Answer:
Orthocentre (intersection of altitudes) is at (37/10, 19/5)
Explanation:
Given three vertices of a triangle
A(-5, -2)
B(7, -5)
C(3, 1)
Solution A by geometry
Slope AB = (yb-ya) / (xb-xa) = (-5-(-2)) / (7-(-5)) = -3/12 = -1/4
Slope of line normal to AB, nab = -1/(-1/4) = 4
Altitude of AB = line through C normal to AB
(y-yc) = nab(x-xc)
y-1 = (4)(x-3)
y = 4x-11 .........................(1)
Slope BC = (yc-yb) / (xc-yb) = (1-(-5) / (3-7)= 6 / (-4) = -3/2
Slope of line normal to BC, nbc = -1 / (-3/2) = 2/3
Altitude of BC
(y-ya) = nbc(x-xa)
y-(-2) = (2/3)(x-(-5)
y = 2x/3 + 10/3 - 2
y = (2/3)(x+2) ........................(2)
Orthocentre is at the intersection of (1) & (2)
Equate right-hand sides
4x-11 = (2/3)(x+2)
Cross multiply and simplify
12x-33 = 2x+4
10x = 37
x = 37/10 ...................(3)
substitute (3) in (2)
y = (2/3)(37/10+2)
y=(2/3)(57/10)
y = 19/5 ......................(4)
Therefore the orthocentre is at (37/10, 19/5)
Alternative Solution B using vectors
Let the position vectors of the vertices represented by
a = <-5, -2>
b = <7, -5>
c = <3, 1>
and the position vector of the orthocentre, to be found
d = <x,y>
the line perpendicular to BC through A
(a-d).(b-c) = 0 "." is the dot product
expanding
<-5-x,-2-y>.<4,-6> = 0
simplifying
6y-4x-8 = 0 ...................(5)
Similarly, line perpendicular to CA through B
<b-d>.<c-a> = 0
<7-x,-5-y>.<8,3> = 0
Expand and simplify
-3y-8x+41 = 0 ..............(6)
Solve for x, (5) + 2(6)
-20x + 74 = 0
x = 37/10 .............(7)
Substitute (7) in (6)
-3y - 8(37/10) + 41 =0
3y = 114/10
y = 19/5 .............(8)
So orthocentre is at (37/10, 19/5) as in part A.