Forms for the equation of a straight line
Suppose that we have the graph of a straight line and that we wish to find its equation. (We will assume that the graph has x and y axes and a linear scale.) The equation can be expressed in several possible forms. To find the equation of the straight line in any form we must be given either:
two points, (x1, y1) and (x2, y2), on the line; or
one point, (x1, y1), on the line and the slope, m; or
the y intercept, b, and the slope, m.
In the first case where we are given two points, we can find m by using the formula:
Once we have one form we can easily get any of the other forms from it using simple algebraic manipulations. Here are the forms:
1. The slope-intercept form:
y = m x + b.
The constant b is simply the y intercept of the line, found by inspection. The constant m is the slope, found by picking any two points (x1, y1) and (x2, y2) on the line and using the formula:
2. The point-slope form:
y − y1 = m (x − x1).
(x1, y1) is a point on the line. The slope m can be found from a second point, (x2, y2), and using the formula:
3. The general form:
a x + b y + c = 0.
a, b and c are constants. This form is usually gotten by manipulating one of the previous two forms. Note that any one of the constants can be made equal to 1 by dividing the equation through by that constant.
4. The parametric form:
The area of polygon MNOPQR = Area of a rectangle that is 15 square units + Area of a rectangle that is 2 square units.
In the given polygon MNOPQR, side MN is parallel to side RQ and the side MR is parallel to side PQ
We will draw a perpendicular line from point O on the side RQ, which will intersect RQ at point S. So, we can now divide the whole polygon into two different rectangles MNSR and OPQS with the areas as A₁ and A₂ respectively.
In rectangle MNSR, length(MN) = 5 units and width (MR) = 3 units
According to the formula for Area of rectangle,
A₁ = (length)×(width)
A₁ = (5 units)×(3 units)
A₁ = 15 square units
Now in rectangle MNSR, side MN= side RS and side MR = side NS,
so RS= 5 units and NS= 3 units
That means, SQ= RQ- RS = 7-5 = 2 units
and OS= NS - NO = 3- 2 = 1 unit
In rectangle OPQS, we have length(SQ) = 2 units and width(OS) = 1 unit
So, A₂ = (length)×(width)
A₂ = (2 units)×(1 unit)
A₂ = 2 square units
So, the area of polygon MNOPQR = (Area of a rectangle that is 15 square units + Area of a rectangle that is 2 square units)
Since you didn't provide the points, we can't find the exact answer. However, I can explain how to do the problem for you.
To test the point, just plug in the x and y values and see it the statement is true or false.
Let's try: (3, 4)
Plug in the points and evaluate:
2(4) = 2 + 3
8 = 6
FALSE
Since this is a false statement, then the point is not on the line.
36, |-12|=12 and |-3|= 3. Using inverse operations, we multiply instead of dividing and get 36.
