25. In AABC, A = (5,-2), B = (6,3) and C = (8,5)
Find the equation of
the median through A.
(ii) the altitude through B.
Answers
Answer:
i. 3x - y - 17 = 0
ii. 3x + 7y - 39 = 0
Step-by-step explanation:
When a median is drawn from A, it bisects BC at D which means DB = DC.
And, D is the mid point of BC.
So,
Coordinates of D = ( { 6 + 8 } / 2 , { 3 + 5 } / 2 )
⇒ Coordinates of D = ( 14 / 2 , 8 / 2 )
⇒ Coordinates of D = ( 7 , 4 )
Now, it is obvious that the median pass throw the points A and D.
Slope of AD = { 4 - ( - 2 ) } / { 7 - 5 } ⇒ Slope of AD = 6 / 2
⇒ Slope of AD = 3
Therefore,
⇒ Equation of AD( using y - y₁ = m( x - x₁ )
⇒ y - 4 = 3( x - 7 )
⇒ y - 4 = 3x - 21
⇒ 3x - y - 17 = 0
Equation of AD( median throw A ) is 3x - y - 17 = 0.
When a perpendicular( altitude ) is drawn from B, it intersects AC at point E. Since BE ⊥ AC, their slopes must of -ve reciprocals of each other.
Slope of AC is { 5 - ( - 2 ) } / { 8 - 5 } ⇒ Slope of AC is ( 5 + 2 ) / 3
⇒ Slope of AC is 7 / 3
Hence, slope of BE must be - 3 / 7.
We are familiar of a point lying on the line BE( that B ), therefore,
⇒ Equation of BE( using the same method )
⇒ y - 3 = (-3/7) ( x - 6 )
⇒ 7( y - 3 ) = - 3( x - 6 )
⇒ 7y - 21 = - 3x + 18
⇒ 3x + 7y - 21 - 18 = 0
⇒ 3x + 7y - 39 = 0
Therefore, equation of altitude through B is 3x + 7y - 39 = 0
