For the triangle shown below, find the circumcenter and orthocenter

Mathematics

1 answer:

natta225 [31]2 years ago

5 0

Answer: Circumcenter = $\bold{\bigg(1,-\dfrac{3}{2}\bigg)}$ Orthocenter = (-4, -6)

Step-by-step explanation for Circumcenter:

Step 1: Find the midpoint of a line: I chose (-4, 3) and (-4, -6)

$\bigg(\dfrac{-4-4}{2},\dfrac{3-6}{2}\bigg)=\bigg(\dfrac{-8}{2},\dfrac{-3}{2}\bigg) = \bigg(-4,-\dfrac{3}{2}\bigg)$

Step 2: Find the perpendicular line that passes through that point:

Since it is a vertical line, the perpendicular line is $y=-\dfrac{3}{2}$

Step 3: repeat Steps 1 and 2 for another line: chose (-4, -6) and (6, -6)

$\bigg(\dfrac{-4+6}{2},\dfrac{-6-6}{2}\bigg)=\bigg(\dfrac{2}{2},\dfrac{-12}{2}\bigg) = (1,-6)$

Since it is a horizontal line, the perpendicular line is: x = 1

Step 4: Find the intersection of the two lines $\bigg(y=-\dfrac{3}{2}\ \text{and}\ x = 1\bigg)$

Their point of intersection is: $\bigg(1, -\dfrac{3}{2}\bigg)$

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Step-by-step explanation for Orthocenter:

Step 1: Find the perpendicular slope of a line: I chose (-4, 3) and (-4, -6)

Slope is undefined. Perpendicular slope is 0.

Step 2: Use the Point-Slope formula to find the equation of the line that passes through the vertex that is opposite of the line from Step 1 and has the perpendicular slope (found in Step 1).

Vertex (6, -6) and m⊥ = 0 ⇒ y + 6 = 0(x - 6) ⇒ y = -6

Step 3: repeat Steps 1 and 2 for another line: chose (-4, -6) and (6, -6)

Slope is 0. Perpendicular slope is undefined (x = __ )

Vertex (-4, 3) and m⊥ = undefined ⇒ x = -4

Step 4: Find the intersection of the two lines $(y=-6\ \text{and}\ x = -4)$

Their point of intersection is: (-4, -6)

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where does the graph of the function y=tan(x) have asymptotes? at the values of x where cos(x)=0 at the values of x where sin(x)

Gnom [1K]

Answer: at the values where cos(x) = 0

Justification:

1) tan(x) = sin(x) / cos(x).

2) functions have vertical asymptotes at x = a if Limit of the function x approaches a is + or - infinity.

3) the limit of tan(x) approaches +/- infinity where cos(x) approaches 0.

Therefore, the grpah of y = tan(x) has asymptotes where cos(x) = 0.

You can see the asympotes at x = +/- π/2 on the attached graph. Remember that cos(x) approaches 0 when x approaches +/- (n+1) π/2, for any n ∈ N, so there are infinite asymptotes.