An altitude of a triangle is a segment that passes through one vertex and is perpendicular to the opposite side. find the standard form of the equation if the line containing each altitude of (triangle)
a(7,10)
b(10,-5)
c(4,-5)

Step-by-step explanation:

A geometry app can make short work of this. The result from one of them is attached. To put the equations into standard form, the leading (x) coefficient must be positive, so the equations shown need to be multiplied by -1.

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A: Segment BC is horizontal, so the altitude through A is the vertical line at the same x-coordinate:

  x = 7

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B: Segment AC has slope Δy/Δx = (10-(-5))/(7-4) = 15/3 = 5/1. The equation of the perpendicular line* through point (h, k) for slope a/b can be written ...

  b(x -h) +a(y -k) = 0

and this can be simplified to ...

  bx +ay = bh +ak

For the altitude through point B, we have a=5, b=1, h=10, k=-5, so the equation is ...

  x +5y = 1·10 +5(-5)

  x + 5y = -15

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C: We note from symmetry, the graph, or from computation that the slope of segment AB is -5. Then the equation of the line through C perpendicular to AB is ...

  1·x -5·y = (1)(4) +(-5)(-5)

  x - 5y = 29

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* The formula shown here is useful for many problems.

        bx +ay = bh +ak

Similarly, ...

        ax -by = ah -bk

d

step-by-step explanation:

kk.bj.

step-by-step explanation:

.mnm.kn.kj