Question

$\Large{\underline{\underline{\bf{Question:-}}}}$
❐ The line segment joining A(6, 3) to B(–1, –4) is doubled in length by having half its length added to each end. Find the coordinates of the new ends.

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Answer 1

Answer in the attachment..

Answer 2

$\huge{\bf{\underline{\red{Answer :}}}}$

Let P and Q be the required new ends.

Coordinates of P

Let AP = K

∴ AB = 2AP = 2k

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and PB = AP + AB = k + 2k = 3k

AP/PB = k/3k = 1/3

∴ P divides AB externally in the ratio 1:3

Using the section formula for the external division, i.e.

$\large(x,y) = ( \frac{m_{1}x_{2} - m_{2}x_{1}}{m_{1} - m_{2}} , \frac{m_{1}y_{2} - m_{2}y_{1}}{m_{1} - m_{2}} )$

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Here $m_{1}= 3$

$(x_{1}, y_{1} = (6,3) \: and \: (x_{2}, y_{2}) = (-1,4)$

Putting the above values in the above formula, we get,

$\large x = \frac{1( - 1) - 3(6)}{1 - 3} ,y = \frac{1( - 4) - (3)}{1 - 3}$

$\large x = \frac{ - 1 - 18}{ - 2} ,y = \frac{ - 4 - 9}{ - 2}$

$\large x = \frac{ - 19}{ - 2} ,y = \frac{ - 13}{ - 2}$

$\boxed{\red{x = \frac{19}{2} ,y = \frac{13}{2} }}$

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Coordinates of P are $\sf{\dfrac{19}{2}, \dfrac{13}{2}}$

$\boxed{\pink{Coordinates\:of\:Q }}$

Q divides AB externally in the ratio 3:1

Again, Using the section formula for the external division, i.e.

$\large (x,y) = \large(\frac{m_{1}x_{2} - m_{2}x_{1}}{m_{1} - m_{2}} , \frac{m_{1}y_{2} - m_{2}y_{1}}{m_{1} - m_{2}} )$

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Here, $m_{1} = 3, m_{2} = 1$

$\sf (x_{1}, y_{1}) = (6, 3) and (x_{2}, y_{2}) = (-1, -4)$

Putting the above values in the above formula, we get

$\large x = \frac{3( - 1) - 1(6)}{3 -1} ,y = \frac{3( - 4) - (3)}{3 - 1}$

$\large x = \frac{ - 3 - 6}{2} ,y = \frac{ - 12 - 3}{2}$

$\large x = \frac{ - 9}{2} ,y = \frac{ - 15}{2}$

Coordinates of Q are $\frac{ - 9}{2} ,\frac{ -15}{2}$