Please help with this

[nathan411]

A(-2, 1), B(6, 5) and C(4, k) are the vertices of a right- angled triangle ABC.
Angle ABC is the right angle.

Find an equation of the line that passes through A and C.
Give your answer in the form ay + bx + c where a, b and c are integers.

 

[Deleted member 4993]

A(-2, 1), B(6, 5) and C(4, k) are the vertices of a right- angled triangle ABC.
Angle ABC is the right angle.

Find an equation of the line that passes through A and C.
Give your answer in the form ay + bx + c where a, b and c are integers.

If a line passes through two points - whose co-ordinates are (x₁,y₁) and (x₂,y₂) - the equation of the line passing through those points is:

(y - y₁)/(x - x₁) = (y₂ - y₁)/(x₂ - x₁)

Continue....

 

[MarkFL]

Hello, and welcome to FMH!

Consider the following diagram:



I have plotted points A and B, and we know point C mut be at the intersection of the vertical line \(x=4\), and the line through B that is perpendicular to the line through A and B. Can you find the equation of this line?

 

[HallsofIvy]

"Find an equation of the line that passes through A(-2,1) and C(4, k).
Give your answer in the form ay + bx + c where a, b and c are integers."
This makes no sense. You are asked to "find an equation" but then told to "give your answer in the form ay+ bx+ c" which is not an equation! I am going to assume that the equation is supposed to be ay+ bx+ c= 0.
At the point A x= -2 and y= 1 so we have a- 2b+ c= 0.
At the point C x= 4 and y= k so we have ka+ 4b+ c= 0.
We have two equations in three unknowns so the solution will not be unique. We can eliminate c by subtracting the first equation from the second: (k-1)a+ 6b= 0. b= (1-k)a/6. Are we to assume that "k" is an integer? If so then so is 1- k and taking a= 6, an integer, b= 1- k, an integer. Putting a= 6, b= 1- k into the equation we got from point A, 6- 2(1- k)+ c= 0 so c= 2- 2k- 6= -4- 2k.

An equation of the line is 6y+ (1-k)x- 4- 2k= 0.

This is a solution. Choosing a to be other non-zero multiples of 6 would give other valid solutions.

 

[Macha]

"Find an equation of the line that passes through A(-2,1) and C(4, k).
Give your answer in the form ay + bx + c where a, b and c are integers."
This makes no sense. You are asked to "find an equation" but then told to "give your answer in the form ay+ bx+ c" which is not an equation! I am going to assume that the equation is supposed to be ay+ bx+ c= 0.
At the point A x= -2 and y= 1 so we have a- 2b+ c= 0.
At the point C x= 4 and y= k so we have ka+ 4b+ c= 0.
We have two equations in three unknowns so the solution will not be unique. We can eliminate c by subtracting the first equation from the second: (k-1)a+ 6b= 0. b= (1-k)a/6. Are we to assume that "k" is an integer? If so then so is 1- k and taking a= 6, an integer, b= 1- k, an integer. Putting a= 6, b= 1- k into the equation we got from point A, 6- 2(1- k)+ c= 0 so c= 2- 2k- 6= -4- 2k.

An equation of the line is 6y+ (1-k)x- 4- 2k= 0.

This is a solution. Choosing a to be other non-zero multiples of 6 would give other valid solutions.

I believe the asker desires an equation with only the variables y and x, not k as well.

 

[Deleted member 4993]

I believe the asker desires an equation with only the variables y and x, not k as well.

MarkFL (response#3) provided the method/process needed to calculate the numerical value of "k".