Math, asked by titasrun, 1 year ago

Pls solve 7,8,9 fast

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Answered by Anonymous

Answer:

7) 72

8) ( 2, -5/3 ) and ( 0, -7/3 )

9) ( -2, 0 )

Hope this helps you.

Step-by-step explanation:

$\displaystyle\mbox{Area} = \tfrac12\left(\left|\begin{array}{cc}-5&7\\-4&-5\end{array}\right|+\left|\begin{array}{cc}-4&-5\\-1&-6\end{array}\right|+\left|\begin{array}{cc}-1&-6\\4&5\end{array}\right|+\left|\begin{array}{cc}4&5\\-5&7\end{array}\right|\right)\\ \\=\tfrac12\left(25+28+24-5-5+24+28+25\right)\\ \\= 25 + 28 + 24 - 5 = 72$

8) A = (4,-1) and B = (-2,-3)

Points of trisection are at:

(2A+B)/3 = ( (8-2)/3, (-2-3)/3 ) = ( 2, -5/3 )
(A+2B)/3 = ( (4-4)/3, (-1-6)/3 ) = ( 0, -7/3 )

9) Let the point be (x,0).

Then equidistant from (-2,5) and (2,-3)

=> (x+2)² + (0-5)² = (x-2)² + (0+3)²

=> x² + 4x + 4 + 25 = x² - 4x + 4 + 9

=> 8x = 9 - 25 = -16

=> x = -2

So the point is ( -2, 0 ).

titasrun: 55Hanks Bro it's very helpful

titasrun: Which formula did u took in 9 th sum

Anonymous: You're welcome. Glad to have helped.

Anonymous: In q9, used the distance formula (which is really just Pythagoras' Theorem). e.g. square of distance from (x,0) to (-2,5) = (diff in x coords)^2 + (diff in y coords)^2 = (x+2)^2+(0-5)^2.

titasrun: Okkk thanks brio

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