A triangle has corners A, B, and C located at #(3 ,4 )#, #(2 ,5 )#, and #(2 , 1 )#, respectively. What are the endpoints and length of the altitude going through corner C?

Question

A triangle has corners A, B, and C located at #(3   ,4   )#, #(2  ,5    )#, and #(2 ,  1   )#, respectively. What are the endpoints and length of the altitude going through corner C?

Jayden Jackson · Accepted Answer

Length of altitude #=2\sqrt2#

End points of altitude from vertex #C# are #(2, 1)# & #(4, 3)# The vertices of #\Delta ABC# are #A(3, 4)#, #B(2, 5)# & #C(2, 1)# The area #\Delta# of #\Delta ABC# is given by following formula #\Delta=1/2|3(5-1)+2(1-4)+2(4-5)|# #=2# Now, the length of side #AB# is given as #AB=\sqrt{(3-2)^2+(4-5)^2}=\sqrt2# If #CN# is the altitude drawn from vertex #C# to the side #AB# then the area of #\Delta ABC# is given as #\Delta =1/2(CN)(AB)# #2=1/2(CN)(\sqrt2)# #CN=2\sqrt2# Let #N(a, b)# be the foot of altitude CN drawn from vertex #C(2, 1)# to the side #AB# then side #AB# & altitude #CN# will be normal to each other i.e. the product of slopes of AB & CN must be #-1# as follows #\frac{b-1}{a-2} imes \frac{5-4}{2-3}=-1# #a=b+1\ ............(1)# Now, the length of altitude CN is given by distance formula #\sqrt{(a-2)^2+(b-1)^2}=2\sqrt2# #(b+1-2)^2+(b-1)^2=(2\sqrt2)^2# #(b-1)^2=4# #b=3, -1# Setting these values of #b# in (1), we get the corresponding values of #a# as follows #a=3+1=4\ \ & \ \ \ a=-1+1=0# #a=4, 0# The endpoints of altitude from vertex #C(2, 1)# are #(4, 3)# & #(0, -1)# But #(0, -1)# is not the end point of altitude. hence, the end points of altitude from vertex #C# are #(2, 1), (4, 3)#

Sadie Adams · Answer

undefined To find the endpoints of the altitude going through corner C, you first need to determine the equation of the line containing side AB. Then, find the perpendicular line passing through C. The intersection point of these two lines will give you the endpoints of the altitude. Finally, calculate the distance between this intersection point and corner C to find the length of the altitude.

1. Calculate the slope of line AB using the formula: slope = (y2 - y1) / (x2 - x1).
   Slope of AB = (5 - 4) / (2 - 3) = 1 / -1 = -1.

2. Use the point-slope form of a line to find the equation of line AB. Choose any point on AB. Using (3, 4):
   y - 4 = -1(x - 3)
   y - 4 = -x + 3
   y = -x + 7.

3. The slope of a line perpendicular to AB is the negative reciprocal of the slope of AB. So, the slope of the altitude is 1.

4. Using the point-slope form of a line with slope 1 passing through point C(2, 1):
   y - 1 = 1(x - 2)
   y - 1 = x - 2
   y = x - 1.

5. Solve the system of equations formed by equating the equations of lines AB and the perpendicular line:
   -x + 7 = x - 1
   2x = 8
   x = 4.

6. Substitute x = 4 into either equation to find y:
   y = 4 - 1 = 3.

7. The intersection point is (4, 3), which is the endpoint of the altitude.

8. Calculate the distance between (4, 3) and C(2, 1) using the distance formula:
   Distance = √((x2 - x1)^2 + (y2 - y1)^2)
            = √((4 - 2)^2 + (3 - 1)^2)
            = √(2^2 + 2^2)
            = √8
            = 2√2.

Therefore, the endpoints of the altitude going through corner C are (4, 3) and the length of the altitude is 2√2.