A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find the equations of the median AD and line parallel to AC passing through the point B.

A(5, 4), B(–3, –2) and C(1, –8) are the vertices of a triangle ABC. Find the equation of median AD

#### Solution 1

AD is the median

∴ CD BD ... (∵ D is the midpoint of BC)

Coordinates of D can be found by using section formula.

Let (x,y) be coordinates of the centre of the circle.

`(x,y)=((1-3)/(1+1),(-8-2)/(1+1))=(-1,-5)`

Coordinates of point D are (-1,-5)

Let m be the slope of AD.

Coordinates of A(5, 4) and D(–1, –5)

`m=(y_2-y_1)/(x_2-x_1)=(4-(-5))/(5-(-1))=9/6=3/2`

Equation of line is y = mx+c, where c is the y int ercept.

`4=(3 xx 5)/2+c ...(" Substituting the coordinates of A ")`

c=-7/2

Equation of line of AD is

`y="3x"/2-7/2`

2y =3x - 7 is equation of the median AD.

The coordinates of A (5,4) and C (1, -8)

Slope of AC = `(y_2-y_1)/(x_2-x_1)=-12/-4=3`

Substituting the coordinates of A(5,4) in the equation y = mx + c

4 = (3x5)+c

c=-11

Line parallel to AC and pas sing through B(-3, -2) has slope 3

Substituting the coordinates of B 3, 2 in the equation y=mx+c

-2=(3x -3)+c

c=7

Equation of line parallel to AC and passing through B(-3,-2) is y=3x+7

#### Solution 2

Let A(5,4) ≡ `(x_1,y_1)`; B(-3, -2) ≡ `(x_2,y_2)` and C(1,-8) ≡ `(x_3,y_3)`

D(x, y) is the midpoint of BC.

∴ the coordinates of `D = ((x_2 + x_3)/2, (y_2+y+_3)/2)`

`= ((-3+1)/2, (-2-8)/2) = (-2/2, -10/2) = (-1,-5)`

Let D(-1,-5) ≡ `(x_4,y_4)`

The equation of median AD is

`(x - x_1)/(x_1-x_4) = (y - y_1)/(y_1 - y_4)`

`:. (x-5)/(5-(-1)) = (y-4)/(4-(-5))`

`:. (x-5)/(5+1) = (y-4)/(4+5)`

`:. (x - 5)/6 = (y-4)/9`

`:. (x-5)/2 = (y -4)/3`

Multiplying both the sides by 6,

3(x - 5) = 2(y - 4) ∴ 3x - 15 = 2y - 8

∴ 3x - 2y - 15 + 8 = 0 ∴ 3x - 2y - 7 = 0