Answer:
d = 13 / 5 units
Step-by-step explanation:
Given:
- Point P = (-2 , 3 )
- line : 3x - 4y + 5 = 0
Find:
- Use projection to show the given formula
- use the formula to find the distance from given point and line.
Solution:
- Let line L be defined as:
a*x + b*y + c = 0
- Choose A as fixed point on the line L whose coordinates are ( m , k ).
- Now the coordinates of any point (x , y ) can be given by the line L:
a*( x - m ) + b*( y - k ) + c = 0
- We did that by subtracting the coordinates (x,y) from (m,k).
- The above line can represented by dot product of constants (a , b ) to vector coordinates between points (x,y) to (m,k):
( a , b ) . ( x - m , y - k) = 0
- For above expression to be true i.e the dot product of two vectors is only zero when the vectors are orthogonal. So vector (a , b ) is always perpendicular to every vector in direction of line L
- The vector:
v = ( x - m , y - k)
- It may represent the line connecting the points (x_1 , y_1) and (m, k).
Then the distance d from the point P_1: (x_1, y_1) to the line L is given by the magnitude of the scalar projection of v onto a, because the latter is the length of the side of a right triangle with two of the vertices being P_1 and
( m , k), and the other vertex lying on L So:
d = | component v along x |
d = | v . ( a , b ) | / | (a , b ) |
d = | (x_1 - m , y_1 - k ) . ( a , b)| / sqrt(a^2 + b^2)
d = | ax_1 + by_1 - (a*m + b*k) | / sqrt(a^2 + b^2)
Where point A (m,k) lies on line hence;
a*m + b*k + c = 0
c = - (a*m + b*k)
Hence,
d = | ax_1 + by_1 + c | / sqrt(a^2 + b^2)
- Given point P (-2,3) and line L : 3x - 4y + 5 = 0
where,
(x_1 , y_1) = (-2 , 3 )
(a , b) = ( 3 , -4 )
c = 5
-Evaluate:
d = | 3*(-2) + -4*3 + 5 | / sqrt(3^2 + 4^2)
d = | - 13 | / 5
d = 13 / 5 units
which is 15!