Question

derive an expression for the co-ordinate of the point that divides the line joining A(x1 y1 z1) b(x2 y2 z2) internally in the ratio m:n​

Answer 1

The expression for the coordinate of the point that divides the line joining A and B in the ratio m:n is

$N(x,y,z)=(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})$

Step-by-step explanation:

To derive an expression for the coordinate of the point that divides the line joining A and B in the ratio m:n

• Let $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ be the given points
• Let R(x, y, z) divide PQ internally in the ratio m : n
• Draw the line segments AP, BQ, CR perpendicular to xy-plane.
•  ∴ AP ∥ BQ ∥ CR

• ∴ AP, BQ, CR lines are lie in one plane.
•  So the points P,Q,and R lie in a straight line
• And the points intersects the plane and xy plane.  
• Through the point R draw a parallel line AB to the line segment PQ. The line AB intersects the line segment LP externally at the point A and the line segment MQ at the point B.

From the figure we have that the triangles ALN and RBQ are similar triangles

So we can write $\frac{AL}{MB}=\frac{LN}{NB}=\frac{m}{n}$

$\frac{AP-PL}{BQ-MQ}=\frac{m}{n}$

$\frac{NR-LP}{QB-NR}=\frac{m}{n}$

$\frac{m}{n}=\frac{z-z_1}{z_2-z}=\frac{m}{n}$

$n(z-z_1)=m(z_2-z)$

$nz-nz_1=mz_2-mz$

$nz+mz=nz_1+mz_2$

$z(n+m)=nz_1+mz_2$

Rewritting the above equation

$z(m+n)=mz_2+nz_1$

$z=\frac{mz_2+nz_1}{m+n}$

Similarly we can find x and y

Therefore $x=\frac{mx_2+nx_1}{m+n}$ and $x=\frac{my_2+ny_1}{m+n}$

Therefore the expression for the coordinate of the point that divides the line joining A and B in the ratio m:n is

$N(x,y,z)=(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n})$