A(-3,0) B(10-2) and C(12.3) are the vertices of AABC. Find the equation of the altitude through A and B.
...
Answers
Answer:
Hope it would help you
Kindly mark as brainliest:-)
stay safe and blessed...
Step-by-step explanation:
Equation of altitude through A
The altitude passing through the vertex A intersect the side BC at D.
AD is perpendicular to BC.
Slope of BC = (y2 - y1)/(x2 - x1)
= (3 - (-2))/(12 - 10)
= (3 + 2)/2
= 5/2
Equation of the altitude passing through the vertex A :
(y - y1) = (-1/m)(x - x1)
A(-3, 0) and m = 5/2
(y - 0) = -1/(5/2)(x - (-3))
y = (-2/5) (x + 3)
5y = -2x - 6
2x + 5y + 6 = 0 is the equation of altitude through A
Equation of altitude through B
Slope of AC = (y2 - y1)/(x2 - x1)
= (3 - 0)/(12 - (-3))
= 3/(12+3)
= 3/15
= 1/5
Equation of the altitude passing through the vertex B :
(y - y1) = (-1/m)(x - x1)
B(10, -2) and m = 1/5
(y - (-2)) = -1/(1/5)(x - 10)
y + 2 = -5(x - 10)
y + 2 = -5x + 50
5x + y + 2 - 50 = 0
5x + y - 48 = 0 is the equation of altitude through B