HOW TO FIND THE EQUATION OF ALTITUDE OF A TRIANGLE

How to Find the Equation of Altitude of a Triangle ?

Here we are going to see, how to find the equation of altitude of a triangle.

How to Find the Equation of Altitude of a Triangle - Questions

Question 1 :

A(-3, 0) B(10, -2) and C(12, 3) are the vertices of triangle ABC . Find the equation of the altitude through A and B.

Solution :

Equation of altitude through A

Solution :

The altitude passing through the vertex A intersect the side BC at D.

AD is perpendicular to BC.

Slope of BC = (y_{2} - y_{1})/(x_{2} - x_{1})

= (3 - (-2))/(12 - 10)

= (3 + 2)/2

= 5/2

Equation of the altitude passing through the vertex A :

(y - y_{1}) = (-1/m)(x - x_{1})

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

Equation of altitude through B

Slope of AC = (y_{2} - y_{1})/(x_{2} - x_{1})

= (3 - 0)/(12 - (-3))

= 3/(12+3)

= 3/15

= 1/5

Equation of the altitude passing through the vertex B :

(y - y_{1}) = (-1/m)(x - x_{1})

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.

Question 2 :

Find the equation of the perpendicular bisector of the line joining the points A(-4,2) and B(6,-4).

Solution :

Perpendicular bisector means, the line will pass through the midpoint of the line segment AB and makes an 90 degree angle.

midpoint = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

= (-4 + 6)/2, (2 - 4)/2

= 2/2, -2/2

= (1, -1)

Slope of AB = (y_{2} - y_{1})/(x_{2} - x_{1})

= (-4-2)/(6+4)

= -6/10

= -3/5

Equation of perpendicular bisector :

(y - y_{1}) = (-1/m)(x - x_{1})

(y + 1) = (5/3)(x - 1)

3(y + 1) = 5(x - 1)

3y + 3 = 5x - 5

5x - 3y - 5 - 3 = 0

3x + 5y - 8 = 0

After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Equation of Altitude of a Triangle".

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