For this case we have that by definition, the equation of the line of the slope-intersection form is given by:
Where:
m: It is the slope of the line
b: It is the cut-off point with the y axis
According to the statement data we have:
Then, the equation is of the form:
We substitute the given point and find "b":
Finally, the equation is of the form:
Answer:
Answer:
P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.
Step-by-step explanation:
P and Q are the intersection points of
x-y+1 = 0 and the circle x^2 + y^2 = 25
sub y = x+1 into the circle
x^2 + (x+1)^2 = 25
x^2 + x^2 + 2x + 1 - 25 = 0
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3 or x = -4
y = 4 or y = -3
so P(3,4) and Q(-4,3) are our two points
Height of triangle.
h = |0 - 0 + 1|/√2 = 1/√2
PQ = √( (-7)^2 + 1^2) = √50 = 5√2
area POQ = (1/2)(1/√2)(5√2) = 5/2 square units
hope this helped
I'm pretty sure to solve this all you do is subtract 77.95 - 75.65, which would be 2.30. Hope that helps!
Not sure what the question is, but I'm guessing you want to find out the area of the rectangle.
The distance between points P and Q is 4.
The distance between points Q and R is 3.
So the area is 3 * 4 = 12.
