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November 10, 2024In coordinate geometry or Euclidean geometry, the distance formula is utilised to calculate the distance between any two points in an XY plane. Also, a point’s x-coordinate, or abscissa, is its distance from the Y-axis. The y-coordinate, or ordinate, of a point, is the amount of distance from the x-axis.
A point on the x-axis has coordinates of the form (x, 0), while a point on the y-axis has coordinates of the form (y, 0). (0, y). The Pythagoras theorem is used to find the distance between any two points in a plane. This is an important topic for pupils in Class 10. Let us take a closer look at the formula and how it was developed.
For coordinate geometry, the distance formula is defined in arithmetic. Students should not misunderstand the formula for speed and distance. Both are entirely distinct subjects. When we talk about speed and distance, we are talking about real-life situations where a person or a vehicle goes at a certain pace to cover a certain distance in a certain amount of time. The speed is, thus, equal to the distance travelled divided by the time taken.
When a human goes from one point to another in a plane in coordinate geometry, the distance between the start point and the endpoint is determined using the distance formula. In the next section, we will look at its formula.
The distance formula is a formula that may be used to calculate the distance between any two points if we know their coordinates. These coordinates could be on the x, y, or both axes. Assume there are two points in an XY plane, P and Q (x1, y1) are the coordinates of point P, and (x1, y1) are the coordinates of point Q (x2,y2). The formula for calculating the distance between two points PQ is as follows.
There are different types of distance formulas in coordinate geometry.
Have you ever considered calculating the distance between a point and a line? How are we going to do it? The point is not on the line. How do you go about doing it? Let us have a look.
Assume there is a line l in the XY plane, and N(x1, y1) is any point d away from line l. Ax + By + C = 0 is the equation for this line. The length of the perpendicular drawn from N to l is the distance of a point from a line, ‘d’. C/A and C/B are the x and y-intercepts, respectively.
The line intersects the Y and the X-axis at points A and B, respectively. The coordinates of those points are A (0, −C/B) and B (−C/A, 0). Therefore we can give area of the triangle by the following.
area (Δ NAB) = ½ base × height = ½ AB × NM,
⇒ NM = 2 area (Δ NAB) / AB … (I)
Also, area (Δ NAB) = ½ |x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|
Or, ½ | x1 (0 + C/B) + (−C/A) (−C/B − y1) +0 (y1 − 0)| = ½ |x1 C/B + y1 C/A + C2/AB|
Or, ½ |C/ (AB) |.|Ax1 + By1 + C|… (II)
Distance of the line AB = ((0 + C/A)2 + (C/B − 0)2)½ = |C| × ((1/A2) + (1/B2))½
Or, Distance, AB = |C/AB| (A2 + B2)½ … (III)
Putting (II) & (III) in (I), we have
NM = d = |Ax1 + By1 + C| / (A2 + B2)½.
When the distance between two lines is constant at any given point, they are said to be parallel. To put it another way, if the slopes of both lines are the same, they will be parallel. Assume that on the XY plane, two parallel lines, l1 and l2, have the same slope = m. The equations for parallel lines are as follows.
y = mx + c1 … (I)
y = mx + c2 … (II)
The distance of a point from a line is used to compute the distance between two parallel lines. It is equivalent to the perpendicular distance between any two points and one of the lines. Let N be the location where the perpendicular or normal from M (c2/m, 0) is drawn to l1. The distance between two lines is known to be as follows.
d =|Ax1 + By1 + C| / (A2 + B2)½.
Here, A = m, B = 1 and C = c1 (comparing y = mx + c1 and Ax + By + C = 0)
And, x1 = − c2/m and y1 = 0
So, d = |m (− c2/m) + 0 + c1| / (m2 + 1)½ = |c1 − c2| / (m2 + 1)½
Generalizing the above, we have, d = |C1 − C2| / (A2 + B2)½
(If l1: Ax + By + C1 = 0 and l2: Ax + By + C2 = 0)
We hope the above article helped you clear your doubts on Derivation of Distance Formula 2D.
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