MHB How Do You Solve a Triangle Problem on a Number Plane?

[ArchieK]

Please just hint me in the right direction, I'm kind of lost with it. Thanks for any help

View attachment 4519

 

[MarkFL]

For part i), we need to know the coordinates of both points $A$ and $B$. This means we need to find the two intercepts of the line:

$$2x+y=8\tag{1}$$

One way to do this is the express the line in the two-intercept form:

$$\frac{x}{a}+\frac{y}{b}=1\tag{2}$$

and then we know the points:

$$A=(a,0)\,\text{and}\,B=(0,b)$$

are on the line. So, can you express (1) in the form of (2)?

 

[MarkFL]

I see this thread has been marked as [SOLVED], and so I will post methods for answering the questions for the benefit of other users who might find this thread.

If we divide (1) by 8, we obtain:

$$\frac{x}{4}+\frac{y}{8}=1$$

And so we now know:

$$A=(4,0)\,\text{and}\,B=(0,8)$$

Now we can answer part i) using the distance formula:

$$\overline{AB}=\sqrt{(0-4)^2+(8-0)^2}=4\sqrt{1^2+2^2}=4\sqrt{5}$$

For part ii), let observe that $$\triangle OBC$$ is a right triangle, since we can show $$\overline{BC}\perp\overline{OC}$$ using the definition of slope and the fact that two lines are perpendicular if the product of their slopes is -1:

$$\frac{8-4}{0-7}\cdot\frac{7-0}{4-0}=-\frac{4}{7}\cdot\frac{7}{4}=-1$$

We also see that:

$$\overline{OC}=\overline{BC}$$, which means $$\triangle OBC$$ is a right isosceles triangle.

Now, we may compute:

$$\angle ABC=\angle OBC-\angle OBA=45^{\circ}-\arctan\left(\frac{1}{2}\right)\approx18^{\circ}$$

For part iii), we need to first find the line passing through $C$, which is perpendicular to $$\overline{AB}$$...using the point-slope formula, we obtain:

$$y=\frac{1}{2}(x-7)+4=\frac{x+1}{2}$$

Now we need to find where this line intersects with:

$$2x+y=8\implies y=8-2x$$

Hence, equating both expressions for $y$, we get:

$$\frac{x+1}{2}=8-2x$$

$$x+1=16-4x$$

$$5x=15$$

$$x=3\implies y=2$$

Thus:

$$N=(3,2)$$