Let A be #(−3,5)# and B be #(5,−10))#. Find:
(1) the length of segment #bar(AB)#
(2) the midpoint #P# of #bar(AB)#
(3) the point #Q# which splits #bar(AB)# in the ratio #2:5#?

Question

Wyatt Anderson · Accepted Answer

(1) the length of the segment #bar(AB)# is #17#
(2) Midpoint of #bar(AB)# is #(1,-7 1/2)#
(3) The coordinates of the point #Q# which splits #bar(AB)# in the ratio #2:5# are #(-5/7,5/7)# If we have two points #A(x_1,y_1)# and #B(x_2,y_2)#, length of #bar(AB)# i.e. distance between them is given by #sqrt((x_2-x_1)^2+(x_2-x_1)^2)# and coordinates of the point #P# that divides the segment #bar(AB)# joining these two points in the ratio #l:m# are #((lx_2+mx_1)/(l+m),(lx_2+mx_1)/(l+m))# and as midpoint divided segment in ratio #1:1#, its coordinated would be #((x_2+x_1)/2,(x_2+x_1)/2)# As we have #A(-3,5)# and #B(5,-10)# (1) the length of the segment #bar(AB)# is #sqrt((5-(-3))^2+((-10)-5)^2)# = #sqrt(8^2+(-15)^2)=sqrt(65+225)=sqrt289=17# (2) Midpoint of #bar(AB)# is #((5-3)/2,(-10-5)/2)# or #(1,-7 1/2)# (3) The coordinates of the point #Q# which splits #bar(AB)# in the ratio #2:5# are #((2xx5+5xx(-3))/7,(2xx(-10)+5xx5)/7)# or #((10-15)/7,(-20+25)/7)# i.e. #(-5/7,5/7)#

Theodore Abad · Answer

undefined 1) The length of segment $ \overline{AB} $ can be found using the distance formula:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Given $ A(-3, 5) $ and $ B(5, -10) $, the length of $ \overline{AB} $ is:

$$ d = \sqrt{(5 - (-3))^2 + ((-10) - 5)^2} $$
$$ d = \sqrt{8^2 + (-15)^2} $$
$$ d = \sqrt{64 + 225} $$
$$ d = \sqrt{289} $$
$$ d = 17 $$

So, the length of segment $ \overline{AB} $ is $ 17 $.

2) The midpoint $ P $ of segment $ \overline{AB} $ can be found by averaging the coordinates of $ A $ and $ B $:

$$ P\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) $$

For $ A(-3, 5) $ and $ B(5, -10) $, the midpoint $ P $ is:

$$ P\left(\frac{-3 + 5}{2}, \frac{5 + (-10)}{2}\right) $$
$$ P\left(\frac{2}{2}, \frac{-5}{2}\right) $$
$$ P(1, -\frac{5}{2}) $$

So, the midpoint $ P $ of segment $ \overline{AB} $ is $ (1, -\frac{5}{2}) $.

3) To find the point $ Q $ which splits $ \overline{AB} $ in the ratio $ 2:5 $, we use the section formula:

$$ Q\left(\frac{2x_2 + 5x_1}{2 + 5}, \frac{2y_2 + 5y_1}{2 + 5}\right) $$

For $ A(-3, 5) $ and $ B(5, -10) $, $ Q $ is:

$$ Q\left(\frac{2(5) + 5(-3)}{2 + 5}, \frac{2(-10) + 5(5)}{2 + 5}\right) $$
$$ Q\left(\frac{10 - 15}{7}, \frac{-20 + 25}{7}\right) $$
$$ Q\left(\frac{-5}{7}, \frac{5}{7}\right) $$

So, the point $ Q $ which splits $ \overline{AB} $ in the ratio $ 2:5 $ is $ \left(\frac{-5}{7}, \frac{5}{7}\right) $.

Let A be #(−3,5)# and B be #(5,−10))#. Find: (1) the length of segment #bar(AB)# (2) the midpoint #P# of #bar(AB)# (3) the point #Q# which splits #bar(AB)# in the ratio #2:5#?