Question

A) Find length of perpendicular from the point p(3,4) on the line 3x+4y-5=0
B) Find the equation of line passing through (1, 7) and having slope 2 units.

Answer 1

Part (1):
Given line is 3x+4y-5=0   .......... (1)
Point is given as,
(x₁,y₁) = (3,4)
The slope is,
x-x₁ / y-y₁ = x-3 / y-4
Now, slope of the line 3x+4y-5=0 is  -3/4
We know that the product of two slopes of perpendicular lines is -1
So,
(x-3 / y-4) (-3/4) = -1
3(x-3) = 4(y-4)
3x-9 = 4y-16
3x-4y = 9-16
3x-4y = -7  
3x-4y+7 = 0  .......... (2)
Now we solve equations (1) and (2).
First add both equations, we get:
6x+2 = 0
6x = -2
x = - 1/3
Put this value in equation (2), we get:
3 (-1/3) -4y+7 = 0
-1 - 4y + 7 = 0
- 4y + 6 = 0
4y - 6 = 0
4y = 6
y = 3/2
Hence, point is (-1/3,3/2) .
Now we apply the distance formula for the two points (3,4) and (-1/3,3/2) to calculate the length.
 d = √ [(-1/3 - 3)² + (3/2 - 4)²]
d = √ [(-10/3)² + (-5/2)²]
d = √ (100/9 + 25/4)
d = √ (400 + 225)/36
d = √ 625/36
d = 25/6
Part (2):
Now we find the equation of line passing through (1, 7) and having slope 2 units.
Using the equation,
y - y₁ = m (x - x₁)
y - 7 = 2 (x - 1)
y - 7 = 2x - 2
2x - y - 2 + 7 = 0
2x - y + 5 = 0
this is the required equation.

Answer 2

Answer:

find the distance between the parallel line 3x+2y-6=0 and 3x+2y-12 =o