3) The value of x = -2 for AB||CD
4) The equation is y = x/2 - 8.
5) The equation of the line is y = 5x + 9
6) The equation of the line is y = -4x/5 + 3.
7) The equation of line l is y = -x/2 + 9/2.
How to write equation of line in slope-intercept form.
3) Given
A(-4,x), B(-3,2),C(1,-13) and D(5,3)
For AB to be parallel to CD their slope must be equal.
Slope = ∆y/∆x
slope AB = 2 - x/-3-(-4)
= 2 - x
slope CD = 3 -(-13)/5-1
= 16/4 = 4
For parallelism
2 - x = 4
x = -2
The value of x = -2 for AB||CD
4) For (4,-6) and slope = 1/2
Equation of a line is given as
y = mx + b
where
m is slope
b is y-intercept
y - y1 = m(x - x1)
y - (-6) = 1/2(x - 4)
y + 6 = x/2 - 2
y = x/2 - 2 - 6
y = x/2 - 8
The equation is y = x/2 - 8.
5) passes through (-2,-1) and parallel to y = 5x + 1
Since the lines are parallel, their slope is equal
slope of the line = 5
using point (-2,-1)
y - (-1) = 5(x - (-2)
y + 1 = 5x + 10
y = 5x + 10 - 1
y = 5x + 9
The equation of the line is y = 5x + 9
6) From the graph
Locate points (0,3) and (5,-1)
slope= -1 -3/5 -0
= -4/5
When x = 0, y = 3
The y-intercept = 3
y = -4x/5 + 3
The equation of the line is y = -4x/5 + 3.
7) Given
A(-1,5) and B(1,9)
Slope of AB = 9 - 5/1 - (-1)
m = 4/2 = 2
let the slope of line l be m1
For perpendicularity
m * m1 = -1
2 * m1 = -1
m1 = -1/2
Using point (-1,5)
y - 5 = -1/2(x + 1)
y - 5 = -x/2 -1/2
y = -x/5 - 1/2 + 5
y = -x/2 + 9/2
The equation of line l is y = -x/2 + 9/2.