Question

Hey i need help finding the orthocenter for the following points, please show the steps you used. thanks

Answer 1

9514 1404 393

Answer:

  (81/43, -31/43) ≈ (1.884, -0.7209)

Step-by-step explanation:

The orthocenter is the intersection point of the altitudes of the triangle. Since this is not a right triangle, you need to find the equations of the altitude lines, then find where those intersect.

If we call the points A, B, C in the order given, the attached figure shows the altitude lines from B perpendicular to AC, and from C perpendicular to AB.

We can find the equation of the perpendicular line using the form ...

  ∆x·x +∆y·y = c

where ∆x and ∆y are the differences in coordinates between the points on the "side" and the value of constant 'c' is determined by evaluating the expression using x and y as the coordinates of the opposite vertex.

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For line AC, we have ...

  (∆x, ∆y) = C - A = (2, -1) -(-3, 0.5) = (2+3, -1-0.5) = (5, -1.5)

Then the equation of the first altitude line is ...

  5x -1.5y = 6(3) -1.5(3) = 10.5

Multiplying by 2 puts this in standard form:

  10x -3y = 21

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For line AB, we have ...

  (∆x, ∆y) = B - A = (3, 3) -(-3, 0.5) = (3+3, 3-0.5) = (6, 2.5)

Then the equation of the second altitude line is ...

  6x +2.5y = 6(2) +2.5(-1) = 9.5

Multiplying by 2 puts this in standard form:

  12x +5y = 19

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Then the orthocenter is the solution to the system of equations ...

  10x -3y -21 = 0

  12x +5y -19 = 0

This set of equations can be solved a number of ways. One is the "cross-multiplication" method.

  1/(10·5 -12·(-3)) = x/(-3·(-19) -5·(-21)) = y/(-21(12) -(-19)(10))

  x = 162/86 = 81/43

  y = -62/86 = -31/43

The orthocenter is (81/43, -31/43) ≈ (1.884, -0.7209).

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Additional comment

The "cross multiply method" is based on cross products when the equations  are written in standard form:

  ax +by +c = 0

  dx +ey +g = 0

Usually, the coefficients are written in two rows of four, with the first column repeated at the end:

  a  b  c  a

  d  e  g  d

Then cross products are formed, where diagonal down is positive and diagonal up is subtracted from that:

  p1 = (ae -db), p2 = (bg -ec), p3 = (cd -ga)

These give the solution to the equations when put in the form ...

  1/p1 = x/p2 = y/p3   ⇒   x = p2/p1   and   y = p3/p1

If you look at videos of this "cross multiply method," they will put the 1/p1 at the end of the string of equalities, and they will use the coefficients in the order starting with b instead of a. I find the arrangement here easier to remember, so less likely to get confused.

Further comment

You can, of course, find the slopes of the lines, and their opposite reciprocals, then use a point-slope form or some other method for writing the equations of the altitude lines. The solution to the equations can proceed by substitution or elimination. No doubt these methods are somewhat familiar. The method used here tries to get to the answer much more quickly with a lot less writing. (Unfortunately, it takes a bit of writing to explain it all.)