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Sia ? 6 years, 3 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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Sia ? 6 years, 4 months ago
The standard algebraic identities are:
- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 – 2ab + b2
- a2 – b2 = (a + b)(a – b)
- (x + a)(x + b) = x2 + (a + b) x + ab
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
- (a + b)3 = a3 + b3 + 3ab (a + b)
- (a – b)3 = a3 – b3 – 3ab (a – b)
- a3 + b3 + c3– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
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Yogita Ingle 6 years, 4 months ago
It is rational because decimal expansion is terminating. Therefore, it can be expressed in p/q form where factors of q are of the form 2n.5m where n and m are non-negative integers.
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Rani Mishra ??? 6 years, 4 months ago
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