Originally Posted by

**novice** Friend, sorry for misreading you question the first time.

Now, let’s go back to square one.

When you are given a line equation $\displaystyle L_1: y = m_1 x + c_1$, to which the perpendicular line equation is $\displaystyle L_2: y = \frac{-1}{m_1} x + c_2 $.

These two lines intersect at a point at the coordinate (a, b). Knowing $\displaystyle m_1$ and coordinate (a, b), you can plug them into $\displaystyle L_2$ and find $\displaystyle c_2$.

$\displaystyle L_2: b = \frac{-1}{m_1}(a)+c_2 \rightarrow c_2 = b + \frac{a}{m_1}$

Then $\displaystyle L_2: y = \frac{-1}{m_1} x + ( b + \frac{a}{m_1})$ <---This is the answer to you part 1 of your assignment.

Next we will find the equation a distance = 1 unit from the origin (0,0):

Distance equation is given by $\displaystyle d= \frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$

Since we know $\displaystyle d =1, A = \frac{-1}{m_1}, B = 1$ and the origin (0,0), we can plug them into the distance equation so that it becomes:

$\displaystyle \pm 1= \frac{|A(0)+B(o)+C|}{\sqrt{(\frac{-1}{m_1})^2+1}}$ and obtain

$\displaystyle C= \pm \sqrt{(\frac{-1}{m_1})^2+1}$

Now we have the line equations perpendicular to L:1 and passing through the point at coordinate (a, b) and 1 unit away from the origin:

$\displaystyle y = \frac{-1}{m_2} x \pm \sqrt{(\frac{-1}{m_1})^2+1}$<---this is the answer to part 2 of your assignment. One line is below the origin, and the other above the origin.

Note: In your assignment, $\displaystyle L:1$ is given as $\displaystyle y=ax + (2-1)$. The (a,b) is (1,2), and the equation $\displaystyle y = -x+3$ is only a decoy that was used to confuse you. It’s only another line that passes through (1,2) that bears no relation with $\displaystyle L_1$