(1,2) (4,y)(x,6)(3,5) are vertices of parallelogram …
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Sia ? 6 years, 6 months ago
Let A → (1, 2), B → (4, y), C→ (x, 6) and D→ (3, 5).
We know that the diagonals of parallelogram bisect each other.
So, Coordinates of the mid-point of diagonal AC
= Coordinates of the mid-point of diagonal BD
{tex}\Rightarrow \left( {\frac{{1 + x}}{2},\frac{{2 + 6}}{2}} \right) = \left( {\frac{{4 + 3}}{2},\frac{{y + 5}}{2}} \right){/tex}
{tex}\Rightarrow \left( {\frac{{1 + x}}{2},4} \right) = \left( {\frac{7}{2},\frac{{y + 5}}{2}} \right){/tex}
{tex}\Rightarrow \frac{{1 + x}}{2} = \frac{7}{2}{/tex}
{tex}\Rightarrow{/tex} 1 + x = 7
{tex}\Rightarrow{/tex} x = 6
and {tex}4 = \frac{{y + 5}}{2}{/tex}
{tex}\Rightarrow{/tex} y + 5 = 8
{tex}\Rightarrow{/tex} y = 3
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