I have a test tomorrow with a few sample questions that are expected to be on the test that I do not know how to properly do.

1)Find the slope intercept equation of the line passing through the intersection of the lines 2x-4y=1 & 3x+4y=4 and parallel to the line 5x+7y+3=0

2)Find the angle between the lines described by y=3x-5 & 4x+2y=12

3) Find the general form of the equation of the line through the intersection of 5x-2y=2 & 2x-3y=3 and perpendicular to the line through (4,-1) and (1,-6)

4)Calculate the vertex, x intercept & y intercept. State if the vertex represents a maximum or minimum and opening upwards or downwards. y=-3x^2 + 6x + 9

5a) Determine the equation of the line parallel to the x axis and passes through (8,-14)

b)Find the slope-intercept form equation between two points (-8,14) and (-7,-3)

c)Calculate the distance between two points

d) Calculate midpoint between the two points

Thanks in advance!

1 answer

#1
I assume you can find the intersection of two lines. In this case, that is at (1,1/4).
The line 5x+7y+3=0 has slope -5/7, so now we have a point and a slope, and the desired solution is
y - 1/4 = -5/7 (x-1)

#2
y=3x-5 has slope 3, so the angle it makes with the x-axis is arctan(3)
4x+2y=12 has slope -1/2, so it makes an angle of arctan(-1/2)

The angle between the lines is thus arctan(3)-arctan(-1/2). That will give you an obtuse angle. There is also an acute angle, which is its supplement.

#3 is just like #1, except the new line's slope is the negative reciprocal of the slope between the two points.

#4
y=-3x^2 + 6x + 9
You know it opens downward because of the negative coefficient for x^2
Now, to get the vertex, complete the square
y = -3(x^2-2x+3) = -3(x-3)(x+1)
Now you can read off the x-intercepts, and the y-intercept (from the original equation) is y=9.

#5
(a) naturally, the line is y = -14
(b) just use the two-point form:
y-14 = (-7-14)/(-3+8) (x+8)
and massage that.
(c) I assume you know the distance formula
(d) the midpoint is just the average of the two points: ((-8 + -7)/2 , (14 + -3)/2)