Question

A(-3,0) B(10,-2) and C(12,3)are the vertices of triangleABC.find the altitude through AandB.​

Answer 1

Altitude through A. Altitude throughB.

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 = (y2 - y1)/(x2 - x1)

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

= (3 + 2)/2

= 5/2

Equation of the altitude passing through the vertex A :

(y - y1) = (-1/m)(x - x1)

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 = (y2 - y1)/(x2 - x1)

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

= 3/(12+3)

= 3/15

= 1/5

Equation of the altitude passing through the vertex B :

(y - y1) = (-1/m)(x - x1)

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.

Answer 2

altitude A is perpendicular on line BC.

means, slope of altitude A × slope of line BC = -1

slope of line BC = (3 +2)/(12 -10) = 5/2

or, slope of altitude A = -1/(5/2) = -2/5 .

now equation of altitude A :

(y - y1) = m(x - x1)

here, (x1, y1) = (-3,0) and m = -2/5

so, (y - 0) = (-2/5)(x + 3)

or, 5y + 2x + 6 = 0

or, 2x + 5y + 6 = 0

similarly, slope of altitude B × slope of line CA = -1

slope of line CA = (3 - 0)/(12 + 3) = 1/5

then, slope of altitude B = -1/(1/5) = -5

now equation of altitude B :

{y - (-2)} = -5(x - 10)

or, (y + 2) + 5(x - 10) = 0

or, 5x + y - 48 = 0