To find the equation of line L1, we need to determine its slope. Since L1 is perpendicular to L2, the slope of L1 will be the negative reciprocal of the slope of L2.
The equation of line L2 is given as 3x - y = 4. To find the slope of L2, let's rewrite this equation in slope-intercept form: y = 3x - 4. Comparing this to the standard slope-intercept form y = mx + b, we can see that the slope of L2 is 3.
As L1 is perpendicular to L2, the slope of L1 will be -1/3 (negative reciprocal of 3).
(i) Equation of L1:
We have the point A(5, 1) on line L1 and its slope as -1/3. Using the point-slope form of a line, the equation of L1 can be written as:
y - y1 = m(x - x1)
where (x1, y1) is the point A(5, 1) and m is the slope,
y - 1 = (-1/3)(x - 5)
y - 1 = (-1/3)x + 5/3
y = (-1/3)x + 5/3 + 1
y = (-1/3)x + 5/3 + 3/3
y = (-1/3)x + 8/3
So, the equation of line L1 is y = (-1/3)x + 8/3.
(ii) Intersection of L1 and L2:
To find the coordinates of the point of intersection of lines L1 and L2, we can solve their equations simultaneously.
Substituting the equation of L1 into L2, we have:
3x - (-1/3)x + 8/3 = 4
(9/3)x + (1/3)x = 4 - 8/3
(10/3)x = 12/3 - 8/3
(10/3)x = 4/3
x = (4/3) * (3/10)
x = 4/10
x = 2/5
Plugging this value of x back into the equation of L1, we can find y:
y = (-1/3)(2/5) + 8/3
y = -2/15 + 40/15
y = 38/15
Therefore, the coordinates of the point of intersection of L1 and L2 are (2/5, 38/15).
(iii) Perpendicular distance from A(5, 1) to line L2:
To find the perpendicular distance from a point to a line, we can use the formula:
Distance = |Ax + By + C| / sqrt(A^2 + B^2)
Considering L2's equation as Ax + By + C = 0, where A = 3, B = -1, and C = -4, and substituting the coordinates of point A(5, 1), the formula becomes:
Distance = |3(5) + (-1)(1) + (-4)| / sqrt(3^2 + (-1)^2)
Distance = |15 - 1 - 4| / sqrt(9 + 1)
Distance = |10| / sqrt(10)
Distance = 10 / sqrt(10)
Therefore, the perpendicular distance from point A(5, 1) to line L2 is 10 / sqrt(10).