Two lines L1: 2y - 3x - 6 = 0 and L2: 3y + x - 20 = 0 interest at a point A.
(a) find the coordinates of A.
(b) A third line L3 is perpendicular to L2 at point A. Find the equation of L3 in the form y = mx + c.
(c) Another line L4 is parallel to L1 and passes through (-1,3). Find the x and y intercepts of L4
I misread the 2nd equation, (should put on glasses)
here is the correct version:
I would use substitution.
from the 2nd equation: x = 20-3y
sub that into the first:
2y - 3x - 6 = 0
2y - 3(20-3y) = 6
2y - 60 + 9y = 6
11y = 66
y = 6
from there find x to get point A
To find the coordinates of point A where lines L1 and L2 intersect:
(a) One way to find the coordinates of point A is by solving the simultaneous equations formed by L1 and L2.
L1: 2y - 3x - 6 = 0
L2: 3y + x - 20 = 0
Solving these equations will give you the values of x and y for point A.
(b) To find the equation of the line L3 that is perpendicular to L2 at point A, we need to determine its gradient.
The gradient of L2 can be found by rearranging its equation into the form y = mx + c, where m represents the gradient.
L2: 3y + x - 20 = 0
Rearranging: 3y = -x + 20
Dividing by 3: y = (-1/3)x + 20/3
Since L3 is perpendicular to L2, its slope (gradient) will be the negative reciprocal of (-1/3). Therefore, the slope of L3 is 3.
Using the coordinates of point A, you can write the equation of L3 in the form y = mx + c.
(c) To find the equation of line L4, which is parallel to L1 and passes through (-1,3), we can use the fact that parallel lines have the same gradient.
The gradient of L1 can be found by rearranging its equation into the form y = mx + c, where m represents the gradient.
L1: 2y - 3x - 6 = 0
Rearranging: 2y = 3x + 6
Dividing by 2: y = (3/2)x + 3
The gradient (m) of L1 is 3/2.
Since L4 is parallel to L1, it will also have a gradient of 3/2.
Using the coordinates (-1,3) and the gradient 3/2, you can write the equation of L4 in the form y = mx + c.
To find the x-intercept of L4, substitute y = 0 into the equation of L4 and solve for x.
To find the y-intercept of L4, substitute x = 0 into the equation of L4 and solve for y.
To find the coordinates of the point of intersection A for lines L1 and L2, we need to solve the system of equations formed by these lines.
(a) The system of equations is:
L1: 2y - 3x - 6 = 0
L2: 3y + x - 20 = 0
To solve this system, we can use substitution or elimination method. Let's use the elimination method to find the solution.
Multiply the first equation (L1) by 3 and the second equation (L2) by 2 to make the coefficients of x in both equations equal:
L1: 6y - 9x - 18 = 0
L2: 6y + 2x - 40 = 0
Subtracting L2 from L1, we get:
-11x + 22 = 0
-11x = -22
x = 2
Substituting the value of x into either equation, let's use L1:
2y - 3(2) - 6 = 0
2y - 6 - 6 = 0
2y = 12
y = 6
Therefore, the coordinates of point A are (2, 6).
(b) To find the equation of line L3, which is perpendicular to L2 at point A, we need to determine the slope of L2 and then find the negative reciprocal of that slope.
The equation of L2 is given as: 3y + x - 20 = 0
Rewriting it in slope-intercept form (y = mx + c), we get:
3y = -x + 20
y = (-1/3)x + (20/3)
The slope of L2 is -1/3. The negative reciprocal of -1/3 is 3.
So, the equation of L3 in the form y = mx + c is:
y = 3x + c
To find the value of c, we substitute the coordinates of A (2, 6) into the equation:
6 = 3(2) + c
6 = 6 + c
c = 6 - 6
c = 0
Therefore, the equation of line L3 is y = 3x + 0, which simplifies to y = 3x.
(c) To find the equation of line L4, which is parallel to L1 and passes through (-1,3), we know that the slope of L4 will be the same as the slope of L1.
The equation of L1 is given as: 2y - 3x - 6 = 0
Rewriting it in slope-intercept form (y = mx + c), we get:
2y = 3x + 6
y = (3/2)x + 3
The slope of L1 is 3/2. Therefore, the slope of L4 will also be 3/2.
Using the point-slope form of a line equation, we can write the equation of L4 as:
y - y1 = m(x - x1), where (x1, y1) is (-1, 3).
Substituting the values into the equation, we get:
y - 3 = (3/2)(x + 1)
To find the x-intercept, we set y to 0:
0 - 3 = (3/2)(x + 1)
-3 = (3/2)(x + 1)
-2 = x + 1
x = -3
Therefore, the x-intercept of L4 is -3.
To find the y-intercept, we set x to 0:
y - 3 = (3/2)(0 + 1)
y - 3 = (3/2)(1)
y - 3 = 3/2
y = 3/2 + 3/2
y = 3
Therefore, the y-intercept of L4 is 3.
Hence, the x-intercept is -3 and the y-intercept is 3 for line L4.
I would use substitution.
from the 2nd equation: y = 20-x
sub that into the first:
3y + x - 20 = 0
3(20-x) + x = 20
60 - 3x + x = 20
-2x = -40
x = 20
from there find y to get point A
slope of 3y + x - 20 = 0 is -1
so the slope of L# has to be +1
and its equation is y = x + b
sub in the values of point A to find b
since L3 is parallel to 2y - 3x - 6 = 0 it must look like
2y - 3x - c = 0 , that is, it will only differ by the constant
plut in (-1,3) to find the value of c
I will assume you know how to find the x and y intercepts.