Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

please help!
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=

Sagot :

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.

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.