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Which point is collinear with points B and C?



A.
(0, 0)

B.
(1, 1)

C.
(1, –5)

D.
(6, –8)

please help Which point is collinear with points B and C A 0 0 B 1 1 C 1 5 D 6 8 class=


Answer :

naǫ
Points are collinear if they lie on the same line.

First find the equation of the line that passes through the points B and C.
[tex]B(4, -3) \\ x_1=4 \\ y_1=-3 \\ \\ C(-4,3) \\ x_2=-4 \\ y_2=3 \\ \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{3-(-3)}{-4-4}=\frac{3+3}{-8}=\frac{6}{-8}=-\frac{3}{4} \\ \\ y=-\frac{3}{4}x+b \\ (-4,3) \\ 3=-\frac{3}{4} \times (-4)+b \\ 3=3+b \\ b=0 \\ \\ y=-\frac{3}{4}x[/tex]

The points lie on the line y=(-3/4)x.
Now plug the coordinates of the given points into the equation and check if they satisfy the equation.

[tex](0,0) \\ x=0 \\ y=0 \\ \Downarrow \\ 0 \stackrel{?}{=} -\frac{3}{4} \times 0 \\ 0 \stackrel{?}{=} 0 \\ 0=0 \\ \hbox{the point lies on the line} \\ \\ (1,1) \\ x=1 \\ y=1 \\ \Downarrow \\ 1 \stackrel{?}{=} -\frac{3}{4} \times 1 \\ 1 \stackrel{?}{=} -\frac{3}{4} \\ 1 \not= -\frac{3}{4} \\ \hbox{the point doesn't lie on the line}[/tex]

[tex] (1,-5) \\ x=1 \\ y=-5 \\ \Downarrow \\ -5 \stackrel{?}{=} -\frac{3}{4} \times 1 \\ -5 \stackrel{?}{=} -\frac{3}{4} \\ -5 \not= -\frac{3}{4} \\ \hbox{the point doesn't lie on the line} \\ \\ (6,-8) \\ x=6 \\ y=-8 \\ \Downarrow \\ -8 \stackrel{?}{=} -\frac{3}{4} \times 6 \\ -8 \stackrel{?}{=} -\frac{9}{2} \\ -8 \not= -\frac{9}{2} \\ \hbox{the point doesn't lie on the line}[/tex]

The answer is A.

Answer:  Option 'A' is correct.

Step-by-step explanation:

Since we have given that

Coordinates of B = (4,-3)

Coordinates of C = (-4,3)

We need to find the collinear  point with B and C.

There is one method to find the collinear point i.e. Slope method.

Slope of BX = Slope of CX =  [tex]\dfrac{y_2-y_1}{x_2-x_1}[/tex]

Let Coordinates of X = (0,0)

So, Slope of BX is given by

[tex]\dfrac{0+3}{0-4}=\dfrac{3}{-4}[/tex]

Slope of CX is given by

[tex]\dfrac{0-3}{0+4}=\dfrac{-3}{4}[/tex]

So, Slope of BX = Slope of CX = [tex]\dfrac{-3}{4}[/tex]

And we can see from the graph (0,0) is the collinear point with B and C too.

Hence, Option 'A' is correct.

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