Leaving Cert Mathematics Co-Ordinate Geometry: The points A(6, -2), B(5, 3) and

The points A(6, -2), B(5, 3) and C(-3, 4) are shown on the diagram - Leaving Cert Mathematics - Question 1 - 2016

Question 1

The points A(6, -2), B(5, 3) and C(-3, 4) are shown on the diagram. (a) Find the equation of the line through B which is perpendicular to AC. (b) Use your answer t... show full transcript

Worked Solution & Example Answer:The points A(6, -2), B(5, 3) and C(-3, 4) are shown on the diagram - Leaving Cert Mathematics - Question 1 - 2016

Step 1

Find the equation of the line through B which is perpendicular to AC.

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Answer

To find the equation of the line through point B that is perpendicular to line AC, we first need the slope of line AC.

Step 1: Determine the coordinates of points A and C

A = (6, -2)
C = (-3, 4)

Step 2: Calculate the slope of AC

The formula for the slope (m) between two points (x1, y1) and (x2, y2) is: $m = \frac{y_2 - y_1}{x_2 - x_1}$ Thus, the slope of AC is: $m_{AC} = \frac{4 - (-2)}{-3 - 6} = \frac{6}{-9} = -\frac{2}{3}$

Step 3: Slope of line through B perpendicular to AC

The slope of a line perpendicular to another is the negative reciprocal, thus: $m_{perpendicular} = \frac{3}{2}$

Step 4: Using point-slope form to determine the equation of the line

Using point B(5, 3) and the slope of \frac{3}{2}, we apply the point-slope formula: $y - y_1 = m(x - x_1)$ Substituting in our values: $y - 3 = \frac{3}{2}(x - 5)$ Expanding this gives: $y - 3 = \frac{3}{2}x - \frac{15}{2}$ Thus, $y = \frac{3}{2}x - \frac{15}{2} + 3$ $y = \frac{3}{2}x - \frac{15}{2} + \frac{6}{2}$ $y = \frac{3}{2}x - \frac{9}{2}$ Finally, converting this into standard form leads to: $3x - 2y = 9$

Step 2

Use your answer to part (a) above to find the co-ordinates of the orthocentre of the triangle ABC.

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Answer

To find the orthocentre of triangle ABC, we will calculate the equations of the altitudes.

Step 5: Find the slope and equation of altitude AB

The slope of line AB: $m_{AB} = \frac{3 - (-2)}{5 - 6} = \frac{5}{-1} = -5$ The slope of the altitude from C (perpendicular) is: $m_{C} = \frac{1}{5}$ Using point C(-3, 4) with the slope: $y - 4 = \frac{1}{5}(x + 3)$ This simplifies to: $5y - 20 = x + 3$ Or: $x - 5y + 23 = 0$

Step 6: Find the slope and equation of altitude BC

The slope of BC: $m_{BC} = \frac{3 - 4}{5 - (-3)} = \frac{-1}{8}$ The slope of the altitude from A (perpendicular) is: $m_{A} = 8$ Using point A(6, -2): $y + 2 = 8(x - 6)$ Expanding gives:

ightarrow 8x - y - 50 = 0$$ ### Step 7: Find the intersection of the two altitudes to determine the orthocentre We need to solve the system of equations: 1. $x - 5y + 23 = 0$ 2. $8x - y - 50 = 0$ Solving these simultaneously: From the first equation, $$y = \frac{1}{5}x + \frac{23}{5}$$ Substituting in the second equation gives: $$8x - \frac{1}{5}x - 10 = 0$$ Multiplying through by 5, $$40x - x - 50 = 0 \\ ightarrow 39x = 50 \\ ightarrow x = \frac{50}{39}$$ Substituting back to find y: $$y = \frac{1}{5}\left(\frac{50}{39}\right) + \frac{23}{5} = \frac{50}{195} + \frac{87}{195} = \frac{137}{195}$$ Thus, the orthocentre is approximately at (7, 6).