The coordinates of the vertices of a triangle are \(A(0,0),\ B(8,0),\) and \(C(0,6)\). What is the location of the orthocenter?
\((0,0)\)
The vertices are \(A(0,0)\), \(B(8,0)\) and \(C(0,6)\). Side \(AB\) lies along the x-axis (both points have \(y=0\)) and side \(AC\) lies along the y-axis (both points have \(x=0\)).
The x-axis and y-axis meet at right angles, so \(AB\perp AC\). The triangle is therefore right-angled, with the right angle at vertex \(A(0,0)\).
The orthocentre is the common point of the three altitudes, where each altitude is the line drawn from a vertex perpendicular to the opposite side.
Consider the altitude from \(C\) to side \(AB\). Since \(AC\) is already perpendicular to \(AB\), the leg \(CA\) itself is that altitude, and it passes through \(A\).
Likewise the altitude from \(B\) to side \(AC\) is the leg \(BA\), which also passes through \(A\). Two altitudes already meet at \(A\), so the third must pass through \(A\) as well.
This illustrates the general property: in any right-angled triangle the orthocentre coincides with the vertex holding the right angle.
Hence the orthocentre is at \((0,0)\).
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Important Questions from Co-ordinate Geometry
In which ratio the point (-3, p) divides the line segment joining the points (-5, -4) and (-2, 3)?
In which quadrant both abscissa and ordinate are negative?
Find the slope of the line joining the points (3, -4) and (5, 2).
A1 (x1, y1), A2 (x2, y2), A3 (x3, y3) and A4 (x4, y4) are 4 points, p1 divides A1A2 in the ratio 1 ∶ 1, p2 divides P1A3 in the ratio 1 ∶ 2, P3 divides p2A4 in the ratio 1 ∶ 3. Then the coordinates of p3 are:
If (3, 2) are coordinates of the centroid of a triangle, whose vertices are (0, 4), (-2, 0) and (p, 2), then find the value of p.
