Question

x coordinates of the point which divide the line joining the points (1,3)and (6,-3) externally in the ratio 3:2 is ​

Answer 1

☆ Given Question:-

The x - coordinates of the point which divides the line segment joining the points (1, 3)and (6, -3) externally in the ratio 3:2 is

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☆Required Answer:-

$\bf \:x \: - coordinate \: = 16$

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☆Explanation:-

☆Given :-

• The line segment joining the points (1,3)and (6,-3) divides by a coordinate externally in the ratio 3:2.

☆To Find :-

• x - coordinate of the point which divides the line segment joining the points (1, 3)and (6, -3) externally in the ratio 3:2.

☆Formula used :-

Section Formula :-

• Let us consider a line segment joining the points A and B and let P(x, y) divides the line segment AB in the ratio m : n externally, then coordinates of P is given by

$\bf \:( x, y) = (\dfrac{mx_2 - nx_1}{m - n} , \dfrac{my_2 - ny_1}{m - n} )$

where A and B coordinates are

$\bf \:(x_1,y_1) \: and \: (x_2,y_2)$

$\bf \: \mathcal\colorbox{blue}{{\color{white}{AŋʂᏯɛཞ࿐}}}$

Let P(x, y) be the required coordinates which line segment joining the points (1, 3)and (6, -3) externally in the ratio 3:2.

☆Using Section Formula

$\bf \:( x, y) = (\dfrac{mx_2 - nx_1}{m - n} , \dfrac{my_2 - ny_1}{m - n} )$

On substituting the values of

$\bf \: x_1 =1 ,y_1 =3 ,x_2= 6,y_2= - 3, m = 3, n = 2$

we get,

$\bf \:( x, y) = (\dfrac{3 \times 6 - 2 \times 1}{3 - 2} , \dfrac{ - 3 \times 3 - 2 \times 3}{3 - 2} )$

$\bf\implies \:(x, y) = (16, - 14)$

$\large{\boxed{\boxed{\bf{Hence, \: x - coordinate = 16}}}}$

Answer 2

Answer:

Step-by-step explanation: