Can you write an equation of these lines? (For pre-calculus people)?

Question

The line with y-intercept 4 and x-intercept -2

The line through (-1,4) and (5,8)

The line through (2,-7) and (2,3)

The line parallel to the line 0.3x - 1.2y = 6.4 and with y-intercept 1.8

The line through (8,-2) perpendicular to the line y = 7-2x

The perpendicular bisector of the segment joining (0,3) and (-4,5)

Given A (2,0) and B (8,4), show that P(3,5) is on the perpendicular bisector of Line AB by these two methods.

Method 1. Show Line AB ┴ Line PM where M is the midpoint of Line AB.
Method 2. Show PA = PB

Triangle PQR has vertices P(4,-1), Q (-2,7) and R (9,9).

a. Find an equation of the median from R.
b. Find an equation of the altitude from R.
c. Are your answers in (a) and (b) the same? Is triangle PQR isosceles?

Answer 1

The first six questions are easy; it's the last two that are most interesting. We'll do those two.

Method 1. The slope of AB = 4/6 = 2/3. The midpoint M is at (5,2). The slope of PM is (5-2)/(3-5) = -3/2, which is the negative reciprocal of the slope of AB. Therefore, PM and AB are perpendicular.

Method 2. PA^2 = 1 +5^2 = 26. PB^2 = (-5)^2 + 1 = 26. Therefore PA = PB; triangle APB is isosceles, and P lies on the perpendicular bisector of AB.

The midpoint of PQ is S = (1,3). The slope of RS is 6/8 = 3/4. Using point-slope, (y-3)/(x-1) = 3/4 ==> 4y - 12 = 3x - 3 ==> 3x - 4y = -9 (standard form of equation for RS) (answer a).

The slope of PQ is 8/(-6) = -4/3, so the slope of the altitude must be 3/4. Using point-slope, the equation of the altitude line through R is (y-9)/(x-9) = 3/4 ==> 4y - 36 = 3x - 27 ==> 3x - 4y = -9 (standard form) (answer b).

Since the equations for the altitude and median are the same, triangle PQR is isosceles (answer c).

Answer 2

Use the formula of line in intercept form
x/a +y/b= 1
Here a=-2 and b=4
x/-2 +y/4=1
-2x +y - 4+0
or 2x -y +4=0
This is one part rest you do your self

Answer 3

i think i can tell for first three
1)The line with y-intercept 4 and x-intercept -2
a)4y-2x=0
2)The line through (-1,4) and (5,8)
a)x/y=-(2/3)
3)The line through (2,-7) and (2,3)
a)-10y=0

Answer 4

hi, there is an worry-loose way of plotting the graph and hence we do it as follows. to charm to a graph you may desire to apply X & Y axes. So enable us to take the valve of the given function to be "y" and hence y = x^2 - 3 - x Now take the values of x and get the values of y and hence plot the standards the standards interior the airplane and you get your graph. shape of the graph is parabola.