A(-1,-1),B(6,1)C(8,8)D(x,y) are the four vertices of a rhombus taken in order. find co-ordinates of point D.
Answers
Therefore.,
Therefore.,
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Answer:
Firstmethod:
\begin{gathered}Given \: A(-1,-1) = (x_{1},y_{1}) , \\B(6,1) = (x_{2},y_{2}),\:C(8,8) = (x_{3},y_{3}) \\and \:D(x,y) \:are \: vertices \:of \:a \: Rhombus.\end{gathered}
GivenA(−1,−1)=(x
1
,y
1
),
B(6,1)=(x
2
,y
2
),C(8,8)=(x
3
,y
3
)
andD(x,y)areverticesofaRhombus.
\begin{gathered}x = x_{1} + x_{3} - x_{2} \\= -1 + 8 - 6 \\= -7 + 8 \\= 1\end{gathered}
x=x
1
+x
3
−x
2
=−1+8−6
=−7+8
=1
\begin{gathered}y = y_{1} + y_{3} - y_{2} \\= -1 + 8 - 1 \\= -2 + 8 \\= 6\end{gathered}
y=y
1
+y
3
−y
2
=−1+8−1
=−2+8
=6
Therefore.,
\begin{gathered}\red{ Co-ordinates \:of \:point \:D(x,y) }\\\green {= (1,6) }\end{gathered}
Co−ordinatesofpointD(x,y)
=(1,6)
\underline { \pink { Second \: method : }}
Secondmethod:
\begin{gathered}In \: a \:Rhombus , \\Mid-point \: AC \\= Mid-point \:of \: diagonal \:BD\end{gathered}
InaRhombus,
Mid−pointAC
=Mid−pointofdiagonalBD
\begin{gathered}\implies \Big( \frac{x+x_{2}}{2} , \frac{y_+y_{2}}{2}\Big) \\= \Big( \frac{x_{1}+x_{3}}{2} , \frac{y_{1}+y_{3}}{2}\Big)\end{gathered}
⟹(
2
x+x
2
,
2
y
+
y
2
)
=(
2
x
1
+x
3
,
2
y
1
+y
3
)
\implies \Big( \frac{x+6}{2}, \frac{y+1}{2}\Big) = \Big( \frac{-1+8}{2}, \frac{-1+8}{2}\Big)⟹(
2
x+6
,
2
y+1
)=(
2
−1+8
,
2
−1+8
)
\implies \Big( \frac{x+6}{2}, \frac{y+1}{2}\Big) = \Big( \frac{7}{2}, \frac{7}{2}\Big)⟹(
2
x+6
,
2
y+1
)=(
2
7
,
2
7
)
\begin{gathered}Now, i ) \frac{x+6}{2} = \frac{7}{2} \\\implies x + 6 = 7 \\\implies x = 7-6 \\\implies x = 1\end{gathered}
Now,i)
2
x+6
=
2
7
⟹x+6=7
⟹x=7−6
⟹x=1
\begin{gathered}Now, ii) \frac{y+1}{2} = \frac{7}{2} \\\implies y+ 1 = 7 \\\implies y = 7-1 \\\implies y = 6\end{gathered}
Now,ii)
2
y+1
=
2
7
⟹y+1=7
⟹y=7−1
⟹y=6
Therefore.,
\begin{gathered}\red{ Co-ordinates \:of \:point \:D(x,y) }\\\green {= (1,6) }\end{gathered}
Co−ordinatesofpointD(x,y)
=(1,6)