Question

27. If the points A(-1,-4) B(b,c) C(5,-1) are collinear and 2b+c=4,find the value of

‘b’ and ‘c’
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Answer 1

GIVEN :

The points A(-1,-4) , B(b,c) and C(5,-1) are collinear and if 2b+c=4, find the value of b and c.

TO FIND :

The values of b and c in the given points

SOLUTION :

Given that the points A(-1,-4) , B(b,c) and C(5,-1) are collinear and 2b+c=4

Since the given points A , B and C are collinear then

⇒ Area of ΔABC=0

Let the points A(-1,-4) be $(x_1,y_1)$ ,

B(b,c) be $(x_2,y_2)$

From 2b+c=4

c=4-2b

The point B becomes (b,4-2b) be $(x_2,y_2)$

and C(5,-1) be $(x_3,y_3)$

The formula for the points be collinear is

$Area=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}=0$

Substitute the points in the formula we get$Area=\frac{1}{2}|(-1)(4-2b-(-1))+b(-1-(-4))+5(-4-(4-2b))|=0$

$\frac{1}{2}|(-1)(4-2b+1)+b(-1+4)+5(-4-4+2b)|=0$

$\frac{1}{2}|(-1)(5-2b)+b(3)+5(-8+2b)|=0$

$\frac{1}{2}|(-1)(5)+(-1)(-2b)+3b+5(-8)+5(2b)|=0$

$\frac{1}{2}|-5+2b+3b-40+10b|=0$

$\frac{1}{2}|15b-45|=0$

$|15b-45|=0$

15b=45

$b=\frac{45}{15}$

⇒ b=3

Substitute the value b=3 in 2b+c=4

2(3)+c=4

6+c=4

c=4-6

c=-2

⇒ c=-2

∴ the values are b=3 and c=-2 in the given points.