In a \triangle A B C, side A B has the equation 2 x + 3 y = 29 and the side A C has the equation x + 2 y = 16. If the mid point of B C is \left(\right. 5 , 6 \left.\right), then the equation of B C is

Engineering

Mathematics

Line of Greatest Slope

Question

In a

△ A B C

, side

A B

has the equation

2 x + 3 y = 29

and the side

A C

has the equation

x + 2 y = 16

. If the mid point of

B C

(5, 6)

, then the equation of

B C

2 x + y = 7

x + y = 1

2 x - y = 17

None of these

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Let co-ordinates of

B

(x_{1}, y_{1})

C

(x_{2}, y_{2})

∴

(5, 6)

is the mid point,

so,

\frac{x_{1} + x_{2}}{2} = 5, \frac{y_{1} + y_{2}}{2} = 6

\Rightarrow x_{1} + x_{2} = 10, y_{1} + y_{2} = 12

B (x_{1}, y_{1})

lies on the line

2 x + 3 y = 29

∴ 2 x_{1} + 3 y_{1} = 29

----(1)

C (x_{2}, y_{2})

lies on the line

x + 2 y = 16,

∴ x_{2} + 2 y_{2} = 16

----(2)

∴

putting

x_{1}, y_{1}

in the form of

x_{2}, y_{2}

in (1)

2 (10 - x_{2}) + 3 (12 - y_{2}) = 29

{

x_{1} = 10 - x_{2}, y_{1} = 12 - y_{2}

}

\Rightarrow 20 - 2 x_{2} + 36 - 3 y_{2} = 29

\Rightarrow 2 x_{2} + 3 y_{2} = 27

----(3)

on subtracting

(3)

and

(2)

\times

2

- y_{2} = - 5

y_{2} = 5

Putting

y_{2}, i n (2)

x_{2} + 2 (5) = 16

x_{2} = 6

x_{1} = 10 - x_{2}

= 4

y_{1} = 12 - 5 = 7

Equation :

\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}}

\Rightarrow \frac{x - 4}{2} = \frac{y - 7}{- 2}

\Rightarrow - x + 4 = y - 7

\Rightarrow x + y = 11

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