4).

TO FIND:

• The cordinates of a point A.

FORMULAE:

$\large { \bold{Mid \ point = \left(\dfrac{x_1 + x_2}{2} \ , \ \dfrac{y_1 + y_2}{2}\right)}}$

EXPLANATION:

Centre is the Mid point of the diameter A and B. So we can use mid point formula to find point A.

Let point A = ( x₁ , y₁ )

Let point B = ( x₂ , y₂ )

We know that ( x₂ , y₂ ) = (1 , 4)

Mid point = Centre = (2 , - 3)

$(2 , - 3)= \left(\dfrac{x_1 + 1}{2} \ , \ \dfrac{y_1 + 4}{2}\right)$

Equate x and y coordinates

$\dfrac{x_1 + 1}{2} = 2 \: \: , \: \: \dfrac{y_1 + 4}{2} = - 3$

$x_1 + 1= 4 \: \: , \: \: y_1 + 4 = - 6$

$x_1 = 3 \: \: , \: \: y_1 = - 10$

HENCE POINT A = ( 3 , - 10)

VERIFICATION:

Substitute ( x₁ , y₁ ) = ( 3 , - 10) and

( x₂ , y₂ ) = ( 1 , 4 ) in midpoint formula.

$Mid \ point = \left(\dfrac{3 + 1}{2} \ , \ \dfrac{-10+4 }{2}\right)$

$Mid \ point = \left(\dfrac{4}{2} \ , \ \dfrac{-6}{2}\right)$

Mid point = Centre = ( 2 , - 3)

HENCE VERIFIED.