2. (a) Using graph paper, draw the triangle with vertices A(-2, 0), B(3, 0) and C(1, 4) - Junior Cycle Mathematics - Question 2 - 2012

To find the area of triangle ABC, we can use the formula:

ext{Area} = rac{1}{2} imes ext{base} imes ext{height}

For base AB, the length is 5 (from -2 to 3). The height can be found by considering the vertical distance from point C(1, 4) to line AB (y=0), which is 4.

Calculating the area:

ext{Area} = rac{1}{2} imes 5 imes 4 = 10

Thus, the area of triangle ABC is 10 square units.

To find the intersection of lines l (2x - 11y = -16) and k (x + 2y = -8), we can solve the equations simultaneously:

1. Rearranging the second equation gives: $$x + 2y = -8 \implies x = -8 - 2y$$
2. Substitute x into the first equation: $$2(-8 - 2y) - 11y = -16$$ Simplifying gives: $$-16 - 4y - 11y = -16 \implies -15y = 0 \implies y = 0$$
3. Substitute y back to find x: $$x = -8 - 2(0) = -8$$

Thus, the intersection point P_b is (-8, 0).

To prove triangle PQR (with points P_b, Q(3, 2), and R(2, -5) is isosceles, we need to show that at least two sides are equal:

1. Calculate the lengths of PQ, QR, and PR.

   • Using the distance formula: $$d = ext{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2)$$

2. For PQ:

   $$d_{PQ} = ext{sqrt}((-8 - 3)^2 + (0 - 2)^2) = ext{sqrt}(121 + 4) = ext{sqrt}(125)$$

3. For QR:

   $$d_{QR} = ext{sqrt}((3 - 2)^2 + (2 + 5)^2) = ext{sqrt}(1 + 49) = ext{sqrt}(50)$$

4. For PR:

   $$d_{PR} = ext{sqrt}((-8 - 2)^2 + (0 + 5)^2) = ext{sqrt}(100 + 25) = ext{sqrt}(125)$$

5. As PQ = PR = $ext{sqrt}(125)$, triangle PQR is isosceles.

To find the perpendicular bisector of segment ST, first determine the midpoint:

1. Midpoint M of ST = $$ M = \left( \frac{-4 + 2}{2}, \frac{-2 + 6}{2} \right) = ( -1, 2)

   $$$$

2. The slope of segment ST is calculated as:

   $$m_{ST} = \frac{6 - (-2)}{2 - (-4)} = \frac{8}{6} = \frac{4}{3}$$

3. The slope of the perpendicular bisector is the negative reciprocal:

   $$m_{bisector} = -\frac{3}{4}$$

4. Use point-slope form for the equation:

   $$y - 2 = -\frac{3}{4}(x + 1)$$

Rearranging yields the equation of the perpendicular bisector.

To verify if the point (-5, 5) lies on the perpendicular bisector:

1. Substitute x = -5 and y = 5 into the bisector equation derived: $$5 - 2 = -\frac{3}{4}(-5 + 1)$$
2. Simplifying, $$3 = -\frac{3}{4}(-4) \implies 3 = 3$$

Thus, (-5, 5) is indeed on the perpendicular bisector.

To find the image of point (-5, 5) under axial symmetry in line segment ST:

1. Identify the midpoint M of segment ST which is already calculated as (-1, 2).
2. Reflect the point (-5, 5). If M is the midpoint between points P and its image P': $$M = \left( \frac{x + (-5)}{2}, \frac{y + 5}{2} \right)$$
3. Setting M = (-1, 2) gives two equations: $$-1 = \frac{x - 5}{2} \quad ext{and} \quad 2 = \frac{y + 5}{2}$$
4. Solving these:
   • For x: $\rightarrow x = -2$.
   • For y: $\rightarrow y = -1$.
5. Therefore, the coordinates of the image are (-2, -1).