Question

Do the points P(1, 0), Q($-$5, 0) and R($-$2, 5) form a triangle ? If so, name the type of triangle formed.

Hint:

Alternatively, you can compute the area of $\Delta\text{PQR}$ using the coordinate area formula.
If the calculated area is non-zero, the points form a triangle.
In this case, Area $= \frac{1}{2} |1(0-5) - 5(5-0) - 2(0-0)| = 15 \neq 0$, which proves a triangle is formed.

Answer 1

Step 1: Understanding the Question:
The question provides the coordinates of three points in the coordinate plane: $P(1, 0)$, $Q(-5, 0)$, and $R(-2, 5)$.
First, we must determine if these three points are collinear or if they form a triangle.
If they form a triangle, we must determine its type (scalene, isosceles, equilateral, or right-angled) by analyzing the lengths of its sides.

Step 2: Key Formula or Approach:
1. The distance $d$ between any two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
2. If the three points are non-collinear, they form a triangle. The Triangle Inequality Theorem states that the sum of the lengths of any two sides must be greater than the length of the third side.
3. To classify the triangle:
- Isosceles: Two sides are equal in length.
- Equilateral: All three sides are equal in length.
- Scalene: All three sides have different lengths.
- Right-angled: The lengths satisfy the Pythagoras theorem ($a^2 + b^2 = c^2$).

Step 3: Detailed Explanation:

• Let us calculate the length of side $PQ$:
\[ PQ = \sqrt{(-5 - 1)^2 + (0 - 0)^2} \]
\[ PQ = \sqrt{(-6)^2 + 0} = \sqrt{36} = 6 \text{ units} \]

• Let us calculate the length of side $QR$:
\[ QR = \sqrt{(-2 - (-5))^2 + (5 - 0)^2} \]
\[ QR = \sqrt{(3)^2 + (5)^2} \]
\[ QR = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \text{ units} \]

• Let us calculate the length of side $RP$:
\[ RP = \sqrt{(1 - (-2))^2 + (0 - 5)^2} \]
\[ RP = \sqrt{(3)^2 + (-5)^2} \]
\[ RP = \sqrt{9 + 25} = \sqrt{34} \approx 5.83 \text{ units} \]

• Check the Triangle Inequality Theorem:
- $PQ + QR = 6 + 5.83 = 11.83 \gt 5.83$ ($PQ + QR \gt RP$)
- $QR + RP = 5.83 + 5.83 = 11.66 \gt 6$ ($QR + RP \gt PQ$)
- $RP + PQ = 5.83 + 6 = 11.83 \gt 5.83$ ($RP + PQ \gt QR$)
Since the sum of any two side lengths is greater than the third side, the points P, Q, and R are non-collinear and indeed form a valid triangle.

• Classify the triangle based on the side lengths:
We observe that $QR = \sqrt{34}$ and $RP = \sqrt{34}$.
Since two sides of the triangle are equal in length ($QR = RP \neq PQ$), $\Delta\text{PQR}$ is an isosceles triangle.


Step 4: Final Answer:
The points P, Q, and R form an isosceles triangle.