(a) A straight line L whose equation is 3y – 2x = -2 meets the x-axis at R. Determine the coordinates of
R.
b). A second line L2 is perpendicular to L1 at R. Find the equation of L2 in the form y=mx + c where
m and c as contract.
c). A third line L3 passes through (-4, 1) and is parallel to L1.Find
i.
The equation of L3 in the form y=mx + c where m and c are constants.
ii.
The coordinators of point S at which L3 intersects L2.
(a)
To find the coordinates of the point R where line L intersects the x-axis, we set y = 0 in the equation of L and solve for x:
3(0) - 2x = -2
-2x = -2
x = 1
So the coordinates of R are (1, 0).
(b)
Since line L2 is perpendicular to line L1 at R, we know that the slopes of L1 and L2 are negative reciprocals. The slope of L1 can be found by rearranging the equation in slope-intercept form:
3y - 2x = -2
3y = 2x - 2
y = (2/3)x - 2/3
So the slope of L1 is 2/3.
The slope of L2 will be the negative reciprocal of 2/3, which is -3/2. So the equation of L2 in the form y = mx + c is:
y = (-3/2)x + c
(c)
(i)
Since line L3 is parallel to line L1, it will have the same slope. The slope of L1 is 2/3, so the equation of L3 in the form y = mx + c is:
y = (2/3)x + c
(ii)
To find the coordinates of the point S where L3 intersects L2, we can set the equations of L2 and L3 equal to each other and solve for x and y:
(-3/2)x + c = (2/3)x + c
(-3/2)x - (2/3)x = c - c
(-9/6)x = 0
x = 0
Substituting x = 0 into the equation of L2, we can solve for y:
y = (-3/2)(0) + c
y = c
So the coordinates of point S are (0, c).