A triangle has corners A, B, and C located at #(4 ,7 )#, #(3 ,2 )#, and #(2 ,4 )#, 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 #(4  ,7  )#, #(3  ,2    )#, and #(2 ,4   )#, respectively. What are the endpoints and length of the altitude going through corner C?

Avery Allen · Accepted Answer

End points of altitude CD #(2,4) & (-7,97/26)#
Length of altitude CD =13.39#

Equation of AB is obtained using coordinates of A & B #(y-7)/(2-7)=(x-4)/(3-4)# #(y-7)/(-5)=(x-4)/(-1)# #y-7=5(x-4)# #y=5x-13# Slope of AB #m1=5# Slope of Altitude to line AB and passing through point C be m2. #m2=-(1/m1)=-(1/5)# Equal of Altitude passing through point C is #(y-4)=-(1/5)*(x-2)# #(-5y)+20=x-2# #x+5y=22# Solving equations of AB & Altitude through C gives base of altitude point D. #y=5x-13# #-5x+y=-13# Eqn (1) #x+5y=22# #5x+25y=110# Eqn (2) Adding (1) & (2), #26y=97# #y=97/26# #-5x+(97/26)=-13# #-5x=-(97+338)/13=-(435/13)=-35# #x=-7# Coordinates of point D(-7,97/26) Length of altitude CD is #=sqrt((2-(-7))^2+(4-(97/26))^2)# #sqrt(81+98.38)=13.39#

Aaliyah Anderson · Answer

undefined To find the endpoints and length of the altitude going through corner C of the triangle, we first need to find the equation of the line containing side AB. Then, we find the equation of the perpendicular line passing through point C. The intersection of these two lines gives us the endpoint of the altitude. Finally, we can calculate the length of the altitude using the distance formula.

1. Find the equation of the line containing side AB using the two given points A(4,7) and B(3,2):
   Slope of AB, $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 7}{3 - 4} = -5$.
   Using point-slope form, the equation of AB is $y - y_1 = m_{AB}(x - x_1)$.
   Substituting the coordinates of point A, we get $y - 7 = -5(x - 4)$ which simplifies to $y = -5x + 27$.

2. Since the altitude passing through C is perpendicular to AB, the slope of the altitude is the negative reciprocal of the slope of AB. Thus, slope of the altitude $m_{\perp} = \frac{1}{5}$.

3. Using point-slope form with point C(2,4), the equation of the altitude line is $y - 4 = \frac{1}{5}(x - 2)$, which simplifies to $y = \frac{1}{5}x + \frac{18}{5}$.

4. To find the intersection point of AB and the altitude, we solve the system of equations formed by equating the equations of AB and the altitude.
   $ -5x + 27 = \frac{1}{5}x + \frac{18}{5} $
   Solving for $x$, we get $x = \frac{21}{2}$. 
   Substituting $x = \frac{21}{2}$ into either equation, we find $y = \frac{27}{2}$.

5. Thus, the endpoint of the altitude is $\left(\frac{21}{2}, \frac{27}{2}\right)$.

6. Finally, we use the distance formula to find the length of the altitude between point C and the endpoint.
   $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
   $d = \sqrt{\left(2 - \frac{21}{2}\right)^2 + \left(4 - \frac{27}{2}\right)^2}$
   $d = \sqrt{\left(-\frac{17}{2}\right)^2 + \left(-\frac{19}{2}\right)^2}$
   $d = \sqrt{\frac{289}{4} + \frac{361}{4}}$
   $d = \sqrt{\frac{650}{4}}$
   $d = \frac{5\sqrt{26}}{2}$.

Therefore, the endpoints of the altitude passing through corner C are $\left(\frac{21}{2}, \frac{27}{2}\right)$ and $\left(2,4\right)$, and its length is $\frac{5\sqrt{26}}{2}$.