Question

10.The coordinates of a point P, where PQ is the diameter of circle whose centre is (2, – 3) and Q is (1, 4) is:

2 points

(3,-10)

(2,-10)

(-3,10)

(-2,10)



​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{One end of the diameter PQ of the circle is Q(1,4)}$

$\textsf{and centre is (2,-3)}$

$\underline{\textsf{To find:}}$

$\textsf{The co ordinates of P}$

$\underline{\textsf{Solution:}}$

$\underline{\textsf{Concept used:}}$

$\boxed{\begin{minipage}{6cm} \textsf{The midpoint of the line joining}\\\\\mathsf{(x_1,y_1)\;\;\&\;\;(x_2,y_2)}\;\textsf{is}\\\\\mathsf{(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})} \end{minipage}}$

$\textsf{Let the coordinates of P be}\;\mathsf{(x_1,y_1)}$

$\textsf{We know that,}$

$\textsf{Centre of the circle is the midpoint of the diameter PQ}$

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

$\implies\mathsf{\dfrac{x_1+1}{2}=2\;\;\&\;\;\dfrac{y_1+4}{2}=-3}$

$\implies\mathsf{x_1+1=4\;\;\&\;\;y_1+4=-6}$

$\implies\mathsf{x_1=4-1\;\;\&\;\;y_1=-6-4}$

$\implies\mathsf{x_1=3\;\;\&\;\;y_1=-10}$

$\therefore\boxed{\textsf{The coordinates of P is(3,-10)}}$

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Answer 2

Answer:

Given:

\textsf{One end of the diameter PQ of the circle is Q(1,4)}One end of the diameter PQ of the circle is Q(1,4)

\textsf{and centre is (2,-3)}and centre is (2,-3)

\underline{\textsf{To find:}}

To find:

\textsf{The co ordinates of P}The co ordinates of P

\underline{\textsf{Solution:}}

Solution:

\underline{\textsf{Concept used:}}

Concept used:

\begin{gathered}\boxed{\begin{minipage}{6cm}$\textsf{The midpoint of the line joining}\\\\\mathsf{(x_1,y_1)\;\;\&\;\;(x_2,y_2)}\;\textsf{is}\\\\\mathsf{(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})}$\end{minipage}}\end{gathered}

\textsf{Let the coordinates of P be}\;\mathsf{(x_1,y_1)}Let the coordinates of P be(x

1

,y

1

)

\textsf{We know that,}We know that,

\textsf{Centre of the circle is the midpoint of the diameter PQ}Centre of the circle is the midpoint of the diameter PQ

\implies\mathsf{(\dfrac{x_1+1}{2},\dfrac{y_1+4}{2})=(2,-3)}⟹(

2

x

1

+1

,

2

y

1

+4

)=(2,−3)

\implies\mathsf{\dfrac{x_1+1}{2}=2\;\;\&\;\;\dfrac{y_1+4}{2}=-3}⟹

2

x

1

+1

=2&

2

y

1

+4

=−3

\implies\mathsf{x_1+1=4\;\;\&\;\;y_1+4=-6}⟹x

1

+1=4&y

1

+4=−6

\implies\mathsf{x_1=4-1\;\;\&\;\;y_1=-6-4}⟹x

1

=4−1&y

1

=−6−4

\implies\mathsf{x_1=3\;\;\&\;\;y_1=-10}⟹x

1

=3&y

1

=−10

\therefore\boxed{\textsf{The coordinates of P is(3,-10)}}∴

The coordinates of P is(3,-10)

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