a. y = 0 at the x intercept,
x = 0 at the y intercept,
Therefore: (3 , 0) , (0 , -2),
Slope = (-2-0) / (0-3) = -2/-3 = 2/3,
y = mx + b,
0 = (2/3)*3 + b,
0 = 6/3 + b,
b = -2.
Eq: y = (2/3)x -2.
b. On a vertical line, x is the same for all values of y.In other words, x is constant. On the given vertical line,x = 5. Therefore, x = 5 for all values of y:
(5 , -1) , (5 , 2) , (5 , 4). All of these points are on the required vertical line.
Eq: x = 5.
c. Given: 3x - 4y = 5, (3 , -6).
m1 = m2 = -A/B = -3/-4 = 3/4,
Y = mx + b,
-6 = (3/4)*3 + b,
-6 = 9/4 + b,
-6 -9/4 = b,
-24/4 -9/4 = b,
b = -33/4,
Eq: y = (3/4)x - 33/4.
d. Given: 4x - 6y = -3.
Slope = -A/B = -4/-6 = 4/6 = 2/3,
Let x = 0:,
4*0 - 6y = -3,
-6y = -3,
y = -3/-6 = 3/6 = 1/2 = y-intercept
.
Write an equation of the line satisfying the given condition.
a) x intercept = 3 y intercept=-2
I am really not sure how to start here.
b) vertical line containing (5,-1)
Would it be y=5x-1?
c) parallel to the line 3x-4y=5 and containing the points (3,-6) .
I know that the line has to have the same slope as the one about.
d) Find the slope and y intercept of the line 4x-6y=-3
To find the y value would i plug in zero for the x and solve? And then I can find the slope?
Thank You for helping!!
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