Please I need help, JUST one problem. I need to show my work please!!
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
A line perpendicular to the given line has a slope that is the negative inverse of the reference line.
Rewrite the given equation in the format of y=mx+b, where mi is the slope and b is the y-intercept (the value of y when x = 0.
2x + 3y = 4
3y=-2x+4
y = -(2/3)X + (4/3)
The reference slope is -(2/3). The negative inverse is (3/2), which will be the slope of a perpendicular line. We can write the new line as:
y = (3/2)x + b
Any value of b will still result in a line that is perpendicular. But we want a value of b that will shift the line so that it intersects the point (-3,-5). Simply enter this point in the above equation and solve for b.
y = (3/2)x + b
-5 = (3/2)(-3) + b
-5 = -(9/2) + b
-5 = -4.5 + b
b = - 0.5
The equation of the line that is perpendicular to 2x + 3y = 4 and includes point (-3,-5) is
y = (3/2)x - 0.5
DGFH is a rhombus where E is the midpoint of the lines GE and GF
please look at the image
please look at the image