The equation
3x2−5xy+2y2=0 can be written as two factors:
(3x−2y)(x−y)=0.
This means the two sides of the triangle are
3x−2y=0 and
x−y=0.
The slope of
3x−2y=0 is
m1=, and the slope of
x−y=0 is
m2=1.
To find where the triangle's heights (altitudes) go from the orthocentre at
(2,1), we need the slope of each altitude. The altitude to
3x−2y=0 will have a slope that is the negative reciprocal of
, which is
−.
The altitude to
x−y=0 will have a slope that is the negative reciprocal of 1 , which is -1 .
Write the equation for the altitude with slope
− that passes through
(2,1) :
y−1=−(x−2)⇒3y−3=−2x+4⇒2x+3y=7Write the equation for the altitude with slope -1 that passes through
(2,1) :
y−1=−1(x−2)⇒y−1=−x+2⇒x+y−3=0Find where each altitude meets the opposite side. First, where
2x+3y=7 (altitude) meets
x−y=0 (side): Since
x=y, substitute it into
2x+3x=7, so
5x=7 which means
x=.
The intersection is at
(,).
From
3x−2y=0,x=y.
Put this in
x+y−3=0 to get
y+y=3⟹y=3⟹y=.
When
y=,x=, so the intersection point is
(,).
The third side of the triangle must be the line connecting these two intersection points:
(,) and
(,).
Find the slope of this line:
m===−2.
The equation for this line, using point-slope form with the point
(,) and slope -2 , is:
y−=−2(x−)⇒5y−7=−10x+14⇒10x+5y−21=0So, the equation of the third side is
10x+5y−21=0.
