2.15: If the points A(6, 1), B(8.2), C(9.4) and D(p. 3) are the vertices of a parallelogram, taken in order, find the value of p​

Question

2.15: If the points A(6, 1), B(8.2), C(9.4) and D(p. 3) are the vertices of a parallelogram, taken in order,
find the value of p​

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Lydia 5 days 2021-09-14T18:13:40+00:00 1 Answer 0 views 0

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    2021-09-14T18:15:29+00:00

    Answer:

    \large{\pink{ \textbf{P = 7}}}

    To Find :

    • value of p = ?

    DIAGRAM :

    \setlength{\unitlength}{1.3cm}\begin{picture}(8,2)\thicklines\put(8,3){\large{A (6,1)}}\put(7.5,0.7){\large{D (p,3)}}\put(11.1,0.9){\large{C (9,4)}}\put(9.9,2.1){\large{O}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{B (8,2)}}\end{picture}

    Step-by-step explanation:

    O is the mid point of AC and BD {Diagonal of parallelogram bisect each other}

    1) Finding mid point of AC

    x coordinate of O = \sf \dfrac{x_1 + x_2}{2}

    Where, \sf x_1=6 and \sf x_2 = 9

    \leadsto\:[tex]\sf \dfrac{6 + 9}{2}[/tex]

    \leadsto\:[tex]\sf \dfrac{15}{2}—-(1)[/tex]

    2) Finding mid point of BD

    x coordinate of O = \sf \dfrac{x_1 + x_2}{2}

    Where, \sf x_1=8 and \sf x_2=p

    \leadsto\:[tex]\sf \dfrac{8+p}{2}—-(2)[/tex]

    Comparing equation (1) and (2), we get :

    \leadsto\:[tex]\sf \dfrac{15}{2} = \dfrac{8+p}{2}[/tex]

    \leadsto\:[tex]\sf 15 = 8+p[/tex]

    \leadsto\:[tex]\sf p = 15 – 8[/tex]

    \leadsto\:[tex]\sf p = 7[/tex]

    Therefore, value of p is 7.

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18:9+8+9*3-7:3-1*13 = ? ( )