The 12 department chairs at Cedar H.S. represent the teaching staff. The principal assigns each chair a number and uses a random
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A line parallel to the given one will have the same slope, 0.5. For the purpose here, it is convenient to start with a point-slope form of the equation, then simplify. For slope m and point (h, k), the equation of the line can be written as
... y = m(x -h) +k
We have m=0.5, (h, k) = (-9, 12), so the equation is ...
... y = 0.5(x +9) +12
... y = 0.5x +16.5
Answer:
Circumcenter =(-1,0)
Orthocenter =(2,-3)
Step-by-step explanation:
Given : Points A = (2,3), B = (-4,-3), C = (2,-3)
Formula used :
→Mid point of two points-
→Slope of two points -
→Perpendicular of a line =
Circumcenter- The point where the perpendicular bisectors of a triangle meets.
Orthocenter-The intersecting point for all the altitudes of the triangle.
To find out the circumcenter we have to solve any two bisector equations.
We solve for line AB and AC
So, mid point of AB =
Slope of AB =
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of the perpendicular bisector = -1
Equation of AB with slope -1 and the coordinates (-1,0) is,
(y – 0) = -1(x – (-1))
y+x=-1………………(1)
Similarly, for AC
Mid point of AC =
Slope of AC =
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of the perpendicular bisector = 0
Equation of AC with slope 0 and the coordinates (2,0) is,
(y – 0) = 0(x – 2)
y=0 ………………(2)
By solving equation (1) and (2),
put y=0 in equation (1)
y+x=-1
0+x=-1
⇒x=-1
So the circumcenter(P)= (-1,0)
To find the orthocenter we solve the intersections of altitudes.
We solve for line AB and BC
So, mid point of AB =
Slope of AB =
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of CF = -1
Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF
(y – 0) = -1(x – (-1))
y+x=-1………………(3)
Similarly, mid point of BC =
Slope of AB =
Slope of the bisector is the negative reciprocal of the given slope.
So, the slope of AD = 0
Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD
(y-(-3)) = 0(x – (-1))
y+3=0
y=-3………………(4)
Solve equation (3) and (4),
Put y=-3 in equation (3)
y+x=-1
-3+x=-1
x=2
Therefore, orthocenter(O)= (2,-3)
