Question

Lines a, b, c, and d are in the same plane. Line a is parallel to line c. Lines b and c are perpendicular to line d. Based on this, tell how lines a and b are related.

Its my last question and I have been stuck on it for some time.

Answer 1

Are you familiar with vectors on analytic geometry?

Answer 2

Not really....

Answer 3

I dont know how to draw out the plane....

Answer 4

Well then, let's then use slopes and traditional geometry.
Let's for instance name the lines with their equations:
\[(a) a_1x+b_1y+k_1=0\]
\[(b) a_2x+b_2y+k_2=0\]
\[(c)a_3x+b_3y+k_3=0\]
\[(d)a_4x+b_4y+k_4=0\]
See I numered the coefficients in order to separate their equations since they are all different lines.
Now, let's begin... We can begin by calculating the slope of line (a) and (c):
\[a_1x+b_1y+k_1=0 \rightarrow y= -\frac{ a_1 }{ b_1 }x-\frac{ k_1 }{ b_1 } \iff m_a=-\frac{ a_1 }{ b_1 } \]
The same will be applied for line (c) obtaining: \(m_c = -\frac{ a_3 }{ b_3 }\). Now because they are parallel, must mean that their slopes are equal due to the condition of parallelism in geometry, therefore: \[m_a = m_c \iff -\frac{ a_1 }{ b_1 }=-\frac{ a_3 }{ b_3 } \iff \frac{ a_1 }{ b_1 }=\frac{ a_3 }{ b_3 }\]
Remember that equality we have just found, because now, the excercise tells us that line (c) and (b) are perpendicular to line (d), so, we will state those two by calculating the slope of each line, and then use that equality we have calculated above to see if we can match something, we can calculate their slope as we did with line (a) and (c):
\(m_b = -\frac{ a_2 }{ b_2 }\) and \(m_d=-\frac{ a_4 }{ b_4 }\) and now we can apply the condition of perpendicularity and see what happens, and we know the condition of perpendicularity states that \(m_1 \perp m_2 \iff m_1=-\frac{ 1 }{ m_2 }\) which is useful since it allows us to do further analysis on this problem:
\[m_c \perp m_d \iff -\frac{ a_3 }{ b_3 }=-(-\frac{ 1 }{ \frac{ a_4 }{ b_4 } }) \]
\[m_b \perp m_d \iff -\frac{ a_2 }{ b_2 }=-(-\frac{ 1 }{ \frac{ a_4 }{ b_4 } })\]
But look, both slopes are equal to \(-(-\frac{ 1 }{ \frac{ a_4 }{ b_4 } })\) which means that \(m_c = m_b\) which we can translate as:
\[m_c=m_b \iff -\frac{ a_3 }{ b_3 }=-\frac{ a_2 }{ b_2 } \rightarrow \frac{ a_3 }{ b_3 }=\frac{ a_2 }{ b_2 }\]
But remember, wasn't \(\frac{ a_1 }{ b_1 }=\frac{ a_3 }{ b_3 }\) ? indeed it was, so we can write it like this:
\[\frac{ a_1 }{ b_1 }=\frac{ a_3 }{ b_3 }=\frac{ a_2 }{ b_2 }\]
which by transitive property of numbers we obtain:
\[\frac{ a_1 }{ b_1 }=\frac{ a_2 }{ b_2 } \iff -\frac{ a_1 }{ b_1 }=-\frac{ a_2 }{ b_2 }\]
\[\iff m_a=m_b\]
look!, we have just found that that line (a) and line (b) have the same slope. Remember what it meant?, that they are parallel and we can write it as a conclusion:
\[m_a=m_b \rightarrow \therefore (a) \parallel (b)\]
We have just concluded, without any doubt, that line (a) and line (b) under this conditions are indeed parallel.

Answer 5

y does dis seem so obsolete