Question

If one end point of diameter of a circle is (-2,3) and centre of circle is(0,3) then find the other end point
of the diameter.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Coordinate\:of\:B=(2,3)}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given: }} \\ \tt: \implies Coordinate \: of \: A = ( - 2.3) \\ \\ \tt: \implies Coordinate \: of \: O = ( 0.3) \\ \\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies Coordinate \: of \: B=?$

• According to given question :

$\circ\:\tt{AB\:is\:the\:diameter\:of\:circle} \\\\ \circ\:\text{O \: is \: the \: midpoint \: of \: diameter \: AB}\\ \\ \circ \: \tt{Let \: coordinate \: of \:B \: ( x_{2}, y_{2}) } \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies x = \frac{ x_{1} + x_{2} }{2} \:\:\:\:(From\:mid-point\:formula)\\ \\ \tt: \implies 0 = \frac{ - 2 + x_{2}}{2} \\ \\ \tt: \implies - 2 + x_{2} = 0 \\ \\ \green{\tt: \implies x_{2} = 2} \\ \\ \bold{For \: y_{2} \: ordinate : } \\ \tt: \implies y= \frac{ y_{1} + y_{2} }{2} \\ \\ \tt: \implies 3 = \frac{ 3+ y_{2}}{2} \\ \\ \tt: \implies 6 = 3 + y_{2}\\ \\ \green{\tt: \implies y_{2} = 3} \\ \\ \green{\tt{\therefore Coordinate \: of \: B = (2,3)}}$

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

Let us assume that , the other end of the diameter be (x , y)

We know that , centre is the midpoint of the diameter , so , by mid point formula

$\large \mathtt{ \fbox{(x , y) = \frac{ x_{1} + x_{2}}{2} ,\frac{ y_{1} + y_{2}}{2}}}$

0 = x + (-2)/2

0 = x - 2

x = 2

and

3 = (y + 3)/2

6 = y + 3

y = 3

Hence , the other end of the diameter is (2 , 3)

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