The Geometry of Concurrency
Finding the Orthocentre
Welcome, future engineer! Today, we are not just solving a coordinate geometry problem; we are uncovering the hidden architecture of a triangle.
We are tasked with finding the orthocentre of ΔABC with vertices A(−1,7), B(−7,1), and C(5,−5). The orthocentre, denoted as H, is the point where the three altitudes of a triangle meet. It is a point of perfect balance, a geometric center of gravity for the altitudes.
Phase 1
The Logic of Perpendicularity
To find the intersection of two lines, we need their equations. An altitude is a line segment from a vertex perpendicular to the opposite side.
Let's start with the altitude from vertex A to side BC. To find its equation, we first need the slope of side BC. Using the slope formula m=x2−x1y2−y1, we substitute the coordinates of B(−7,1) and C(5,−5):
mBC=5−(−7)−5−1=12−6=−21
Now, here is the magic. Since our altitude AD is perpendicular to BC, their slopes must satisfy the condition mAD⋅mBC=−1.
Thus, the slope of AD is the negative reciprocal of −21, which is 2. With a point A(−1,7) and a slope m=2, we use the point-slope form y−y1=m(x−x1):
y−7=2(x−(−1))⇒y−7=2x+2⇒2x−y+9=0
Phase 2
The Second Altitude
We repeat this elegant process for the altitude from vertex B to side AC. First, we calculate the slope of AC:
mAC=5−(−1)−5−7=6−12=−2
Since the altitude BE is perpendicular to AC, its slope mBE must be the negative reciprocal of −2, which is 21. Now, using point B(−7,1) and slope 21:
y−1=21(x−(−7))⇒2(y−1)=x+7⇒x−2y+9=0
Phase 3
The Intersection
We now have a system of two linear equations:
1) 2x−y+9=0
2) x−2y+9=0
From the first equation, we can write y=2x+9. Substituting this into the second equation:
Finally, substituting x=−3 back into y=2x+9, we get y=2(−3)+9=3.
The orthocentre H is at (−3,3). You have successfully navigated the geometry and the algebra to find the heart of the triangle. Keep this systematic approach in your toolkit, and no coordinate geometry problem will ever stand in your way!