What is the distance between two points with coordinates at (8, 40) and (20, 5)?
2 answers:
You might be interested in
Answer:
√65/2 square units
Step-by-step explanation:
Find the diagram attached
Area of the triangle = 1/2 * base * height
To get the base and height, we will use the distance formula;
Area of the triangle LMN = 1/2 * LM * MN
For MN
Given the coordinate M(3, 2) and N(1, 3)
MN = √(3-2)²+(1-3)²
MN = √1²+(-2)²
MN = √1+4
MN = √5
For LM;
Given the coordinate L(1, -1) and M(3, 2)
LM = √(2-(-1))²+(3-1)²
LM = √(2+1)²+(2)²
LM = √3²+(2)²
LM = √9+4
LM = √13
Area of the triangle LMN = 1/2 * √13 * √5
Area of the triangle LMN = √65/2 square units
we have
This is a vertical parabola open upward
The axis of symmetry is equal to the x-coordinate of the vertex
The vertex is the point (h,k)
the equation of the axis of symmetry is
In this problem
so the x-coordinate of the vertex is
<u>Convert the quadratic equation in vertex form</u>
Group terms that contain the same variable, and move the constant to the opposite side of the equation
Factor the leading coefficient
Complete the square. Remember to balance the equation by adding the same constants to each side
Rewrite as perfect squares
remember that
so
substitute in the equation
therefore
<u>the answer is</u>