Question

please answer this question ​

Answer 1

Given:-

• Three Vertices of Paralleogram ABCD are A(1,-2), B(3,6) and C(5,10)

Find:-

• Its fourth vertex.

Solution:-

As we know that diagonal of a paralleogram are bisectors of each other. So, let P be the mid-point of AC and BD.

Let, D = (x,y)

Now, by using

$\underline{\boxed{\sf \text{\sf Mid-point} = \bigg\lgroup \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup}}$

FOR BD

$\begin{cases}\sf x_1 = 3\\ \sf x_2 = x\\ \sf y_1 = 6\\ \sf y_2 = y\end{cases}$

Substituting these values:-

$\dashrightarrow\sf BD = \bigg\lgroup \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup$

$\dashrightarrow\sf BD = \bigg\lgroup \dfrac{3+x}{2},\dfrac{6+y}{2}\bigg\rgroup$

Now, for AC

$\begin{cases}\sf x_1 = 1\\ \sf x_2 = 5\\ \sf y_1 = 10\\ \sf y_2 = - 2\end{cases}$

Substituting these values:-

$\dashrightarrow\sf AC = \bigg\lgroup \dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg\rgroup$

$\dashrightarrow\sf AC = \bigg\lgroup \dfrac{1+5}{2},\dfrac{10+( - 2)}{2}\bigg\rgroup$

$\dashrightarrow\sf AC = \bigg\lgroup \dfrac{1+5}{2},\dfrac{10 - 2}{2}\bigg\rgroup$

$\dashrightarrow\sf AC = \bigg\lgroup \dfrac{6}{2},\dfrac{8}{2}\bigg\rgroup$

$\dashrightarrow\sf AC = \big\lgroup 3,4\big\rgroup$

Now,

Mid-point of AC = Mid-point of BD

$\sf where\begin{cases}\sf AC = \big\lgroup 3,4\big\rgroup\\ \sf BD = \bigg\lgroup \dfrac{3+x}{2},\dfrac{6+y}{2}\bigg\rgroup\end{cases}$

Substituting these values:-

$\dashrightarrow\sf \big\lgroup 3,4\big\rgroup = \sf \bigg\lgroup \dfrac{3+x}{2},\dfrac{6+y}{2}\bigg\rgroup$

$\begin{gathered} \sf 3 = \dfrac{3 + x}{2} \end{gathered} \qquad ,\quad \begin{gathered} \sf 4 = \dfrac{6 + y}{2} \end{gathered}$

$\begin{gathered} \sf 3 \times 2 = 3 + x \end{gathered} \qquad ,\quad \begin{gathered} \sf 4 \times 2 = 6 + y \end{gathered}$

$\begin{gathered} \sf 6= 3 + x \end{gathered} \qquad ,\quad \begin{gathered} \sf 8= 6 + y \end{gathered}$

$\begin{gathered} \sf 6 - 3=x \end{gathered} \qquad ,\quad \begin{gathered} \sf 8 - 6=y \end{gathered}$

$\begin{gathered} \sf 3=x \end{gathered} \qquad ,\quad \begin{gathered} \sf 2=y \end{gathered}$

$\begin{gathered}\sf x = 3 \end{gathered},\begin{gathered}\sf y = 2 \end{gathered}$

So, vertex of D(x,y) = (3,2)

$\underline{\boxed{\sf \therefore 4^{th} \: vertex \: is\:(3,2)}}$