Q.A line is such that its segment between the lines $5x - y + 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1, 5)$. Obtain its equation.

Step 1 — Coordinates of $P$.

Step 2 — Set up the intercepts. Let the line meet the $x$-axis at $A(a,0)$ and the $y$-axis at $B(0,b)$. "Starting from the $x$-axis" means $P$ divides $AB$ with $AP:PB=1:2$.

Step 3 — Section formula (ratio $m:n=1:2$ measured from $A$):

Setting these equal to $(4,3)$:

$\frac{2a}{3}=4\ \Rightarrow\ a=6,\qquad \frac{b}{3}=3\ \Rightarrow\ b=9.$ …

Step 1 — Locate the dividing point

By the section formula, the point $P$ dividing the segment from $A(-1,2)$ to $B(3,6)$ in ratio $1:3$ internally is

Step 2 — Find the line $L_2$ through $(3,-1)$ and $(0,3)$

Slope: …

Step 1 — Standard form

Divide $x^2 + 3y^2 = 12$ throughout by $12$:

Comparing with $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$:

Step 2 — Which is the major axis?

Since $a^2 = 12 > b^2 = 4$, the major axis lies along the $x$-axis (semi-major $a = 2\sqrt3$, semi-minor $b = 2$). This step matters: getting it backwards flips the formula.

Step 3 — The concept behind the latus-rectum formula

The latus rectum is the focal chord perpendicular to the major axis. Put $x = ae$ (a focus) into the ellipse:

$\frac{a^2e^2}{a^2} + \frac{y^2}{b^2} = 1 \Rightarrow y^2 = b^2(1 - e^2) = b^2\cdot\frac{b^2}{a^2} \Rightarrow y = \pm\frac{b^2}{a}$ …

Concept & Intuition

When a line cuts the axes, its intercepts are the points where it meets the axes. Here, A is on the x‑axis, so A = (a, 0); B is on the y‑axis, so B = (0, b). The point P lies on segment AB and divides it internally in the ratio 5:3. The section formula lets us relate the coordinates of P to a and b using that ratio. Solving for a gives the x‑intercept directly.

Step‑by‑Step Solution

1. Set up the intercepts

   Let A = (a, 0) be the x‑intercept and B = (0, b) be the y‑intercept. The line passes through A and B, and P(–4, 3) lies on segment AB.

2. Apply the internal section formula

   If a point P(x, y) divides the segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n, then

Here the ratio is 5 : 3, with P closer to B? The problem says “divided internally in the ratio 5:3 by the point P”. That means AP : PB = 5 : 3. So m = 5 (from A to P) and n = 3 (from P to B).

3. Plug in coordinates A = (a, 0), B = (0, b), P = (–4, 3). Using the formula:

4. Solve for a and b From the x‑coordinate: $$$$

We are given two points:

$A(3,5,-1)$ and $B(6,3,-2)$.

A point P lies on the line joining them, and its y-coordinate is 2. We need its x-coordinate.

Concept & Intuition

Any point on the line through A and B can be written as

where $t$ is a real number. This is the parametric form of a line in 3D.

When $t = 0$, we get A; when $t = 1$, we get B. For other t, we slide along the line.

We are told the y-coordinate of P is 2, so we can solve for t, then plug back to find x.

Step-by-step solution

1. Find the direction vector from A to B:

2. Write the parametric equations for any point P on the line:

So the coordinates are:

3. Use the given y-coordinate to find t: We know $y = 2$, so

$t = \frac{3}{2}$ …

Concept & Intuition

When a point lies on a line segment joining two given points, its coordinates are a weighted average of the endpoints’ coordinates. But here, a quick observation saves work: the y-coordinate of both endpoints is exactly the same (2). That means the entire segment is parallel to the xz-plane at a fixed y = 2. Any point on that segment, no matter where, will share that same y-coordinate. So the answer is immediate.

Step-by-step reasoning

1. Identify the endpoints

   Endpoint A: $(3, 2, -1)$

   Endpoint B: $(6, 2, -2)$

2. Notice the constant y-coordinate

   Both A and B have $y = 2$. This is not a coincidence — it tells us the segment is horizontal in the y-direction.

3. Interpret what this means for point P

   Since P lies on the segment, its coordinates must be a convex combination of A and B:

For the y-coordinate:

So regardless of $t$, $y_P = 2$.

4. Check the given x-coordinate (optional) …

Concept: parametric (section) form of the segment joining two points in 3-D.

Any point P on the segment joining A(3, 2, -1) and B(6, 2, -2) is

P = A + t(B - A) = (3 + 3t, 2 + 0t, -1 - t), 0 <= t <= 1.

Notice the y-coordinates of A and B are both 2, so B - A has zero y-component. Hence EVERY point of the segment has y = 2, regardless of t. …

Concept: section formula in vector form. If a point P divides AB internally in the ratio m : n, then

OP = (m * OB + n * OA)/(m + n).

Here the point whose position vector is 'a' divides AB in the ratio 2 : 3, and OA = a + 2b.

So a = (2 * OB + 3 * (a + 2b)) / (2 + 3)

=> 5a = 2 OB + 3a + 6b

=> 2 OB = 5a - 3a - 6b = 2a - 6b

=> OB = a - 3b. …

$A=(2,7,0),\ B=(1,2,4),\ C=\left(\tfrac{9}{5},\tfrac{30}{5},\tfrac{4}{5}\right)=(1.8,\,6,\,0.8).$

Let $C$ divide $AB$ in ratio $k:1$, so $C=\dfrac{A+kB}{1+k}$.

• $x$: $\dfrac{2+k}{1+k}=1.8\Rightarrow 2+k=1.8+1.8k\Rightarrow0.2=0.8k\Rightarrow k=\tfrac14.$
• $z$: $\dfrac{0+4k}{1+k}=\dfrac{1}{1.25}=0.8.$ ✓ …

The key idea: to find the distance from a point to a line in 3D, you take any point on the line, form the vector from that point to the given point, and then find the component of that vector perpendicular to the line’s direction. The magnitude of that perpendicular component is the shortest distance.

Why does this work? The line is a straight path; the shortest distance from an external point to it is along a perpendicular. So we need the length of the perpendicular from the point to the line. We can get this by subtracting the projection (the component along the line) from the full vector — what remains is the perpendicular part.

Let’s do it step by step.

1. Identify a point on the line and the direction vector. The line is given in symmetric form:

This tells us the line passes through $A(3, 3, -5)$ and has direction vector $\vec{d} = (2, 3, 6)$.

2. Form the vector from the point on the line to the given point. The given point is $P(1, 2, -4)$. So

3. Find the projection of $\vec{AP}$ onto $\vec{d}$. The projection gives the component of $\vec{AP}$ along the line. Its magnitude is

Compute the dot product:

The magnitude of $\vec{d}$ is

So the projection length is

4. Use the Pythagorean theorem to get the perpendicular distance. The vector $\vec{AP}$ has magnitude

The distance $d$ from $P$ to the line is the length of the perpendicular component: …