Position of Two Points with Respect to a Line

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.

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

Recommended Video Based on Position of Two Points with respect to a Line

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)