Q.Find equation of the line passing through the point $(2, 2)$ and cutting off intercepts on the axes whose sum is $9$.

The key idea here is understanding what the $yz$-plane actually is. In 3D coordinate geometry, the $yz$-plane is the plane where $x = 0$. Every point on this plane has an $x$-coordinate of zero. So the distance from any point to the $yz$-plane is just how far it is from that plane along the $x$-axis — measured perpendicularly.

Think of it like this: the $yz$-plane is a flat vertical wall that passes through the origin and contains the $y$ and $z$ axes. Your point $P$ is somewhere in space. The shortest distance from $P$ to that wall is simply the horizontal distance along the $x$-direction. That distance is the absolute value of the $x$-coordinate of $P$, because the $x$-coordinate tells you exactly how far left or right the point is from the $yz$-plane.

Now let’s apply this to the given point.

1. Identify the $x$-coordinate of $P(-3, 4, 5)$. It is $-3$.

2. Distance is always positive — it’s a length, not a signed quantity. So we take the absolute value: $| -3 | = 3$. …

Concept

An intercept is where a line meets an axis. For $Ax + By + C = 0$ the $x$-intercept is $-C/A$ and the $y$-intercept is $-C/B$. "Equal in length but opposite in sign" means the new line's intercepts are the negatives of the reference line's intercepts, giving one equation per axis.

Solution

1. Intercepts of $2x - 3y + 6 = 0$:
   • $y = 0 \Rightarrow 2x + 6 = 0 \Rightarrow x = -3$.
   • $x = 0 \Rightarrow -3y + 6 = 0 \Rightarrow y = 2$. So they are $(-3, 2)$.
2. Required intercepts: negatives of these, i.e. $x$-intercept $= 3$, $y$-intercept $= -2$.
3. Intercepts of $ax - by + 8 = 0$:
   • $y = 0 \Rightarrow x = -\dfrac{8}{a}$.
   • $x = 0 \Rightarrow y = \dfrac{8}{b}$. …

Step 1 — The concept.

If a line makes angles $\alpha,\beta,\gamma$ with the $X$, $Y$, $Z$ axes, its direction cosines are $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$, and because $(l,m,n)$ is a unit vector along the line,

Step 2 — Substitute the given angles.

Step 3 — Solve for $\gamma$. …

Step 1 — Why signs alone decide the octant

The three coordinate planes ($XY$, $YZ$, $ZX$) cut space into $2\times 2\times 2 = 8$ regions called octants. Which region a point lies in depends only on whether each coordinate is positive or negative — the magnitudes are irrelevant. The conventional numbering is:

Note the pattern: octants I–IV are the four quadrants of the $XY$-plane lifted above it ($z>0$), and octants V–VIII are the same four quadrants pushed below it ($z<0$).

Step 2 — Read the signs of the given point

For $(2,\,-4,\,-7)$: …

Step 1 — The concept

The three coordinate planes ($XY$, $YZ$, $ZX$) cut space into eight regions called octants. Each octant is fixed by the sign triple $(\text{sign}\,x,\ \text{sign}\,y,\ \text{sign}\,z)$. The first four octants are the ones above the $XY$-plane ($z > 0$), numbered anticlockwise as seen from the $+z$ axis; octants V–VIII are their mirror images below it ($z < 0$).

Step 2 — The standard table (NCERT convention)

Notice the useful shortcut: octants I–IV all have $z > 0$; octants V–VIII all have $z < 0$.

Step 3 — Read the signs of the given point

For $(1,\ -3,\ 4)$:

Sign triple $= (+,\,-,\,+)$.

Step 4 — Match …