Question

Vertices of a triangle are A(2,3) B(4, 1− ) C(1, 2). Find the equation of altitude AD on side BC

Answer 1

Step-by-step explanation:

Given: Vertices of a triangle are A(2,3) B(4, -1 ) C(1, 2).

To find:Find the equation of altitude AD on side BC.

Solution:

we know that if a line is passing through( x1, y1) having slope m ,then equation of line is

$\boxed{\bold{y - y_1 = m(x - x_1)}}$

Here points of passing of altitude is A,i.e.(2,3)

Step1: Find the slope:

AD is perpendicular to line BC.

if two lines are perpendicular, then their slopes must satisfy the following condition

$\boxed{m_1m_2 = - 1}$

Step 2: Find the slope of CB:

Slope of a line passing through two points (x1,y1) and (x2,y2)

$\boxed{m = \frac{y_2 - y_1}{x_2 - x_1} }$

Let the slope of CB is m1: B(4,-1) and C(1,2)

$m_1 = \frac{2 +1}{1 - 4} \\ \\ m_1 = \frac{3}{ - 3} \\ \\ m_1 = -1$

Step3: Find slope of altitude AD:

Let the slope of AD is m2, so

$m_2 = \frac{ - 1}{m_1} \\ \\ m_2 = 1$

Step 4:Find the line AD:

A(2,3),slope 1

Equation of line is

$y - 3 = 1(x - 2) \\ \\ y - 3= x - 2 \\ \\ y - x - 3+ 2 = 0 \\ \\ y - x -1= 0 \\ \\ x - y +1 = 0$

Thus,

Equation of altitude AD is x-y+1=0

Hope it helps you.

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Answer 2

Answer:

x-y+1=0

Step-by-step explanation:

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