A line is given by equation $$x-5=5-y, z=5$$ Is the direction ratio of the line 1,1,5?
As per my reasoning all points on the line are its direction ratio, so a,b,5 will the direction ratio provided a and b lies on the line (5 is constant and will always lie on the line as per the equation). However putting x=1 and y=1 the first equation does not satisfy i.e. is $$1-5=5-1 \implies -4=4$$ in the canonical form of line's equation i.e. $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$ we know that the direction ratios are a,b,c but I am not able to reason for the above case. Need help in understanding it both from analytical and geometrical point of view if possible.
All the points on a line are its direction ratio only when the line goes through center.
In 2d a line is written as $y = mx+c$ , so, $(y_2-y_1)=m(x_2-x_1)$. See here the points are not direction ratios unless $c=0$. It is true for higher dimension also. In 3d we get $(z_2-z_1)=m_2(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2})$, where $m_2$ points the angle between the line and the xy plane (as m points the angle between the line and x axis earlier). You can draw the picture and convince yourself also furthur you can derive the direction ratios from this in 3d (by putting the value of $y_2-y_1$ what we got earlier).
In your case $x+y=10$ , so what is m here, also when $z_2-z_1=0$ what is the direction ratio for z. I mean what you need to multiply with $x_2-x_1$ to get $z_2-z_1$ here.
Let $(y_2-y_1)=m(x_2-x_1)$ and we claim that all points are its direction ratio. So then $y_2=mx_2$ and also $y_1=mx_1$ for any/all $(x_2,y_2)$ and $(x_1,y_1)$ that indicates $y=mx$ and $c = 0$, conversely if $c \ne 0$, $y\ne mx$ as $y= mx+c$.