logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Find the coordinates of the point which divides the line segment joining the points $(-2,3,5)$ and $(1,-4,6)$ in the ratio $2:3$ internally.

$\begin{array}{1 1}\big(-\large\frac{4}{5},\frac{1}{5},\frac{27}{5}\big) \\(-8,17,3) \\ \big(\large\frac{8}{5},\frac{17}{5},\frac{27}{5}\big) \\ \big(-\large\frac{8}{5},\frac{1}{5},\frac{27}{5}\big)\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Section formula: The coordinates of the point $C$ that divides the segment joining the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $l:m$ internally is given by $C\big(\large\frac{lx_2+mx_1}{l+m},\frac{ly_2+my_1}{l+m},\frac{lz_2+mz_1}{l+m}\big)$
Given two points are $A(-2,3,5)$ and $B(1,-4,6)$
We know that from section formula the coordinates of the point $C$ that divides
the segment joining the points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ in the ratio $l:m$ internally is given by
$C\big(\large\frac{lx_2+mx_1}{l+m},\frac{ly_2+my_1}{l+m},\frac{lz_2+mz_1}{l+m}\big)$
$\therefore $ the coordinates of the point $C$ which divides $AB$ in the ratio $2:3$ internally is
$C\big(\large\frac{2.1+3.(-2)}{2+3},\frac{2.(-4)+3.3}{2+3},\frac{2.6+3.5}{2+3}\big)$
$=C\big(-\large\frac{4}{5},\frac{1}{5},\frac{27}{5}\big)$
answered May 5, 2014 by rvidyagovindarajan_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...