The points (3,4) and (2,-6) are situated on the __ of the line 3x-4y-8=0.

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(3,4)

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Concept understanding — Distance From Point To Line

Distance from a Point to a Line

The distance from a point to a line is the shortest distance — the length of the perpendicular dropped from the point onto the line. In 3D we compute it with vectors and the cross product.

Let the line be $\vec{r}=\vec{a}+\lambda\vec{b}$ (a point $A$ with position vector $\vec{a}$ , direction $\vec{b}$ ), and let $P$ be the given point with position vector $\vec{p}$ .

The idea

Look at the triangle formed by $A$ , $P$ and the foot of the perpendicular $M$ . The segment $\overrightarrow{AP}=\vec{p}-\vec{a}$ is the hypotenuse, and the perpendicular distance $d=PM$ is the side opposite the angle $\theta$ between $\overrightarrow{AP}$ and the line:

d = |\overrightarrow{AP}|\sin\theta.

But the cross product already contains $\sin\theta$ : $|\overrightarrow{AP}\times\vec{b}| = |\overrightarrow{AP}|\,|\vec{b}|\sin\theta$ . Dividing by $|\vec{b}|$ isolates the distance.

d = \frac{|(\vec{p}-\vec{a})\times\vec{b}|}{|\vec{b}|}

Example

Distance of $P(1,2,3)$ from the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2\hat{i}-\hat{j}+2\hat{k})$ .

Here $\vec{a}=(1,1,0)$ , $\vec{b}=(2,-1,2)$ , and $\overrightarrow{AP}=\vec{p}-\vec{a}=(0,1,3)$ .

$\overrightarrow{AP}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&3\\2&-1&2\end{vmatrix} = (2+3)\hat{i} - (0-6)\hat{j} + (0-2)\hat{k} = 5\hat{i}+6\hat{j}-2\hat{k}.$ …

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To determine if points are on the same or opposite sides of a line $Ax+By+C=0$ , evaluate $Ax+By+C$ for each point. If the signs are the same, they are on the same side; if different, they are on opposite sides. The given points are on opposite sides of the line.

When we talk about points being on one side or the other of a line, we're essentially asking which of the two half-planes created by the line each point falls into. A straight line $Ax+By+C=0$ divides the entire Cartesian plane into two distinct regions.

The key insight here is that for any point $(x,y)$ not on the line, the expression $Ax+By+C$ will have a specific sign – either positive or negative. All points in one of the half-planes will yield a positive value for $Ax+By+C$ , while all points in the other half-plane will yield a negative value. The line itself is where $Ax+By+C=0$ .

This means if two points $(x_1, y_1)$ and $(x_2, y_2)$ are on the same side of the line, then $Ax_1+By_1+C$ and $Ax_2+By_2+C$ will have the same sign. If they are on opposite sides, then $Ax_1+By_1+C$ and $Ax_2+By_2+C$ will have opposite signs.

Tip

A quick way to check this is to multiply the two values:

If $(Ax_1+By_1+C)(Ax_2+By_2+C) > 0$ , the points are on the same side.
If $(Ax_1+By_1+C)(Ax_2+By_2+C) < 0$ , the points are on opposite sides.
If $(Ax_1+By_1+C)(Ax_2+By_2+C) = 0$ , at least one point lies on the line itself.

Let's apply this concept to the given problem.

Identify the line equation:

The equation of the line is given as $3x-4y-8=0$ .

We can define a function $f(x,y) = 3x-4y-8$ .
Evaluate for the first point $(3,4)$ : …

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Q.The points (3,4)(3,4)(3,4) and (2,−6)(2,-6)(2,−6) are situated on the ____ of the line 3x−4y−8=03x-4y-8=03x−4y−8=0.

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Q.The points $(3,4)$ and $(2,-6)$ are situated on the ____ of the line $3x-4y-8=0$ .

Unlock everything free for 14 days