Question

distance of a point from a line derivation of class 11​

Answer 1

Answer:

What?

Step-by-step explanation:

Didnt understood the esution

Answer 2

Answer:

Suppose a line l in XY−plane and N(x1, y1) is any point at a distance d from the line l. This line is represented by Ax + By + C = 0. The distance of point from a line, ‘d’ is the length of the perpendicular drawn from N to l. The x and y-intercepts are −C/A and −C/B respectively.

Distance of Point From a Line

Step-by-step explanation:

The line meets the y and the x axes at points A and B respectively. The coordinates of the points are A (0, −C/B) and B (−C/A, 0). The area of the triangle is given by

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.

It is interesting to find out the distance between two parallel