find orghicentre of triangle whose vertices are (5,-2),(-1,2),(1,4)
Answers
Answer:
Given, the vertices of the triangle,
A = (5, -2) B = (-1, 2) C = (1, 4)
Orthogonal center is the cross section of altitudes of the triangle.
Slope of AB = y2−y/x2−x1 = 2 + 2 / -1 -5 = -2/3 Altitude from C to AB is perpen
dicular to AB. = Perpendicular slope of
AB = −1/Slope of AB = 3/2 The equation of CF is given as
, (F is the point on
AB) y – y1 = m(x – x1)
y - 4 = 3/2(x – 1)
2y – 8 = 3x - 3 3x - 2y = -5
——————————– (1)
SlopeGiven, the vertices of the triangle,
A = (5, -2) B = (-1, 2) C = (1, 4)
Orthogonal center is the cross section of altitudes of the triangle.
Slope of
AB = y2−y/x2−x1 = 2 + 2 / -1 -5 = -2/3
Altitude from C to AB is perpendicular to AB. = Perpendicular slope of
AB = −1/Slope of AB = 3/2 The equation of CF is given as,
(F is the point on
AB) y – y1 = m(x – x1) y - 4 = 3/2(x – 1)
2y – 8 = 3x - 3 3x - 2y = -5
——————————– (1)
Slope of BC = y2–y /
x2–x1 = 4 – 2 / 1 + 1 = 1 Slope of AD (AD is altitude)
Perpendicular slope of
BC = −1/Slope of BC = −1 The equation of AD is given as, y – y1 = m(x – x1) y + 2 = -1(x – 5) x + y = 3 ——————————– (2)
Subtracting equation (1) and 3*(2), 3x - 2y = -5 3x + 3y = 9 —————— -5y = --14 y = 14/5 Substituting the value of y in equation
(2), X = 3 – 14/5 = 1/5r-of-the-equation
-5-2-1-2-and-1-4