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Parents Pricing Home Scottish Highers Maths Straight Line Triangle ABC has vertices A(−5,−12), B(11,−8) and C(−3,6)
Triangle ABC has vertices A(−5,−12), B(11,−8) and C(−3,6) - Scottish Highers Maths - Question 1 - 2019 Question 1
View full question Triangle ABC has vertices A(−5,−12), B(11,−8) and C(−3,6).
(a) Find the equation of the median BD.
(b) Find the equation of the altitude AE.
(c) Find the coordina... show full transcript
View marking scheme Worked Solution & Example Answer:Triangle ABC has vertices A(−5,−12), B(11,−8) and C(−3,6) - Scottish Highers Maths - Question 1 - 2019
Find the equation of the median BD. Only available for registered users.
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To determine the equation of median BD, we first find the midpoint E of AC.
Calculate the Midpoint of AC:
Midpoint E can be calculated using the formula:
E = ( x 1 + x 2 2 , y 1 + y 2 2 ) = ( − 5 + ( − 3 ) 2 , − 12 + 6 2 ) = ( − 4 , − 3 ) E = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-5 + (-3)}{2}, \frac{-12 + 6}{2} \right) = \left( -4, -3 \right) E = ( 2 x 1 + x 2 , 2 y 1 + y 2 ) = ( 2 − 5 + ( − 3 ) , 2 − 12 + 6 ) = ( − 4 , − 3 )
Find the gradient of BD:
The coordinates of B are (11, -8) and E are (-4, -3). The gradient m can be found by:
m B D = y 2 − y 1 x 2 − x 1 = − 3 − ( − 8 ) − 4 − 11 = 5 − 15 = − 1 3 m_{BD} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - (-8)}{-4 - 11} = \frac{5}{-15} = -\frac{1}{3} m B D = x 2 − x 1 y 2 − y 1 = − 4 − 11 − 3 − ( − 8 ) = − 15 5 = − 3 1
Determine the equation of BD:
Using the point-slope form of the line equation,
y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) \ y − y 1 = m ( x − x 1 )
where (x_1, y_1) is the point B (11, -8):
y − ( − 8 ) = − 1 3 ( x − 11 ) y - (-8) = -\frac{1}{3}(x - 11) y − ( − 8 ) = − 3 1 ( x − 11 )
This simplifies to:
y + 8 = − 1 3 x + 11 3 y + 8 = -\frac{1}{3}x + \frac{11}{3} \ y + 8 = − 3 1 x + 3 11
Rearranging gives:
3 y + 24 = − x + 11 3 y + x = − 13 3y + 24 = -x + 11 \ 3y + x = -13 3 y + 24 = − x + 11 3 y + x = − 13
Thus, the equation of median BD isx + 3 y = − 13 x + 3y = -13 x + 3 y = − 13
Find the equation of the altitude AE. Only available for registered users.
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To find the equation of the altitude AE from A to BC, we need to:
Calculate the gradient of BC:
The coordinates of B are (11, -8) and C are (-3, 6). So, the gradient m of BC is:
m B C = 6 − ( − 8 ) − 3 − 11 = 14 − 14 = − 1 m_{BC} = \frac{6 - (-8)}{-3 - 11} = \frac{14}{-14} = -1 m BC = − 3 − 11 6 − ( − 8 ) = − 14 14 = − 1
Use the property of perpendicular lines:
The gradient of AE will be the negative reciprocal of m_{BC}, som A E = 1 m_{AE} = 1 m A E = 1
Determine the equation of AE:
Given A is (−5, −12), using the point-slope form:
y + 12 = 1 ( x + 5 ) y + 12 = 1(x + 5) \ y + 12 = 1 ( x + 5 )
This simplifies to:
y + 12 = x + 5 y = x − 7 y + 12 = x + 5 \ y = x - 7 y + 12 = x + 5 y = x − 7
Therefore, the equation of altitude AE isy = x − 7 y = x - 7 y = x − 7
Find the coordinates of the point of intersection of BD and AE. Only available for registered users.
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To find the intersection of the two lines, we solve the equations:
Set Equations Equal:
From BD:x + 3 y = − 13 x + 3y = -13 x + 3 y = − 13
From AE:y = x − 7 y = x - 7 y = x − 7
Substitute y from AE into BD:
x + 3(x - 7) = -13 \
Simplifying gives:x + 3 x − 21 = − 13 4 x − 21 = − 13 4 x = 8 x = 2 x + 3x - 21 = -13 \ 4x - 21 = -13 \ 4x = 8 \ x = 2 x + 3 x − 21 = − 13 4 x − 21 = − 13 4 x = 8 x = 2
Find y Coordinate:
Substituting x = 2 x = 2 x = 2 y = 2 − 7 = − 5 y = 2 - 7 = -5 y = 2 − 7 = − 5
Therefore, the coordinates of the intersection point are (2, -5).
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