# Techniques in Answering Common MMC Oral Questions-Part 2

Hello guys! For the second part of this *Techniques in Answering Common MMC Oral Questions* series, we now tackle arguably the most common type of Algebra problem in MMC oral competitions – finding the equations of ** parallel** and

**.**

*perpendicular lines*Perhaps most of the aspiring MMC contestants out there already know, if not, has already devised their own ways of solving such problems quickly. But before I share the technique, let us first delve on the derivation of the ‘technique’.

**THE DERIVATION:**

Suppose that you are given the equation of the line in the form ax + by + c = 0 [which is by the way the general form for those who are still confused with it]. Now, you have to find the equation of the line passing through the point (x_{1} , y_{1}) and is **parallel** to the given line. What we could actually use here is the *point-slope form* of a line, which is:

But of course, we still need to find *m* or the slope of the line. Now, we let the slope of the given line as *m _{1}* and that of the missing line be

*m*. By virtue of the concept of parallel lines, we know that

_{2}*m*=

_{1}*m*. Also,

_{2}*m*can be solved by using the formula:

_{1}Thus, . Now, proceeding to the *point-slope form*, we have the equation of the missing line as:

Hence, we have the equation as the equation of the parallel line

Now, suppose that we are to find the equation, this time, of a perpendicular line to ax + by + c = 0 and passing through the point (x_{1} , y_{1}). We know that by virtue of perpendicular lines, . Hence, . Plugging this into the *point-slope form*, we have:

Hence, for summary, in finding the equations of parallel or perpendicular lines to any line ax + by + c = 0, and passing through the point (x_{1} , y_{1}), the following conditions hold:

**Parallel Line: **

**Perpendicular Line: **

**SAMPLE PROBLEM 1: **Find the equation of the line parallel to 2x – 3y – 3 = 0 and passing through the point (1, 4).

**Solution: **We all know from the derivations that the parallel line is in the form Since *a* = 2, *b* = -3, we have the equation of the line as:

Hence, the equation of the parallel line is 2x – 3y + 10 = 0.

**Alternative Solution: **Now, if you are having difficulties memorizing the formula, put it this way: the parallel line to 2x – 3y – 3 = 0 shall be in the form 2x – 3y + c = 0. Now, we just have to solve for c. Using the given point (1,4),

Plugging this back, we have the parallel line as 2x – 3y + 10 = 0, which is precisely what we got earlier.

**SAMPLE PROBLEM 2: **Find the equation of a line perpendicular to 5x + 12y + 13 = 0 and passing through the point .

**Solution: **We know that the perpendicular line is in the form . Since *a* = 5 and *b* = 12, then:

Hence, the equation of the perpendicular line is 12x – 5y + 56 = 0.

**Alternative Solution: **The perpendicular line is actually in the form bx – ay + c = 0. In other words, just interchange the coefficients of x and y in the first equation then change the operation in between. Then, don’t forget the constant *c* that you must add to complete the equation. Hence, since the first equation is 5x + 12y + 13 = 0, just interchange 5 and 12 then change the ‘+’ operation into a minus sign. That is, 12x – 5y + c = 0. Now, solving for c:

Plugging this back, we shall have the equation of the perpendicular line as 12x – 5y + 56 = 0, which is precisely what we got previously.

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