Question 1203459
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Answer:   <font color=red>y = 11/5 = 2.2</font>



Explanation


Compute the slope of line BC
{{{m = slope = rise/run = matrix(1,3,"change","in","y")/matrix(1,3,"change","in","x")}}}


{{{m = (y[2] - y[1])/(x[2] - x[1])}}}


{{{m = (2 - (-8))/(-3 - 1)}}}


{{{m = (2 + 8)/(-3 - 1)}}}


{{{m = (10)/(-4)}}}


{{{m = -5/2}}}
The slope of -5/2 means "go down 5 units every time we go 2 units to the right".
In short: "down 5, right 2".


The negative reciprocal slope will have us do two things:
flip the fraction
flip the sign


-5/2 becomes 2/5


The altitude line through point A has a slope of 2/5
Apply point-slope form
{{{y - y[1] = m(x - x[1])}}}


{{{y - y[1] = (2/5)(x - x[1])}}} Plug in perpendicular slope


{{{y - 3 = (2/5)(x - 2)}}} Plug in coordinates of point A


{{{y - 3 = (2/5)x + (2/5)(-2)}}}


{{{y - 3 = (2/5)x - 4/5}}}


{{{y = (2/5)x - 4/5 + 3}}}


{{{y = (2/5)x - 4/5 + 3*(5/5)}}}


{{{y = (2/5)x - 4/5 + 15/5}}}


{{{y = (2/5)x + (-4 + 15)/5}}}


{{{y = (2/5)x + 11/5}}}
This is the equation of the altitude through point A. It is perpendicular to side BC.


Plug in x = 0 to find that y = 11/5 = 2.2 which is the final answer.


Here's the diagram
{{{
drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,-1000),

circle(2,3,0.1),circle(2,3,0.12),circle(2,3,0.14),circle(2,3,0.16),circle(1,-8,0.1),circle(1,-8,0.12),circle(1,-8,0.14),circle(1,-8,0.16),circle(-3,2,0.1),circle(-3,2,0.12),circle(-3,2,0.14),circle(-3,2,0.16),circle(0,2.2,0.1),circle(0,2.2,0.12),circle(0,2.2,0.14),circle(0,2.2,0.16),

locate(2.3,2.7,"A"),locate(1.3,-8.3,"B"),locate(-2.7-1,1.7+1,"C"),locate(0.3,1.9,"D"),

line(2,3,1,-8),line(1,-8,-3,2),line(-3,2,2,3),

line(-2.21034,1.31586,-2.03241,0.87103),line(-2.03241,0.87103,-2.47724,0.6931),

blue(line(2,3,-2.65517,1.13793))
)
}}}
The blue line is the altitude through point A. 
It intersects the y axis at (0, 11/5) = (0,2.2) which is marked as point D.
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