The triangle ABC is shown on the co-ordinate grid below - Junior Cycle Mathematics - Question 4 - 2016

To find the equations of the lines, we need the slopes and points on each line.

1. Line AB:

   • Slope (m) = 0 (horizontal line)
   • Equation: $y = 3$ (since it passes through B and A which share the same y-coordinate)

2. Line AC:

   • Slope (m) = rac{8 - 3}{1 - 1} = ext{undefined} (vertical line)
   • Equation: $x = 1$ (since it passes through points A and C)

3. Line BC:

   • Slope (m) = rac{8 - 3}{1 - 5} = -rac{5}{4}
   • Using point-slope form, using point B: y - 3 = -rac{5}{4}(x - 5).
   • Rearranging gives: $5x + 4y - 43 = 0$.

Using the tangent ratio: $\tan(\angle ABC) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{4}$

Calculating the angle: $\angle ABC = \tan^{-1}\left(\frac{5}{4}\right) \approx 51.34^{\circ}$ Rounded to two decimal places, we have: $51.34^{\circ}$.

To find the length of BC, we use the distance formula: $|BC| = \sqrt{(5 - 1)^{2} + (3 - 8)^{2}} = \sqrt{4^{2} + (-5)^{2}} = \sqrt{16 + 25} = \sqrt{41}$.

The radius of the circle is the distance from the center to any of the points A, B, or C. Using the diameter formula from points A and C:

• Radius (r) = rac{|BC|}{2} = \frac{\sqrt{41}}{2}.

• Area of the circle: $\text{Area} = \pi r^{2} = \pi \left(\frac{\sqrt{41}}{2}\right)^{2} = \pi \frac{41}{4} = \frac{41\pi}{4}.$

The slope of line BC is $m_{BC} = -\frac{5}{4}$, so the slope of the perpendicular line (m) would be the negative reciprocal: $m = \frac{4}{5}.$ Using point-slope form with point A (1,3):

$y - 3 = \frac{4}{5}(x - 1)$ Simplifying gives: $5y - 15 = 4x - 4 \Rightarrow 4x - 5y + 11 = 0.$