Determine co-ordinates of points B?

[ai93]

I have an equation of a line question

a) Find the equation of the straight line with gradient 2 passing through point A (-4,3)

I worked out the equation of the line, which is, y=2x+11.
But having trouble with question b) and c)

b) if the line in part a) intersects the line y=x+8 at point B, determine the co-ordinates of point B.

c) Find
i) the length
ii) the gradient

 

[MarkFL]

a) This is the correct line. (Yes)

b) Okay, you have two lines:

\(\displaystyle y=2x+11\tag{1}\)

\(\displaystyle y=x+8\tag{2}\)

To find the coordinates of point $B$, where the two lines intersect, you must solve the simultaneous system above. Since we have both lines in function form, we can just equate the two:

\(\displaystyle 2x+11=x+8\)

Solve this for $x$, and then substitute the resulting value for $x$ into either (1) or (2) to get the $y$-coordinate.

For part c), I am assuming you are to find the distance between $A$ and $B$, and the gradient or slope between the two points.

Distance formula:

\(\displaystyle d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Slope formula:

\(\displaystyle m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\)

Can you proceed?

 

[ai93]

a) This is the correct line. (Yes)

b) Okay, you have two lines:

\(\displaystyle y=2x+11\tag{1}\)

\(\displaystyle y=x+8\tag{2}\)

To find the coordinates of point $B$, where the two lines intersect, you must solve the simultaneous system above. Since we have both lines in function form, we can just equate the two:

\(\displaystyle 2x+11=x+8\)

Solve this for $x$, and then substitute the resulting value for $x$ into either (1) or (2) to get the $y$-coordinate.

For part c), I am assuming you are to find the distance between $A$ and $B$, and the gradient or slope between the two points.

Distance formula:

\(\displaystyle d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Slope formula:

\(\displaystyle m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\)

Can you proceed?

Thanks!

\(\displaystyle 2x+11=x+8\)

\(\displaystyle 2x-x=8-11\)

\(\displaystyle \therefore x=-3\)

sub \(\displaystyle x=-3 into y=2x+11\)

= y=5

for c) Why is finding the distance necessary? Since we have to use the gradient/slope formula?

Nevertheless

m=\(\displaystyle \frac{5-3}{(-3)-(-4)}\)

m=2

:D

 

[MarkFL]

Yes, everything looks correct. :D

You asked why do we need the distance formula...well, you originally posted that you need the length, and I assume you are being asked to find the length of line segment $\overline{AB}$.

 

[CellCoree]

iii) the y-intercept

b) To determine the coordinates of point B, we need to solve the system of equations formed by the two lines in part a) and y=x+8. This can be done by setting the equations equal to each other and solving for x and y.

2x+11 = x+8
x = -3

Substituting x=-3 into either equation, we get y=5. Therefore, the coordinates of point B are (-3,5).

c) i) To find the length of the line segment AB, we can use the distance formula:

d = √[(x2-x1)^2 + (y2-y1)^2]

Where (x1,y1) = (-4,3) and (x2,y2) = (-3,5)

d = √[(-3-(-4))^2 + (5-3)^2]
d = √[1+4]
d = √5

Therefore, the length of AB is √5.

ii) The gradient of a line is given by the formula m = (y2-y1)/(x2-x1). Using the coordinates of points A and B, we can find the gradient of the line AB:

m = (5-3)/(-3-(-4))
m = 2/-1
m = -2

Therefore, the gradient of line AB is -2.

iii) To find the y-intercept of the line AB, we can substitute the coordinates of point B into the equation of the line found in part a):

y = 2x+11
5 = 2(-3)+11
5 = -6+11
5 = 5

Therefore, the y-intercept of line AB is 5.