Area of triangle

Question 1: Area of a Triangle and Collinearity

The Question: Choose the correct statement based on the three points $P(0, 10)$, $Q(-20, -30)$, and $R(10, 30)$.

• The given points form a triangle of area 5 square units
• The given points form a triangle of area 15 square units
• The given points do not form a triangle
• None of the above

Core Concepts: Area of a Triangle Formula & Collinearity

To find the area of a triangle when you know the coordinates of its three vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, you can use the following formula:

The absolute value bars $|…|$ are used because area must always be a positive value.

A crucial related concept is collinearity. Three points are said to be collinear if they all lie on the same straight line. If points are collinear, they cannot form a triangle, and the area calculated using the formula will be zero.

Detailed Solution:

1. Assign the coordinates:

   • Let $P(0, 10)$ be $(x_1, y_1)$.
   • Let $Q(-20, -30)$ be $(x_2, y_2)$.
   • Let $R(10, 30)$ be $(x_3, y_3)$.

2. Substitute the values into the area formula:

   $$Area = \frac{1}{2} |0(-30 - 30) + (-20)(30 - 10) + 10(10 - (-30))|$$

3. Simplify the expression inside the absolute value:

   $$Area = \frac{1}{2} |0(-60) + (-20)(20) + 10(10 + 30)|$$
   $$Area = \frac{1}{2} |0 - 400 + 10(40)|$$
   $$Area = \frac{1}{2} |-400 + 400|$$
   $$Area = \frac{1}{2} |0|$$
   $$Area = 0$$

4. Interpret the result: Since the calculated area is 0, the points P, Q, and R are collinear. Therefore, they lie on a single straight line and cannot form a triangle.

Final Answer: The correct statement is “The given points do not form a triangle”.

Question 2: Area of a Triangle Formed by Midpoints

The Question: The area of the triangle formed by the midpoints of line segments $PQ$, $QR$, and $RP$, where the coordinates of $P$, $Q$, and $R$ are $(0, 0)$, $(3, 0)$, and $(3, 4)$ respectively, is __________.

Core Concepts: Midpoint Formula and Area Properties

1. Midpoint Formula: The midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$.
2. Area of a Triangle Formed by Midpoints: A key theorem in geometry states that the area of a triangle formed by joining the midpoints of the sides of another triangle is exactly one-fourth (1/4) of the area of the original triangle.

Detailed Solution:

We can solve this in two ways.

Method 1: The Shortcut (Using the Theorem)

1. Find the area of the original triangle PQR.

   • P(0,0), Q(3,0), R(3,4).
   • Notice that the line segment PQ lies on the x-axis and the line segment QR is a vertical line. This means PQR is a right-angled triangle.
   • The length of the base PQ is the distance between (0,0) and (3,0), which is 3 units.
   • The height is the vertical distance from the base to R, which is 4 units.
   • Area of PQR = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6$ square units.

2. Calculate the area of the midpoint triangle.

   • Area of midpoint triangle = $\frac{1}{4} \times (\text{Area of PQR})$
   • Area = $\frac{1}{4} \times 6 = 1.5$ square units.

Method 2: Direct Calculation (Finding Midpoints First)

1. Find the midpoints of the sides:

   • Midpoint of PQ (let’s call it D) = $\left(\frac{0+3}{2}, \frac{0+0}{2}\right) = (1.5, 0)$.
   • Midpoint of QR (let’s call it E) = $\left(\frac{3+3}{2}, \frac{0+4}{2}\right) = (3, 2)$.
   • Midpoint of RP (let’s call it F) = $\left(\frac{3+0}{2}, \frac{4+0}{2}\right) = (1.5, 2)$.

2. Calculate the area of triangle DEF using the area formula:

   • $D(1.5, 0) \rightarrow (x_1, y_1)$
   • $E(3, 2) \rightarrow (x_2, y_2)$
   • $F(1.5, 2) \rightarrow (x_3, y_3)$
   • Area = $\frac{1}{2} |1.5(2 - 2) + 3(2 - 0) + 1.5(0 - 2)|$
   • Area = $\frac{1}{2} |1.5(0) + 3(2) + 1.5(-2)|$
   • Area = $\frac{1}{2} |0 + 6 - 3|$
   • Area = $\frac{1}{2} |3| = 1.5$ square units.

Final Answer: The area is 1.5 square units.

Question 3: Identifying the Area Formula

The Question: Which of the following gives the area (A) of a triangle whose vertices are $P(x_1, y_1)$, $Q(x_2, y_2)$, $R(x_3, y_3)$?

Core Concept: The Area of a Triangle Formula (Shoelace Formula variation)

This question is a direct test of your knowledge of the standard coordinate geometry formula for the area of a triangle. The formula follows a specific cyclical pattern.

Detailed Solution:

The correct formula is:

Let’s break down the pattern to see why:

• The entire expression is multiplied by $\frac{1}{2}$ and is inside an absolute value.
• There are three terms added together.
• The first term starts with $x_1$. It is multiplied by the difference of the other two y-coordinates, $(y_2 - y_3)$.
• The second term starts with $x_2$. It is multiplied by the difference of the other two y-coordinates in a cycle, $(y_3 - y_1)$.
• The third term starts with $x_3$. It is multiplied by the difference of the other two y-coordinates in a cycle, $(y_1 - y_2)$.

Now, let’s examine the options given in the image:

1. $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
   • This option perfectly matches the standard formula and its cyclical pattern. This is correct.
2. The other options break this pattern. For example, the second option starts with $x_1(y_3 - y_1)$, which is incorrect. The others have similar structural errors.

Final Answer: The first option is the correct formula for the area of a triangle.