Answer :
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.
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.
