The vertices of a triangle are A(3,2), B(2,1) and C(1,6). Find
the equation of the line containing a) side BC b)median AD
Answers
Answer:
(a) \bold{ 6x+y-12 = 0}6x+y−12=0
(b) \bold{2x-3y = 0}2x−3y=0
(c) Mid point of BC = \bold{ (\dfrac{3}{2 } , \: 3 )}(
2
3
,3)
Mid point of AB = \bold{ [\dfrac{5}{2 } , \: 2]}[
2
5
,2]
Explanation:-
Given:-
vertices of a triangle are A(3,4), B(2,0) and C(1,6).
To Find:-
(a) Eq. side BC
(b) Eq. the median AD
(c) the mid points of sides AB and BC
Solution:-
(a) Eq. side BC :-
B = (2,0)
C = (1,6)
From Two-Point form
\boxed{\bold{\dfrac{x - x_1}{x_2-x_1 } = \dfrac{y-y_1}{y_2-y_1}}}
x
2
−x
1
x−x
1
=
y
2
−y
1
y−y
1
{\rightarrow}\dfrac{x - _2}{1-2 } = \dfrac{y-0}{6-0}→
1−2
x−
2
=
6−0
y−0
{\rightarrow}\dfrac{x - _2}{-1} = \dfrac{y}{6}→
−1
x−
2
=
6
y
{\rightarrow} 6(x-2) = -y→6(x−2)=−y
{\rightarrow} 6x-12 = -y→6x−12=−y
\bold{{\rightarrow} 6x+y-12 = 0}→6x+y−12=0
(b) Eq. the median AD:-
w.k.t,
Median bisects the side of triangle
Hence, the point where median intersects the side is midpoint of the side
let, the midpoint of BC be D
From Midpoint formula:-
\boxed{\bold{M = [\dfrac{x_1 +x_2}{2 } ,\: \dfrac{y_1+y_2}{2}]}}
M=[
2
x
1
+x
2
,
2
y
1
+y
2
]
{\rightarrow}M = [\dfrac{2+1}{2 } , \:\dfrac{6+0}{2}]→M=[
2
2+1
,
2
6+0
]
{\rightarrow}M = [\dfrac{3}{2 } , \:\dfrac{6}{2}]→M=[
2
3
,
2
6
]
{\rightarrow}M = [\dfrac{2+1}{2 } , \: 3 ]→M=[
2
2+1
,3]
We have
A = (3,4)
D = ((\dfrac{3}{2 } , \: 3 )(
2
3
,3)
From Two-point form,
{\rightarrow}\dfrac{x - \dfrac{3}{2}}{3-\dfrac{3}{2}} = \dfrac{y-3}{4-3}→
3−
2
3
x−
2
3
=
4−3
y−3
{\rightarrow}\dfrac{ \dfrac{2x-3}{2}}{\dfrac{6-3}{2}} = \dfrac{y-3}{1}→
2
6−3
2
2x−3
=
1
y−3
{\rightarrow}\dfrac{ 2x-3}{3} = \dfrac{y-3}{1}→
3
2x−3
=
1
y−3
{\rightarrow} 2x-3 = 3(y-1)→2x−3=3(y−1)
{\rightarrow} 2x-3 = 3y-3→2x−3=3y−3
\bold{{\rightarrow} 2x-3y = 0}→2x−3y=0
(c) the mid points of sides AB and BC:-
We already have the Mid-point of BC
Let us, find midpoint of AB
From Midpoint Formula
{\rightarrow}M = [\dfrac{3+2}{2 } , \:\dfrac{4+0}{2}]→M=[
2
3+2
,
2
4+0
]
{\rightarrow}M = [\dfrac{5}{2 } , \:\dfrac{4}{2}]→M=[
2
5
,
2
4
]
\bold{{\rightarrow}M = [\dfrac{5}{2 } , \: 2]}→M=[
2
5
,2]