Question

Given points A (-3,-4) and B (2,0), points P (-1,-12/5) partitions line AB in the ratio

Answer 1

`\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ &A&(~ -3 &,& -4~) &P&(~ -1 &,& -(12)/(5)~) \end{array} \\\\\\ d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ AP=\sqrt{[-1-(-3)]^2+[-(12)/(5)-(-4)]^2} \\\\\\ AP=\sqrt{(-1+3)^2+(-(12)/(5)+4)^2}\implies AP=\sqrt{2^2+\left( (8)/(5) \right)^2} \\\\\\ AP=\sqrt{4+(64)/(25)}\implies AP=\sqrt{\cfrac{164}{25}}\implies AP=\cfrac{√(164)}{√(25)} \\\\\\ \boxed{AP=\cfrac{2√(41)}{5}}`




`\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ &P&(~ -1 &,& -(12)/(5)~) &B&(~ 2 &,& 0~) \end{array} \\\\\\ PB=\sqrt{[2-(-1)]^2+[0-\left(-(12)/(5) \right)]^2} \\\\\\ PB=\sqrt{(2+1)^2+\left( (12)/(5) \right)^2}\implies PB=\sqrt{9+(144)/(25)}\implies PB=\sqrt{\cfrac{369}{25}} \\\\\\ PB=\cfrac{√(369)}{√(25)}\implies \boxed{PB=\cfrac{3√(41)}{5}}`

so, let's check the ratio of AP:PB


`\bf AP:PB\implies \cfrac{AP}{PB}\implies \cfrac{(2√(41))/(5)}{(3√(41))/(5)}\implies \cfrac{2\underline{√(41)}}{\underline{5}}\cdot \cfrac{\underline{5}}{3\underline{√(41)}}\implies \cfrac{2}{3}`