Question

Find the equations of the altitudes of the triangle whose vertices are A(2,- 7), B(-2,1) and C(4,3).​

Answer 1

Answer:

Given triangle is triangle ABC with the altitudes AD , BE , CF

Equation of AD :

first of all find the slope of side BC

(x1,y1)= B (-2,1) , (x2,y2)=C(4,3)

Slope of BC = (y2-y1)/(x2-x1)

= (3-1)/(4-(-2)) =2/6 = 1/3

we know , if the two lines are perpendicular to each other then their product of slope is -1

therefore ,

(Slope of AD)(Slope of BC)=-1

Slope of AD = -1/(1/3) = -3 = m

we know,

• slope point form =

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

Here, (x1,y1)=A(2,-7)

therefore, (y-(-7))=-3(x-2)

(y+7)=-3x+6 , 3x+y+1

Equation of altitude AD is 3x+y+1=0

Equation of BE :

first of all find the slope of side AC

A(2,-7)=(x1,y1) C(4,3)=(x2,y2)

Slope of AC = (-3-(-7))/(4-2)

= 4/2= 2

Now,

(slope of BE) (slope of AC) = -1

slope of BE = -1/2 = m

By slope point form,(y-y1)=m(x-x1)

here, (x1,y1)=(-2,1)

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

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

2y+2=-x-2 , x+2y+4=0

Equation of altitude BE is x+2y+4=0

Equation of CF :

first of all find the slope of side AB

A(2,- 7)=(x1,y1) ,B(-2,1)=(x2,y2)

Slope of AB = (y2-y1)/(x2-x1)

= (1-(-7))/(-2-2) = 8/-4 = -1/2

Now

(slope of CF) (Slope of AB) = -1

Slope of CF = 2 = m

By slope point form,(y-y1)=m(x-x1)

here, (x1,y1)= C(4,3)

(y-3)=2(x-4)

(y-3) = 2x-8

y-2x-3+8 = 0 , 2x-y-5=0

Equation of altitude CF is 2x-y-5=0