Question

The orthocentre of the triangle formed by the points \((0,3),(0,0)\) and \((1,0)\) is (1) \(\left(\frac{1}{3}, 1\right)\) (2) \(\left(\frac{1}{2}, \frac{3}{2}\right)\) (3) \((0,0)\) (4) \(\left(0, \frac{3}{2}\right)\)

Short answer

(3) (0,0)

Step-by-step solution

Determine the vertices and sides of the triangle

Let's name the vertices of the triangle: \(A(0,3)\), \(B(0,0)\) and \(C(1,0)\). The sides of the triangle are: \(\overline{AB}, \overline{AC},\) and \(\overline{BC}\).

Find the equations of the altitudes

To find the equations of the altitudes, we need to find the slopes of the sides of the triangle, their negatives reciprocal slopes, and the feet of the altitudes. The slope of \(\overline{AB}\), \(m_{AB} = \frac{0 -3}{0-0} = undefined\) (This is a vertical line with the equation: \(x = 0\)) The slope of \(\overline{AC}\), \(m_{AC} = \frac{0 -3}{1-0} = -3\) The slope of \(\overline{BC}\), \(m_{BC} = \frac{3 -0}{0-1} = -3\) (Slope negative reciprocal: \(\frac{1}{3}\)) Now, let's find the feet of the altitudes: Altitude from vertex \(A\): It's perpendicular to side \(\overline{BC}\). Therefore, its slope is \(\frac{1}{3}\). Hence, the equation of the altitude is \(y-3 = \frac{1}{3}(x-0) \implies y = \frac{1}{3}x +3\) Altitude from vertex \(B\): It's perpendicular to side \(\overline{AC}\). Since \(\overline{AB}\) is vertical and passes through \((0,0)\), the equation of the altitude is \(x = 0\). This is the same as the side \(\overline{AB}\). Altitude from vertex \(C\): It's perpendicular to side \(\overline{AB}\). Since \(\overline{AB}\) is vertical, the altitude would be a horizontal line passing through \(C(1,0)\). The equation is \(y = 0\).

Find the point of intersection

To find the orthocentre, we need to find the point that satisfies the three altitude equations simultaneously. Let's analyze them: 1. Altitude from vertex A: \(y = \frac{1}{3}x +3\) 2. Altitude from vertex B: \(x = 0\) 3. Altitude from vertex C: \(y = 0\) We can notice that the orthocentre lies at the intersection of the altitude from vertex A and the altitude from vertex B (the corner of the triangle). Let's find the intersection point: In altitude 1, substitute the equation of altitude 2: \(y = \frac{1}{3}(0) + 3\) \(y = 3\) Thus, the orthocentre is \((0, 3)\), but this is not present in the given options. However, considering the triangle's vertices, we can notice that the triangle is a right triangle with the 90-degree angle at vertex B. For any right-angled triangle, the orthocentre is the vertex where the 90-degree angle is located. Hence, options (1), (2), and (4) are incorrect, and the orthocentre of the triangle is: (3) \((0,0)\)

Key concepts

Altitude of a Triangle

In geometry, the altitude of a triangle is a straight line through a vertex and perpendicular to (i.e., forming a right angle with) the opposite side or the extension of the opposite side. Altitudes are important when determining the characteristics of a triangle, such as its area. In fact, for any given triangle, there are exactly three altitudes, which may all lie inside the triangle, on a triangle side, or outside the triangle, depending on the type of triangle.

Take, for instance, the triangle in our exercise with vertices at \(A(0,3)\), \(B(0,0)\), and \(C(1,0)\). The altitudes from vertices A, B, and C are essential for finding the orthocentre—the point where all three altitudes intersect. The altitude from vertex B coincides with side \(AB\), which is an extraordinary case because the triangle happens to be right-angled with the right angle at vertex B.

It's particularly important to note that the procedure for finding altitudes involves knowing how to determine the slope of a line in coordinate geometry first, and then finding the perpendicular slope to that line. This perpendicular line will represent the altitude from the corresponding vertex.

Slope of a Line in Coordinate Geometry

Understanding the slope of a line in coordinate geometry is fundamental for solving many types of mathematical problems. The slope is a measure of the steepness of a line and is defined as the ratio of the vertical change (rise) to the horizontal change (run) between two points on a line. If you have two points, \((x_1,y_1)\) and \((x_2,y_2)\), the slope \(m\) is calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For vertical lines, however, the slope is undefined, and for horizontal lines, the slope is 0. In our exercise, we can see these concepts in action. The line \(AB\), being vertical, has an undefined slope, meaning it does not have a 'rise' over 'run' relationship. Conversely, \(AC\) and \(BC\) both have slopes that we can calculate using the points that form these lines.

When finding the altitude of a triangle, as in our exercise, we must use the concept of the slope to derive the equations of the altitudes. The slope of the altitude is the negative reciprocal of the slope of the side it's perpendicular to. For example, since the slope of side \(AC\) is \(-3\), the slope of the altitude from vertex B, which is perpendicular to \(AC\), would be \(\frac{1}{3}\).

Right-Angled Triangle Properties

Properties of right-angled triangles are useful in a wide range of geometric problems. A right-angled triangle is one in which one of the angles is exactly 90 degrees. This allows for the use of trigonometric ratios and the famous Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
\[ a^2 + b^2 = c^2 \]
In our exercise, the right angle is located at \(B(0,0)\). A key property that applies here is that in a right-angled triangle, the orthocentre falls exactly on the vertex that has the right angle. Therefore, without performing complex calculations, we can conclude that the orthocentre is \((0,0)\) for the given right-angled triangle. This serves as a quick check and is a great example of how understanding fundamental properties can simplify calculations and offer faster routes to solutions.

Moreover, the altitudes in a right-angled triangle have special relations. One altitude aligns with the hypotenuse, while the other two are the legs (sides that form the right angle) themselves. Thus, in some cases, unveiling the nature of a triangle simplifies the problem significantly, as illustrated in finding the orthocentre of a right-angled triangle.