Plot the points A(3,1), B(0,4), and C(-2,-1) on the grid below - Junior Cycle Mathematics - Question 6 - 2014

To calculate |AC| and |BC|, use the distance formula:

$|AC| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

1. For |AC|:

   • A(3, 1) and C(-2, -1)
   • $|AC| = \sqrt{((-2 - 3)^2 + (-1 - 1)^2)}$
   • $= \sqrt{(-5)^2 + (-2)^2}$
   • $= \sqrt{25 + 4} = \sqrt{29}$

2. For |BC|:

   • B(0, 4) and C(-2, -1)
   • $|BC| = \sqrt{((-2 - 0)^2 + (-1 - 4)^2)}$
   • $= \sqrt{(-2)^2 + (-5)^2}$
   • $= \sqrt{4 + 25} = \sqrt{29}$

Thus, |AC| = |BC|.

Triangle ABC is classified as an Isosceles triangle because two of its sides, |AC| and |BC|, are equal in length, both measuring ( \sqrt{29} ).

To find point D, the midpoint of line segment AB, use the midpoint formula:

$D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

• A(3, 1) and B(0, 4)
• $D = \left( \frac{3 + 0}{2}, \frac{1 + 4}{2} \right)$
• $D = \left( \frac{3}{2}, \frac{5}{2} \right)$

Therefore, the co-ordinates of D are (1.5, 2.5).

Draw a straight line from point C(-2, -1) to point D(1.5, 2.5) on the previous diagram.

To prove that triangles ΔADC and ΔBDC are congruent, we will use the SSS (Side-Side-Side) rule:

1. Sides of ΔADC:

   • |AD| = |BD| (as D is the midpoint of AB)
   • From part (ii), we know |AC| = |BC| = √29
   • |CD| is common to both triangles.

2. Thus, by SSS: All three corresponding sides of triangles ΔADC and ΔBDC are equal, confirming that the triangles are congruent.