Question

find the orthocenter of a triangle with vertices (5,-2)(-1,2)(1,4)
plz I need answer​

Answer 1

Step-by-step explanation:

Now let's find slopes of AB, AC and BC

mAB = (-2-(-1))/(5-2), mAB = - 1/3

mAC = (-2 -4)/(5–1), mAC = -3/2

mBC = (4-2)/(1-(-1)), mBC = 1

So after finding slopes of these line segments, we can make equation of perpendicular of these segments by using reciprocal of BC and coordinates A, reciprocal of AC and coordinates B.

Y = mX + b, - 2 = -1 × 5 + b, b = 3

(E)BC = Y = -X + 3..(1)

2 = 2/3×(-1) + b, b = 2 + 2/3, b = 8/3

(E)AC = Y = 2X/3 + 8/3..(2)

Subtracting first equation from second equation, we can get coordinates of intersection point of these two perpendicular lines, which is the orthocentre of triangle ABC.

2X/3+ X + 8/3 -3 = 0, 5X/3 = 3 - 8/3, 5X/3 = 1/3 , X = 1/5

Y = - 1/5 + 3, Y = (-1+15)/5, Y = 14/5

So the orthocentre of triangle is (1/5,14/5) or

(0.2,2.8)

Answer 2

Answer:

Orthocentre=14/5,1/5

Step-by-step explanation:

Given,

Three vertices of triangle

A=(5,-2)

B=(-1,2)

C=(1,4)

Orthogonal centre is the cross section of altitudes of 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

The equation of AD is given as

y-y1=m(x-x1)

y+2=-1(x-5)

x+y-3=0.................› 2

subtracting 1 and 3*(2)

3x-2y= -5

3x+3y=9

——————

-5y=-14

y=14/5

substituting y value in equation 2

X=3-14/5

=1/5

ortho centre=(14/5,1/5)

hope it is useful