# Distance of a Point from a Line

The shortest distance of a point form a line (not passing through the point) is the perpendicular distance. We shall now derive an expression for the perpendicular distance of from the line

Let AB be the given line and be the given point (not on the line). Draw PD âŠ¥ AB. PD is the required distance. Draw OM âŠ¥ AB.

Let and

Using normal form of line, equation of AB is

But AB is given to be .

Since (1) and (2) represent the same line, the corresponding coefficients (of and constant) must be proportional.

Corollary: The length of the perpendicular from the origin to

Note:
Two points having same distance from a given line on either side of the line will have the distances numerically same; but opposite in signs. Since distance is a scalar quantity, we take only the numerical value Similarly the distance of (0, 0) from is numerically equal to .

# Distance between two parallel lines

Examples 1:
Show that the locus of a point whose distances from are equal is a straight line. Also show that this line bisects angle between the original line.

Solution:
Let be the point equidistance from .

Both (1) and (2) are straight lines. To show that bisects the angle between and .
Slope of is
Slope of is

Similarly let the angle between

Hence the result.

Note: The line  is the angle bisector of angle between which are two perpendicular lines {Recall locus theorem which you have learnt in earlier classes}

Example 2
a) What are the points on X-axis whose perpendicular distance from the straight line is 4?
b) Find the points on Y-axis whose perpendicular distance from the straight line is 3

Solution:
a) Let us take a point on the X-axis as . Its perpendicular distance from the line i.e.,is which is given to be 4 units

âˆ´ The points are (8, 0) and (âˆ’ 2, 0) which are at a distance of 4 units from the line
b) Any point on the Y-axis will have co-ordinate as zero
Let the point be . Perpendicular distance of P from the line is 3 units.

âˆ´ The points are (0, 1) and (0, âˆ’ 9)

Example 3:
Find the distance between the parallel lines

Solution:
Distance between parallel lines is . We have to make the coefficients of in the two equations the same.

Multiplying by 2,
and
Distance between (1) & (2) is

The condition for three straight lines to be concurrent:
Let the three lines be:

Solving (1) & (2), we get the point of intersection as

substituting for in equation (3),

in determinant form

Example 1:
If are concurrent, show that

Solution:
Solving

Substituting for and in

Example 2:
Find the co-ordinates of the circumcentre of the triangle whose sides are given by the equations

Solution:
Solving the given equations taking two at a time, we get the vertices of the triangle. A(3,1), B(2,2) and C(2,0)

Mid-point D of BC is (2,1)
Slope of BC is
slope of perpendicular bisector of BC = 0
Equation of SD:
............(1)
Mid-point of CA is
Slope of AC =
Slope of SE = 1
Equation of SE is

(1) and (2) are perpendicular bisectors of two sides of ,they meet at the circumcentre.
we get

Another Method:

If S is the circumcentre, then SA = SB = SC. Let S be

So,

Solving (3) & (4), we get

The circumcentre is (2, 1).

Example 3:
Find the co-ordinate of the orthocentre of the triangle formed by the straight lines .

Solution:

Orthocentre is the point of concurrence of the altitudes of a triangle. Let AB, BC, CA have equations .Obviously, C is the origin (0, 0).

[Any equation in which the constant term is zero is a line passing through the origin. Here both BC and AC pass through the origin. Hence they meet at the origin.]

Solving

and solving

slope of BC is âˆ’ 3