2. To find the intersection point of the lines, we convert them to slope-intercept form then equate them to each other.
3x - 2y - 7 = 0
2y = 3x - 7
y = (3/2)x - 7/2
and
4x - 3y + 2 = 0
3y = 4x + 2
y = (4/3)x + 2/3
Equating the two:
(3/2)x - 7/2 = (4/3)x + 2/3
(1/6)x = 25/6
x = 25
So, the two lines intersect at x = 25. Substituting x = 25 into the first equation,
3(25) - 2y - 7 = 0
75 - 2y - 7 = 0
68 = 2y
y = 34
This gives (25, 34).
The gradient of the line passing through (25, 34) and (1, 4) is:
(34 - 4)/(25 - 1) = 5/4
Writing the point-slope form:
y - 4 = (5/4)(x - 1)
This gives:
y - 4 = (5/4)x - 5/4
y - 11/4 - (5/4)x = 0
3. The slope of the parabola y = 4 - 3x - x^2 is found by taking the derivative.
y' = -2x - 3
For a line to be tangent, it has to touch the parabola and have a slope equal to the slope of the parabola at the point of intersection.
The point of intersection of the line and parabola can be found by equating the two equations:
x + 8 = 4 - 3x - x^2
x^2 + 4x + 4 = 0
(x + 2)^2 = 0
This gives x = -2, substituting this into the equation of the slope of the parabola:
y' = -2(-2) - 3 = 1
Since the slope of the parabola is 1 at the point of intersection, and the slope of y = x + 8 is clearly 1, the line y = x + 8 is a tangent to the parabola.