# Position of a Point Relative to a Line

We will learn how to find the position of a point relative to a line and also the condition for two points to lie on the same or opposite side of a given straight line.

Let the equation of the given line AB be ax + by + C = 0…………….(i) and let the coordinates of the two given points P (x$$_{1}$$, y$$_{1}$$) and Q (x$$_{2}$$, y$$_{2}$$).

I: When P and Q are on opposite sides:

Let us assumed that the points P and Q are on opposite sides of the straight line.

The coordinate of the point R which divides the line joining P and Q internally in the ratio m : n are

($$\frac{mx_{2} + nx_{1}}{m + n}$$, $$\frac{my_{2} + ny_{1}}{m + n}$$)

Since the point R lies on ax + by + C = 0 hence we must have,

a ∙ $$\frac{mx_{2} + nx_{1}}{m + n}$$ + b ∙ $$\frac{my_{2} + ny_{1}}{m + n}$$ + c = 0

⇒ amx$$_{2}$$ + anx$$_{1}$$ + bmy$$_{2}$$ + bny$$_{1}$$ + cm + cn = 0

⇒ m(ax$$_{2}$$ + by$$_{2}$$ + c )= - n(ax$$_{1}$$ + by$$_{1}$$ + c)

⇒ $$\frac{m}{n} = - \frac{ax_{1} + by_{1} + c}{ax_{2} + by_{2} + c}$$………………(ii)

II: When P and Q are on same sides:

Let us assumed that the points P and Q are on same side of the straight line. Now join P and Q. Now assume that the straight line, (produced) intersects at R.

The coordinate of the point R which divides the line joining P and Q externally in the ratio m : n are

($$\frac{mx_{2} - nx_{1}}{m - n}$$, $$\frac{my_{2} - ny_{1}}{m - n}$$)

Since the point R lies on ax + by + C = 0 hence we must have,

a ∙ $$\frac{mx_{2} - nx_{1}}{m - n}$$ + b ∙ $$\frac{my_{2} - ny_{1}}{m - n}$$ + c = 0

⇒ amx$$_{2}$$ - anx$$_{1}$$ + bmy$$_{2}$$ - bny$$_{1}$$ + cm - cn = 0

⇒ m(ax$$_{2}$$ + by$$_{2}$$ + c )= n(ax$$_{1}$$ + by$$_{1}$$ + c)

⇒ $$\frac{m}{n} = \frac{ax_{1} + by_{1} + c}{ax_{2} + by_{2} + c}$$………………(iii)

Clearly, $$\frac{m}{n}$$ is positive; hence, the condition (ii) is satisfied if (ax$$_{1}$$+ by$$_{1}$$ + c) and (ax$$_{2}$$ + by$$_{2}$$ + c) are of opposite signs. Therefore, the points P (x$$_{1}$$, y$$_{1}$$) and Q (x$$_{2}$$, y$$_{2}$$) will be on opposite sides of the straight line ax + by + C = 0 if(ax$$_{1}$$+ by$$_{1}$$ + c) and (ax$$_{2}$$ + by$$_{2}$$ + c) are of opposite signs.

Again, the condition (iii) is satisfied if (ax$$_{1}$$+ by$$_{1}$$ + c) and (ax$$_{2}$$ + by$$_{2}$$ + c) have the same signs. Therefore, the points P (x$$_{1}$$, y$$_{1}$$) and Q (x$$_{2}$$, y$$_{2}$$will be on the same side of the line ax + by + C = 0 if (ax$$_{1}$$+ by$$_{1}$$ + c) and (ax$$_{2}$$ + by$$_{2}$$ + c) have the same signs.

Thus, the two points P (x$$_{1}$$, y$$_{1}$$) and Q (x$$_{2}$$, y$$_{2}$$) are on the same side or opposite sides of the straight line ax + by + c = 0, according as the quantities (ax$$_{1}$$+ by$$_{1}$$ + c) and (ax$$_{2}$$ + by$$_{2}$$ + c) have the same or opposite signs.

Remarks: 1. Let ax + by + c = 0 be a given straight line and P (x$$_{1}$$, y$$_{1}$$) be a given point. If ax$$_{1}$$+ by$$_{1}$$ + c is positive, then the side of the straight line on which the point P lies is called the positive side of the line and the other side is called its negative side.

2. Since a ∙ 0 + b ∙ 0 + c = c, hence it is evident that the origin is on the positive side of the line ax + by + c = 0 when c is positive and the origin is on the negative side of the line when c is negative.

3. The origin and the point P (x$$_{1}$$, y$$_{1}$$) are on the same side or opposite sides of the straight line ax + by + c = 0, according as c and (ax$$_{1}$$+ by$$_{1}$$ + c) are of the same or opposite signs.

Solved examples to find the position of a point with respect to a given straight line:

1. Are the points (2, -3) and (4, 2) on the same or opposite sides of the line 3x - 4y - 7 = 0?

Solution:

Let Z = 3x - 4y - 7.

Now the value of Z at (2, -3) is

Z$$_{1}$$ (let) =3 × (2) - 4 × (-3) - 7

= 6 + 12 - 7

= 18 - 7

= 11, which is positive.

Again, the value of Z at (4, 2) is

Z$$_{2}$$ (let) = 3 × (4) - 4 × (2) - 7

= 12 - 8 - 7

= 12 - 15

= -3, which is negative.

Since, z$$_{1}$$ and z$$_{2}$$, are of opposite signs, therefore the two points (2, -3) and (4, 2) are on the opposite sides of the given line 3x - 4y - 7 = 0.

2. Show that the points (3, 4) and (-5, 6) lie on the same side of the straight line 5x - 2y = 9.

Solution:

The given equation of the straight line is 5x - 2y = 9.

⇒ 5x - 2y - 9 = 0 ……………………… (i)

Now find the value of 5x - 2y - 9 at (3, 4)

Putting x = 3 and y = 4 in the expression 5x - 2y - 9 we get,

5 × (3) - 2 × (4) - 9 = 15 - 8 - 9 = 15 - 17 = -2, which is negative.

Again, putting x = 5 and y = -6 in the expression 5x - 2y - 9 we get,

5 × (-5) - 2 × (-6) - 9 = -25 + 12 - 9 = -13 - 9 = -32, which is negative.

Thus, the value of the expression 5x - 2y - 9 at (2, -3) and (4, 2) are of same signs. Therefore, the given two points (3, 4) and (-5, 6) lie on the same side of the line given straight line 5x - 2y = 9.

The Straight Line