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:

\(\displaystyle 2x+y=8\tag{1}\)

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

\(\displaystyle \frac{x}{a}+\frac{y}{b}=1\tag{2}\)

and then we know the points:

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

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:

\(\displaystyle \frac{x}{4}+\frac{y}{8}=1\)

And so we now know:

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

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

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

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

\(\displaystyle \frac{8-4}{0-7}\cdot\frac{7-0}{4-0}=-\frac{4}{7}\cdot\frac{7}{4}=-1\)

We also see that:

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

Now, we may compute:

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

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

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

Now we need to find where this line intersects with:

\(\displaystyle 2x+y=8\implies y=8-2x\)

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

\(\displaystyle \frac{x+1}{2}=8-2x\)

\(\displaystyle x+1=16-4x\)

\(\displaystyle 5x=15\)

\(\displaystyle x=3\implies y=2\)

Thus:

\(\displaystyle N=(3,2)\)