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in Straight Lines by (25 points)
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Find the orthocenter of the triangle with the following vertices (5, - 2) , (- 1, 2) and (1, 4).

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2 Answers

+2 votes
by (24.8k points)

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)

+3 votes
by (31.3k points)

Equation of line AD is

y + 2 \(=\frac{-1}{\text{Slope of line BC}} (x - 5)\)

\(\Rightarrow\) y + 2 \(=\cfrac{-1}{\frac{4 - 2}{1 + 1}}(x - 5)\)

\(\Rightarrow\) y + 2 = -2/2 (x - 5)

\(\Rightarrow\) x + y = 5 - 2

\(\Rightarrow\) x + y = 3 ....(1)

Equation of line CF is

y - 4 \(=\frac{-1}{\text{Slope of line AB}} (x - 1)\)

\(\Rightarrow\) y - 4 \(=\cfrac{-1}{\frac{2 + 2}{1 + 1}}(x - 1)\)

\(\Rightarrow\) y - 4 = 6/4 (x - 1)

\(\Rightarrow\) 2y - 8 = 3x - 3

\(\Rightarrow\) -3x + 2y = 8 - 3

\(\Rightarrow\) -3x + 2y = 5 ......(2)

Equation (1) × 3 + Equation (2), we get

5y = 14

\(\Rightarrow\) y = 14/5

put y = 14/5 in equation (1), we get

x = 3 - 14/5 = 1/5.

Hence, orthocenter of the triangle is (1/5, 14/5).

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