MHB 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:

$$y=2x+11\tag{1}$$

$$y=x+8\tag{2}$$

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:

$$2x+11=x+8$$

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:

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

Slope formula:

$$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$$

Can you proceed?

 

[ai93]

a) This is the correct line. (Yes)

b) Okay, you have two lines:

$$y=2x+11\tag{1}$$

$$y=x+8\tag{2}$$

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:

$$2x+11=x+8$$

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:

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

Slope formula:

$$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$$

Can you proceed?

Thanks!

$$2x+11=x+8$$

$$2x-x=8-11$$

$$\therefore x=-3$$

sub $$x=-3 into y=2x+11$$

= y=5

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

Nevertheless

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

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 $\overline{AB}$.