Question

The vertices of triangle are A(-2 , 1 ) , B ( 2 , 2 ) and C ( 6 , -2 ) . Find the equation of the altitude of AD drawn from A to BC.

Please help!!:)​

Answer 1

Answer:

Given,

ABC is triangle with vertices A(-2,1), B(2,2) and C(6, -2)

Slope of BC(m_1) =$\frac{y_2-y_1}{x_2-x_1}=\frac{6-2}{-2-2}$

$=\frac{4}{-4}=-1$

Since AD is perpendicular to BC

let slope of BC be ($m_2$).

We have for perpendicular

$m_1*m_2=-1$

$-1*m_2=-1$

$m_2=1$

Now equation of line passing through point A(-2,1) is

$y-y_1=m_2(x-x_1)$

y-1=1(x-(-2))

y-1=x+2

y=x+2+1

y=x+3 which is a required equation of the altitude of AD drawn from A to BC.

Answer 2

The required equation of the altitude of AD drawn from A to BC is y=x+3

Calculations and Parameters:

Given that ABC is triangle with vertices

• A(-2,1), B(2,2)
• C(6, -2)

Given the slope of BC(m_1)

= y2-y1/x2-x1

= -1.

For perpendicular,

m1 * m2= -1

m2= 1.

Now equation of line passing through point A(-2,1) is

• y-1=1(x-(-2))
• y-1=x+2
• y=x+2+1
• y=x+3

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