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

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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 determine the slope of line AC.

  1. Calculate the slope of line AC:

    • Coordinates of A: (6, -2)
    • Coordinates of C: (-3, 4)
    • Slope formula: slope=y2−y1x2−x1=4−(−2)−3−6=6−9=−23.\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-2)}{-3 - 6} = \frac{6}{-9} = -\frac{2}{3}.

  2. Determine the slope of the line perpendicular to AC:

    • The slope of the perpendicular line is the negative reciprocal: slope=32.\text{slope} = \frac{3}{2}.

  3. Use point-slope form to find the equation of the line through B (5, 3):

    • Point-slope form: y−y1=m(x−x1)y - y_1 = m(x - x_1)
    • Substituting point B (5, 3) and the slope 32\frac{3}{2}: y−3=32(x−5).y - 3 = \frac{3}{2}(x - 5).

  4. Rearranging to slope-intercept form:

    • Distributing: y−3=32x−152y - 3 = \frac{3}{2}x - \frac{15}{2}
    • Adding 3 to both sides: y=32x−152+62y = \frac{3}{2}x - \frac{15}{2} + \frac{6}{2}
    • Final equation: y=32x−92.y = \frac{3}{2}x - \frac{9}{2}.

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 need to determine the point of intersection of the altitudes.

  1. Find the slope of line AB:

    • Coordinates of A: (6, -2), B: (5, 3)
    • Slope: slope=3−(−2)5−6=5−1=−5.\text{slope} = \frac{3 - (-2)}{5 - 6} = \frac{5}{-1} = -5.

  2. Determine the slope of the altitude from C to AB (perpendicular slope):

    • Perpendicular slope: mC=15.m_{C} = \frac{1}{5}.

  3. Use point-slope form through C (−3, 4): y−4=15(x+3)y - 4 = \frac{1}{5}(x + 3)

    • Rearranging: y=15x+35+4=15x+235.y = \frac{1}{5}x + \frac{3}{5} + 4 = \frac{1}{5}x + \frac{23}{5}.

  4. Find the altitude from point B to line AC:

    • Slope of AC (from part a) is −23−\frac{2}{3}. Perpendicular slope from B (5, 3) is: 32.\frac{3}{2}.
    • Use point-slope form: y−3=32(x−5)y - 3 = \frac{3}{2}(x - 5)
    • Rearranging gives: y=32x−32.y = \frac{3}{2}x - \frac{3}{2}.

  5. Find the intersection of the two altitudes:

    • Set equations equal: 15x+235=32x−32.\frac{1}{5}x + \frac{23}{5} = \frac{3}{2}x - \frac{3}{2}.
    • Multiplying through by 10 to eliminate fractions gives: 2x+46=15x−15.2x + 46 = 15x - 15.
    • Rearranging yields: 13x=61⇒x=6113=4.69.13x = 61\Rightarrow x = \frac{61}{13} = 4.69.
    • Substitute xx back to find yy: y=15(4.69)+235=4.69+235=27.695=5.54.y = \frac{1}{5}(4.69) + \frac{23}{5} = \frac{4.69 + 23}{5} = \frac{27.69}{5} = 5.54.

  6. Orthocentre:

    • Thus, the orthocentre of triangle ABC is approximately: (4.69,5.54).(4.69, 5.54).

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