# Points P, Q, R, S in that order which divide the line segment joining points A(2,5) and B(5,-5) in 5 equal parts. Find the coordinates of P, Q, R and S.

Consider the figure, Here, use the section formula which states that the coordinates of the point R which divides the line segment joining the two points P(x1,y1) and Q(x2,y2) internally in the ratio m:n is given by $\left(\frac{m{x}_{2}+n{x}_{1}}{m+n},\frac{m{y}_{2}+n{y}_{1}}{m+n}\right)$

Now, the point P divides the line segment AB in the ratio 1:4. So,
$m=1\phantom{\rule{0ex}{0ex}}n=4\phantom{\rule{0ex}{0ex}}{x}_{1}=2\phantom{\rule{0ex}{0ex}}{y}_{1}=5\phantom{\rule{0ex}{0ex}}{x}_{2}=5\phantom{\rule{0ex}{0ex}}{y}_{2}=-5$

Let the coordinates of the point P be (x,y). So, substitute the above values in the section formula.
$x=\frac{m{x}_{2}+n{x}_{1}}{m+n}\phantom{\rule{0ex}{0ex}}x=\frac{\left(1\right)\left(5\right)+\left(4\right)\left(2\right)}{1+4}\phantom{\rule{0ex}{0ex}}x=\frac{5+8}{5}\phantom{\rule{0ex}{0ex}}x=\frac{13}{5}$

Also,
$y=\frac{m{y}_{2}+n{y}_{1}}{m+n}\phantom{\rule{0ex}{0ex}}y=\frac{\left(1\right)\left(-5\right)+\left(4\right)\left(5\right)}{1+4}\phantom{\rule{0ex}{0ex}}y=\frac{-5+20}{5}\phantom{\rule{0ex}{0ex}}y=\frac{15}{5}\phantom{\rule{0ex}{0ex}}y=3$

Thus, the coordinates of point P is $\left(\frac{\mathbf{13}}{\mathbf{5}}\mathbf{,}\mathbf{3}\right)$
Now, as the line segment is divided into 5 equal parts, so R(x3,y3) is the mid point of PB. So, using the mid point formula we get
${x}_{3}=\frac{x+{x}_{2}}{2}\phantom{\rule{0ex}{0ex}}{x}_{3}=\frac{\frac{13}{5}+5}{2}\phantom{\rule{0ex}{0ex}}{x}_{3}=\frac{\frac{13+25}{5}}{2}\phantom{\rule{0ex}{0ex}}{x}_{3}=\frac{38}{10}\phantom{\rule{0ex}{0ex}}{x}_{3}=\frac{19}{5}$

Also,
${y}_{3}=\frac{y+{y}_{2}}{2}\phantom{\rule{0ex}{0ex}}{y}_{3}=\frac{3-5}{2}\phantom{\rule{0ex}{0ex}}{y}_{3}=\frac{-2}{2}\phantom{\rule{0ex}{0ex}}{y}_{3}=-1$

Thus, the coordinates of point R is $\left(\frac{\mathbf{19}}{\mathbf{5}}\mathbf{,}\mathbf{-}\mathbf{1}\right)$
Similarly, S(x4,y4) is the mid point of RB. So,
${x}_{4}=\frac{{x}_{3}+{x}_{2}}{2}\phantom{\rule{0ex}{0ex}}{x}_{4}=\frac{\frac{19}{5}+5}{2}\phantom{\rule{0ex}{0ex}}{x}_{4}=\frac{\frac{13+25}{5}}{2}\phantom{\rule{0ex}{0ex}}{x}_{4}=\frac{44}{10}\phantom{\rule{0ex}{0ex}}{x}_{4}=\frac{22}{5}$

Also,
${y}_{4}=\frac{{y}_{3}+{y}_{2}}{2}\phantom{\rule{0ex}{0ex}}{y}_{4}=\frac{-1-5}{2}\phantom{\rule{0ex}{0ex}}{y}_{4}=\frac{-6}{2}\phantom{\rule{0ex}{0ex}}{y}_{4}=-3$

Thus, the coordinates of point S is $\left(\frac{\mathbf{22}}{\mathbf{5}}\mathbf{,}\mathbf{-}\mathbf{3}\right)$
Similarly, Q(x5,y5) is the mid point of PR. So,
${x}_{5}=\frac{x+{x}_{3}}{2}\phantom{\rule{0ex}{0ex}}{x}_{5}=\frac{\frac{13}{5}+\frac{19}{5}}{2}\phantom{\rule{0ex}{0ex}}{x}_{5}=\frac{\frac{13+19}{5}}{2}\phantom{\rule{0ex}{0ex}}{x}_{5}=\frac{32}{10}\phantom{\rule{0ex}{0ex}}{x}_{5}=\frac{16}{5}$

Also,
${y}_{5}=\frac{y+{y}_{3}}{2}\phantom{\rule{0ex}{0ex}}{y}_{5}=\frac{3-1}{2}\phantom{\rule{0ex}{0ex}}{y}_{5}=\frac{2}{2}\phantom{\rule{0ex}{0ex}}{y}_{5}=1$

Thus, the coordinates of point Q is $\left(\frac{\mathbf{16}}{\mathbf{5}}\mathbf{,}\mathbf{1}\right)$

• 20
What are you looking for?