A triangle has corners A, B, and C located at #(5 ,2 )#, #(2 ,5 )#, and #(8 ,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 #(5  ,2  )#, #(2  ,5    )#, and #(8 ,4   )#, respectively. What are the endpoints and length of the altitude going through corner C?

Mason Adams · Accepted Answer

End points of Altitude CD are #(8,4)# and #(11/2,3/2)#
Length of Altitude #sqrt50/2# Let CD be the altitude drawn perpendicular to line AB of the triangle from point C. The slope of line AB # (5-2)/(2-5)=3/-3=-1 :.# the slope of altitude CD is #(-1)/(-1)=1#(condition of perpendicularity). The equation of line AB is #y-2= (-1)(x-5) = 5-x or y+x = 7 (1)# The equation of line CD is #y-4= 1(x-8) =x-8 or y-x = -4 (2)# Solving equation (1) & (2) we get #x=11/2 ; y= 3/2# So end points of Altitude CD are #(8,4)# and #(11/2,3/2)# Length of Altitude # sqrt((8-11/2)^2+(4-3/2)^2)=sqrt(25/4+25/4)=sqrt50/2#[Ans]

Sadie Adams · Answer

undefined To find the endpoints and length of the altitude going through corner C of the triangle, we first need to determine the equation of the line containing the side AB of the triangle. Then, we can find the point where this line intersects the line passing through point C and perpendicular to side AB, which will give us one endpoint of the altitude.

Let's start by finding the equation of the line passing through points A and B.

The slope of the line AB is given by:

$$m_{AB} = \frac{y_B - y_A}{x_B - x_A}$$

Substitute the coordinates of points A and B:

$$m_{AB} = \frac{5 - 2}{2 - 5} = \frac{3}{-3} = -1$$

Using point-slope form, the equation of the line passing through A and B is:

$$y - y_A = m_{AB}(x - x_A)$$

Substitute the coordinates of point A:

$$y - 2 = -1(x - 5)$$
$$y - 2 = -x + 5$$
$$y = -x + 7$$

Now, we need to find the equation of the line perpendicular to AB passing through point C. The slope of this line will be the negative reciprocal of the slope of AB, which is $m_{AB} = -1$.

Using point-slope form with point C, we have:

$$y - y_C = -\frac{1}{m_{AB}}(x - x_C)$$

Substitute the coordinates of point C:

$$y - 4 = -\frac{1}{-1}(x - 8)$$
$$y - 4 = x - 8$$
$$y = x - 4$$

Now, we solve the system of equations formed by the line AB and the line passing through C:

$$y = -x + 7$$
$$y = x - 4$$

By solving this system, we find the point of intersection which represents one endpoint of the altitude:

$$x - 4 = -x + 7$$
$$2x = 11$$
$$x = \frac{11}{2}$$

Substitute $x = \frac{11}{2}$ into either equation to find $y$:

$$y = \frac{11}{2} - 4$$
$$y = \frac{11}{2} - \frac{8}{2}$$
$$y = \frac{3}{2}$$

So, one endpoint of the altitude is $\left(\frac{11}{2}, \frac{3}{2}\right)$.

To find the other endpoint, we use point C, which is $(8, 4)$. Therefore, the other endpoint of the altitude is $(8, 4)$.

Finally, we calculate the length of the altitude using the distance formula between the two endpoints:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Substitute the coordinates of the two endpoints:

$$d = \sqrt{\left(\frac{11}{2} - 8\right)^2 + \left(\frac{3}{2} - 4\right)^2}$$
$$d = \sqrt{\left(\frac{11}{2} - \frac{16}{2}\right)^2 + \left(\frac{3}{2} - \frac{8}{2}\right)^2}$$
$$d = \sqrt{\left(-\frac{5}{2}\right)^2 + \left(-\frac{5}{2}\right)^2}$$
$$d = \sqrt{\frac{25}{4} + \frac{25}{4}}$$
$$d = \sqrt{\frac{50}{4}}$$
$$d = \sqrt{\frac{25}{2}}$$
$$d = \frac{\sqrt{25}}{\sqrt{2}}$$
$$d = \frac{5}{\sqrt{2}}$$

Therefore, the length of the altitude going through corner C is $\frac{5}{\sqrt{2}}$ units.