can anyone please help me with the following. I need to find possible equations of a line whose length of a segment is 5 between the lines with the following equations:
$$x+2y+1=0$$
$$x+2y-1=0$$
And it contains a point P(-5,4). So far I've written this:
$$y-4 = k(x+5)$$
But I am stuck there, I've created a system of equations but messed something up and can't get a solution, so any help would be much appreciated, cheers 🙂
Find equation of line given length of a segment and point
analytic geometrygeometry
Related Solutions
An alternative method even though the question has been answered:
Begin by creating a vector $\mathbf{\overrightarrow{\text{BA}}}$ by subtracting $\mathbf{A}$ from $\mathbf{B}$: $\langle 3, 3, 3 \rangle - \langle -2, -2, -2 \rangle = \langle 5, 5, 5 \rangle$. Then, normalize this vector by dividing $\mathbf{\overrightarrow{\text{BA}}}$ by its length, $ \frac{\mathbf{\overrightarrow{\text{BA}}}}{\lVert \mathbf{\overrightarrow{\text{BA}}} \rVert} = \frac{\langle 5, 5, 5 \rangle}{\sqrt{5^2+5^2+5^2}} = \langle 0.577, 0.577, 0.577 \rangle$. This new unit vector can then be scaled and added to $\mathbf{A}$ to find the point in space at the desired distance. In this case: $\mathbf{A} + 3\langle 0.577, 0.577, 0.577 \rangle = \langle -0.268, -0.268, -0.268 \rangle$.
Hope this is useful/correct!
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Can we approach the problem without using the cosine rule though? I didn't learn about that yet, I only studied sin and cos in right triangles so far.
Also, when asked if you were familiar with using the sine function to find the triangle area, e.g., $S_{\triangle ABC}=\frac{1}{2}(ab\sin(\angle C))$, you stated in your commment that "Unfortunately no".
Thus, this solution uses only the the sin and cos in specific right triangles, specifically $30^{\circ}-60^{\circ}-90^{\circ}$ ones having side lengths in ratios of $1$, $\sqrt{3}$ and $2$, as well as isosceles and similar triangle properties, so I trust these techniques are more familiar to you.
As shown in the diagram below, extend $AC$ to $G$ so that $\lvert AG\rvert = \lvert AB\rvert = 84$, so $\lvert CG\rvert = 84 - 28 = 56$. Also, add the line segment $BG$. In addition, extend $AD$ to meet $BG$ at $E$, and add the vertical line segment of $CF$, with $F$ on $BG$.
Since $\triangle BAG$ is isosceles (with $\measuredangle EBA = \measuredangle EGA$), and $AE$ is an angle bisector, then $\measuredangle AEB = \measuredangle AEG = 90^{\circ}$ (e.g., as explained in Geometric proof - bisector of triangle). Thus, since $\measuredangle BAE = \measuredangle GAE = 60^{\circ}$, we also have $\measuredangle ABE = \measuredangle AGE = 30^{\circ}$, so
$$\lvert AE \rvert = \frac{84}{2} = 42, \;\; \lvert BE \rvert = \lvert GE \rvert = 84\left(\frac{\sqrt{3}}{2}\right) = 42\sqrt{3} \tag{1}\label{eq1A}$$
With $\triangle GCF$ also being a $30^{\circ}-60^{\circ}-90^{\circ}$ one, then
$$\lvert CF \rvert = \frac{56}{2} = 28, \;\; \lvert FG \rvert = 56\left(\frac{\sqrt{3}}{2}\right) = 28\sqrt{3} \tag{2}\label{eq2A}$$
Using \eqref{eq1A}, we get
$$\begin{equation}\begin{aligned} & \lvert EF\rvert = \lvert EG\rvert - \lvert FG\rvert = 42\sqrt{3} - 28\sqrt{3} = 14\sqrt{3} \\ & \lvert BF\rvert = \lvert BE\rvert + \lvert EF\rvert = 42\sqrt{3} + 14\sqrt{3} = 56\sqrt{3} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Next, $\triangle BDE$ being similar to $\triangle BCF$ means that
$$\frac{\lvert DE\rvert}{\lvert CF\rvert} = \frac{\lvert BE\rvert}{\lvert BF\rvert} \;\to\; \frac{\lvert DE\rvert}{28} = \frac{42\sqrt{3}}{56\sqrt{3}} \;\to\; \lvert DE\rvert = 28\left(\frac{3}{4}\right) = 21 \tag{4}\label{eq4A}$$
Finally, using this and \eqref{eq1A}, we get
$$\lvert AD\rvert = \lvert AE\rvert - \lvert DE\rvert = 42 - 21 = 21 \tag{5}\label{eq5A}$$
Best Answer
Note that the distance between the two given parallel lines is $d=\frac2{\sqrt5}$ and the slope normal to them is $m=2$. Let $m’$ be the slope of the unknown line and $\theta$ the angle between the given lines and the unknown line. Then, $\cos\theta=\frac d5=\frac2{5\sqrt5}$ and
$$\tan\theta = \pm \frac{11}2 = \frac{m-m’}{1+mm’}$$
which yields $m’ = -\frac7{24},\>-\frac34$. Use the point-slope formula below to obtain the equations of the lines
$$y-4=m’(x+5)$$