30 March 20201. The straight line L1 passes through the point A (5,1) and is perpendicular to the line L2 whose equation is 3x - y = 4.Find (i) the equation of L1 (ii) the coordinates of the point of intersection of L1 and L2 (iii) the perpendicular distance from the point A (5,1) to the line L2

A line perpendicular to 3x - y = 4 must have the form

x + 3y = C
plug in the point (5,1) ---> 5 + 3 = C = 8

new line: x + 3y = 8

intersection: from the first, y = 3x-4
plug into 2nd,
x + 3(3x-4) = 8
x+9x-12 = 8
10x = 20
x = 2 , then y = 2 <----- intersect at (2,2)

distance between (2,2) and (5,1)
= √(3^2 + (-1)^2) = √10

check using distance from a point to a line formula
line: 3x-y-4=0 to point (5,1)
= |3(5) - 1 - 4|/√(1+9)
= 10/√10 = √10

L2: slope = -A/B = -3/-1 = 3.

(I). L1: A(5, 1), slope = -(1/3) = -1/3.
Y = mx + b
1 = (-1/3)5 + b
b = 8/3.
Y = -x/3 + 8/3
x + 3y = 8.

(ii). Multiply Eq1 by 3 and add:
9x - 3y = 12
x + 3y = 8
sum: 10x = 20. X = 2.
2 + 3y = 8. Y=2.

(iii). (2, 2), (5, 1).
d^2 = (5-2)^2 + (1-2)^2 = 10. d = 3.2.

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).

To find the equation of line L1, you 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.

First, let's find the slope of L2. The equation of L2 is given as 3x - y = 4. Rewrite it in the slope-intercept form (y = mx + c), where m represents the slope:

3x - y = 4
-y = -3x + 4
y = 3x - 4

The slope of L2 is 3.

Since L1 is perpendicular to L2, the slope of L1 will be -1/3 (negative reciprocal of 3).

Now that we have the slope of L1 (-1/3) and a point on L1 (A with coordinates (5,1)), we can find the equation of L1 using the point-slope form (y - y1 = m(x - x1)).

(i) Equation of L1:
Let's use point A (5,1) to find the equation:

y - 1 = (-1/3)(x - 5)

Simplifying this equation gives us the equation of L1.

(ii) Coordinates of point of intersection of L1 and L2:
To find the point of intersection, we need to solve the system of equations formed by L1 and L2. In other words, we need to find the x and y values that satisfy both L1 and L2 simultaneously.

To do this, substitute the equation of L1 (found in step (i)) into the equation of L2 and solve for x.

3x - 4 = (-1/3)(x - 5)

Simplify and solve for x. Once you have the value of x, substitute it back into either L1 or L2 to find the corresponding y-coordinate.

(iii) Perpendicular distance from point A to line L2:
The perpendicular distance between a point and a line can be found using the formula:

Distance = |Ax + By + C| / sqrt(A^2 + B^2)

In this case, the equation of L2 can be written in the standard form: 3x - y - 4 = 0. Comparing this with the distance formula (Ax + By + C), we have A = 3, B = -1, and C = -4. The point A has coordinates (x1, y1) = (5, 1).

Substitute these values into the distance formula and calculate the distance.

This way, you can find (i) the equation of L1, (ii) the coordinates of the point of intersection of L1 and L2, and (iii) the perpendicular distance from the point A(5,1) to the line L2.