The vertices of a triangle are A(5,6) B(1,-4) C(-4,0) then the length of the altitude through the vertex is ?
Answers
the length of the altitude through the vertex A is 10.3
Step-by-step explanation:
The vertices of a triangle are A(5,6) B(1,-4) C(-4,0)
Let say AD is altitude passing through A
BC slope = ( 0 -(-4))/(-4 - 1) = -4/5
y = -4x/5 + c
0 = 16/5 + c
=> c = -16/5
=> y = -4x/5 - 16/5
=> 5y = -4x - 16
=> 4x + 5y = - 16 Eq1
Slope of AD = -1/(-4/5) = 5/4
y = 5x/4 + c
6 = 25/4 + c
=> c = - 1/4
y = 5x/4 -1/4
=> 4y = 5x - 1
=> 5x - 4y = 1 Eq2
intersection of AD & BC
4 * eq1 + 5 * eq2
=> 41x = -59
=> x = -59/41
4x + 5y = - 16
=> 4(-59/41) + 5y = -16
=> 5y = -420/41
=> y = -84/41
D = ( -59/41 , -84/41) A = (5 , 6)
AD² = (-59/41 - 5)² + ( - 84/41 - 6)²
=> AD² = (-264/41)² + (-330/41)²
=> AD² = (1,78,596/41²)
=> AD = 422.6/41
=> AD = 10.3
Another method
Area of Δ ABC
= (1/2) | 5(-4 - 0) + 1 (0 - 6) -4(6 -(-4)) |
= (1/2) | -20 - 6 - 40 |
= (1/2) | - 66|
= 66/2
= 33
Altitude through A = AD
(1/2) BC * AD = 33
BC = √41
=> AD = 66/√41 = 10.3
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