Position of Two Points with Respect to a Line

The position of two points concerning a line is a fundamental concept in geometry and analytical mathematics. It has many applications used in real life from physics, robotics, etc. The position of the points on the line determines whether the position is on the line, same side of the line, or on the opposite side.

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Position of two points with respect to a line

Two given points $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ lies on the same side of a line $a x+b y+c=0$ when $\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\mathrm{ax}_2+\mathrm{by}_2+\mathrm{c}}>0$ and points lie on the opposite side when $\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\mathrm{ax}_2+\mathrm{by}_2+\mathrm{c}}<0$.

Note:

1. The side of the line where origin lies is known as the origin side.
2. A point $(p, q)$ will lie on the origin side of the line $a x+b y+c=0$ if $\frac{a p+b q+c}{a .0+b .0+c}>0$, meaning ap $+\mathrm{bq}+\mathrm{c}$ and c will have the same sign.
3. A point $(p, q)$ will lie on the non-origin side of the line $a x+b y+c$ $=0$, if $\frac{a p+b q+c}{a .0+b .0+c}<0$, meaning $\mathrm{ap}+\mathrm{bq}+\mathrm{c}$ and c will have the opposite sign.

Position of a point which lies inside a triangle

Let $P\left(x_1, y_1\right)$ be the point that lies inside the triangle

The equations of sides of a triangle are

$\begin{aligned} & A B: a_1 x+b_1 y+c_1=0 \\ & B C: a_2 x+b_2 y+c_2=0 \\ & C A: a_3 x+b_3 y+c_3=0\end{aligned}$

First find the coordinates of vertices of triangle ABC

Let $A=\left(x^{\prime}, y^{\prime}\right), \quad B=\left(x^{\prime \prime}, y^{\prime \prime}\right)$ and $C=\left(x^{\prime \prime \prime}, y^{\prime \prime \prime}\right)$

And if coordinates of vertices of triangle ABC is given then find equation of sides of triangle ABC.

If point P lies inside the triangle, then P and A must be same side of BC, P and B must be same side of AC and P and C must be same side of AB, then.

$\begin{aligned} & \frac{a_2 x_1+b_2 y_1+c_2}{a_2 x^{\prime}+b_2 y^{\prime}+c_2}>0 \\ & \frac{a_3 x_1+b_3 y_1+c_3}{a_3 x^{\prime \prime}+b_3 y^{\prime \prime}+c_3}>0 \\ & \frac{a_1 x_1+b_1 y_1+c_1}{a_1 x^{\prime \prime \prime}+b_1 y^{\prime \prime \prime}+c_3}>0\end{aligned}$

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Solved Examples Based on Position of Two Points with Respect to a Line:

Example 1: if $P\left(p, p^2\right)$ lies inside the triangle having sides along the lines $2 x+3 y=1, x+2 y-3=0,6 y=5 x-1$ then the value of p?
1) $p \epsilon(1 / 2,1) \cup(-3 / 2,-1)$
2) $p \epsilon(1 / 2,1) \cup(-2,-1)$
3) $p \epsilon(1,2) \cup(-3 / 2,-1)$
4) None of these

Solution

First find the coordinates of vertices of triangle ABC

Let $A=\left(x^{\prime}, y^{\prime}\right), \quad B=\left(x^{\prime \prime}, y^{\prime \prime}\right)$ and $C=\left(x^{\prime \prime \prime}, y^{\prime \prime \prime}\right)$

And if coordinates of vertices of triangle ABC is given then find equation of sides of triangle ABC.

If point P lies inside the triangle, then P and A must be same side of BC, P and B must be same side of AC and P and C must be same side of AB, then.

$\begin{aligned} & \frac{a_2 x_1+b_2 y_1+c_2}{a_2 x^{\prime}+b_2 y^{\prime}+c_2}>0 \\ & \frac{a_3 x_1+b_3 y_1+c_3}{a_3 x^{\prime \prime}+b_3 y^{\prime \prime}+c_3}>0 \\ & \frac{a_1 x_1+b_1 y_1+c_1}{a_1 x^{\prime \prime \prime}+b_1 y^{\prime} 1^{\prime}+c_3}>0 \\ & 2 x+3 y=1 \ldots(i) \\ & x+2 y=3 \ldots(\text { ii }) \\ & 5 x-6 y=1 . .(i i i) \\ & \text { equation (i)-2*equation(ii) } \\ & \text { PointA(-7, 5) } \\ & 5^* \text { equation (ii)-equation(iii) } \\ & \text { pointB(5/4,7/8) } \\ & 5^* \text { equation(i)-2*equation(iii) } \\ & \text { Point } C(1 / 3,1 / 9)\end{aligned}$

So $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be vertices of the triangle.
$
\begin{aligned}
& \mathrm{A} \equiv(-7,5), \mathrm{B} \equiv(5 / 4,7 / 8) \\
& \mathrm{C} \equiv(1 / 3,1 / 9)
\end{aligned}
$

If P lies in-side the $\triangle \mathrm{ABC}$, then sign of P will be the same as sign of a w.r.t. the line BC
$
\Rightarrow \quad 5 p-6 p^2-1<0 \Rightarrow\left(-\infty, \frac{1}{3}\right) U\left(\frac{1}{2}, \infty\right)
$

Similarly $2 p+3 p^2-1>0 \Rightarrow(-\infty,-1) U\left(\frac{1}{3}, \infty\right)$
And, $\quad p+2 p^2-3<0 \Rightarrow\left(\frac{-3}{2}, 1\right)$.
Solving, (1), (2) and (3) for $p$ and then taking intersection, We get $p \in(1 / 2,1) \cup(-3 / 2,-1)$.

Example 3: Let the point $\mathrm{P}(\alpha, \beta)$ be at a unit distance from each of the two lines $\mathrm{L}_1: 3 \mathrm{x}-4 \mathrm{y}+12=0$, and $\mathrm{L}_2: 8 \mathrm{x}+6 \mathrm{y}+11=0$. If P lies below $\mathrm{L}_1$ and above $\mathrm{L}_2$, then $100(\alpha+\beta)$ is equal to
1) -14
2) 42
3) -22
4) 14

Solution

$
\begin{aligned}
& \mathrm{P} \text { and origin lie on same side of } \mathrm{L}_1 \\
& \Rightarrow \frac{3 \alpha-4 \beta+12}{12}>0 \\
& \Rightarrow 3 \alpha-4 \beta+12>0
\end{aligned}
$

Similarly for $\mathrm{L}_2$
$
\begin{aligned}
& \frac{8 \alpha+6 \beta+11}{11}>0 \\
& \Rightarrow 8 \alpha+6 \beta+11>0
\end{aligned}
$

Also distance from $\mathrm{L}_1$ and $\mathrm{L}_2=1$
$
\begin{aligned}
& \Rightarrow \frac{|3 \alpha-4 \beta+12|}{5}=1 \\
& \Rightarrow 3 \alpha-4 \beta+12=5 \quad(\text { using (ii)) } \\
& \text { And } \frac{|8 \alpha+6 \beta+11|}{10}=1 \\
& 8 \alpha+6 \beta+11=10
\end{aligned}
$

Solving these 2 equations
$
\begin{aligned}
& \alpha=\frac{-23}{25}, \quad \beta=\frac{53}{50} \\
& \therefore 100\left(\alpha_1 \beta\right)=14
\end{aligned}
$

Hence, the correct option is 4.

Example 4: Find the range of $\theta$ in the interval $(0, \pi)$ such that the points $(3,5)$ and $(\sin \theta, \cos \theta)$ lie on the same side of the line $\mathrm{x}+\mathrm{y}-1=0$.
1) $0<\theta<\pi / 6$
2) $0<\theta<3 \pi / 4$
3) $\frac{\pi}{2}<\theta<\pi$
4) $0<\theta<\pi / 2$

Solution
$
\begin{aligned}
& 3+5-1=7>0 \\
& \therefore \sin \theta+\cos \theta-1>0 \\
& \Rightarrow \sin (\pi / 4+\theta)>1 / \sqrt{ } 2 \Rightarrow \pi / 4<\pi / 4+\theta<3 \pi / 4 \Rightarrow 0<\theta<\pi / 2
\end{aligned}
$

Hence, the answer is the option (4).

Example 5: Find the range of $\theta$ in the interval $(0, \pi)$ such that the points $(3,5)$ and $(\sin \theta, \cos \theta)$ lie on the same side of the line $\mathrm{x}+\mathrm{y}-1=0$.
1) $0<\theta<\frac{\pi}{6}$
2) $0<\theta<3 \frac{\pi}{4}$
3) $\pi<\theta<\pi$
4) $0<\theta<\frac{\pi}{2}$

Solution:
$
\begin{aligned}
& 3+5-1=7>0 \\
& \therefore \sin \theta+\cos \theta-1>0
\end{aligned}
$
$\Rightarrow \sin (\pi / 4+\theta)>1 / \sqrt{ } 2 \Rightarrow \pi / 4<\pi / 4+\theta<3 \pi / 4 \Rightarrow 0<\theta<\pi / 2$.Hence, the answer is the option (4)