Question

A(-3,0)B(10,-2) andC(12,3) are the vertices of triangleABC. Find the altitud through A andB.​

Answer 1

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

Answer 2

Step-by-step explanation:

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