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

Wyatt Anderson · Accepted Answer

endpoints of the altitude are #(2,3) and (4,2)#
length of altitude #=sqrt5#

Given #A(2,3), B(3,5) and C(4,2)#
#=> AB=sqrt((3-2)^2+(5-3)^2)=sqrt5#
#BC=sqrt((4-3)^2+(2-5)^2)=sqrt10#
#AC=sqrt((4-2)^2+(2-3)^2)=sqrt5#
As #AB^2+AC^2=BC^2#,
#=> DeltaABC# is a right-angle triangle, as shown in the diagram.
So line#AC# is the altitude perpendicular to line#AB# from point #C#
#=># endpoints of altitude #= (2,3), and (4,2)#
#=># length of altitude #=AC=sqrt5#

Aurora Ayala · Answer

undefined To find the altitude going through corner C, we first need to determine the equation of the line containing side AB. Then, we can find the perpendicular line passing through corner C, which represents the altitude. Finally, we find the intersection point of these two lines, which will give us the endpoint of the altitude.

The equation of the line passing through points A(2,3) and B(3,5) can be found using the slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept. The slope (m) of AB is (5 - 3) / (3 - 2) = 2.

Using point A(2,3) in the slope-intercept form equation: 3 = 2(2) + b. Solving for b gives us b = -1. Thus, the equation of line AB is y = 2x - 1.

The altitude passing through point C(4,2) must be perpendicular to line AB. Therefore, the slope of the altitude is the negative reciprocal of the slope of line AB, which is -1/2.

Using point C(4,2) and the slope-intercept form, we find the equation of the altitude: 2 = (-1/2)(4) + b. Solving for b gives us b = 4.

Thus, the equation of the altitude passing through point C is y = (-1/2)x + 4.

To find the endpoint of the altitude, we need to find the intersection point of the altitude and line AB. We solve the system of equations formed by the equations of the altitude and line AB.

Substituting y = 2x - 1 into y = (-1/2)x + 4, we get:

2x - 1 = (-1/2)x + 4. Solving for x gives us x = 2.

Substituting x = 2 into the equation of line AB gives us y = 2(2) - 1 = 3.

Therefore, the endpoint of the altitude is (2,3).

To find the length of the altitude, we calculate the distance between point C(4,2) and the endpoint (2,3) using the distance formula:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
          = sqrt((2 - 4)^2 + (3 - 2)^2)
          = sqrt((-2)^2 + (1)^2)
          = sqrt(4 + 1)
          = sqrt(5).

So, the length of the altitude going through corner C is sqrt(5).