Question

8. The vertices of a triangle are A(3,4), B(2,0) and C(1,6). Find the equations of the line containing
(a) side BC (b) the median AD
(c) the mid points of sides AB and BC​

Answer 1

Answer:-

(a) $\bold{ 6x+y-12 = 0}$

(b) $\bold{2x-3y = 0}$

(c) Mid point of BC = $\bold{ (\dfrac{3}{2 } , \: 3 )}$

Mid point of AB = $\bold{ [\dfrac{5}{2 } , \: 2]}$

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

$\boxed{\bold{\dfrac{x - x_1}{x_2-x_1 } = \dfrac{y-y_1}{y_2-y_1}}}$

${\rightarrow}\dfrac{x - _2}{1-2 } = \dfrac{y-0}{6-0}$

${\rightarrow}\dfrac{x - _2}{-1} = \dfrac{y}{6}$

${\rightarrow} 6(x-2) = -y$

${\rightarrow} 6x-12 = -y$

$\bold{{\rightarrow} 6x+y-12 = 0}$

(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:-

$\boxed{\bold{M = [\dfrac{x_1 +x_2}{2 } ,\: \dfrac{y_1+y_2}{2}]}}$

${\rightarrow}M = [\dfrac{2+1}{2 } , \:\dfrac{6+0}{2}]$

${\rightarrow}M = [\dfrac{3}{2 } , \:\dfrac{6}{2}]$

${\rightarrow}M = [\dfrac{2+1}{2 } , \: 3 ]$

We have

• A = (3,4)
• D = ($(\dfrac{3}{2 } , \: 3 )$

From Two-point form,

${\rightarrow}\dfrac{x - \dfrac{3}{2}}{3-\dfrac{3}{2}} = \dfrac{y-3}{4-3}$

${\rightarrow}\dfrac{ \dfrac{2x-3}{2}}{\dfrac{6-3}{2}} = \dfrac{y-3}{1}$

${\rightarrow}\dfrac{ 2x-3}{3} = \dfrac{y-3}{1}$

${\rightarrow} 2x-3 = 3(y-1)$

${\rightarrow} 2x-3 = 3y-3$

$\bold{{\rightarrow} 2x-3y = 0}$

(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

${\rightarrow}M = [\dfrac{3+2}{2 } , \:\dfrac{4+0}{2}]$

${\rightarrow}M = [\dfrac{5}{2 } , \:\dfrac{4}{2}]$

$\bold{{\rightarrow}M = [\dfrac{5}{2 } , \: 2]}$

Answer 2

Step-by-step explanation:

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]