Question

(5,-2) (-1,2) (1.4) find the orthocentre of the triangle with the following verices​

Answer 1

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 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 = 1/5

Ortho center = (14/5,1/5)

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