Question

The coordinates of the three vertices of a triangle are (0, 3), (2, -3)
and (3, 4) in the given order. Find the equations of the perpendiculars
drawn from the vertices to the opposite sides.​

Answer 1

Step-by-step explanation:

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Answer 2

Perpendicular at AB is 3x+y =3

Perpendicular at BC is 7y+x =21

Perpendicular at AC is 3y-x = 9

•Coordinates vertices of a triangle

are (0, 3), (2, -3)

•Perpendicular at AB

•Slope of AB =(Y2-Y1)/(X2-X1)

=(4-3)/(3-0)

=1/3

•also , slope of AB× slope of

perpendicular= -1

•slope of perpendicular = -3

•Now, Perpendicular passes through (2,-3)

•Equation of perpendicular=>

y-Y1 =m(x-X1)

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

y+3 = -3x+6

3x +y =3 ____________(1)

•Perpendicular at BC

•Slope of BC =(Y2-Y1)/(X2-X1)

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

=7

•also , slope of AB× slope of

perpendicular= -1

•slope of perpendicular = -1/7

•Now, Perpendicular passes through (0,3)

•Equation of perpendicular=>

y-Y1 =m(x-X1)

y-(3) =(-1/7)(x-0)

7(y-3) = -x

7y+x =21 ____________(2)

•Perpendicular at AC

•Slope of AC =(Y2-Y1)/(X2-X1)

=(-3-3)/(2-0)

=-3

•also , slope of AB× slope of

perpendicular= -1

•slope of perpendicular = 1/3

•Now, Perpendicular passes through (3,4)

•Equation of perpendicular=>

y-Y1 =m(x-X1)

y-(4) =(1/3)(x-3)

3(y-4) = x-3

3y-x=9 ____________(3)