# 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 (x1 , y1) and is parallel to the given line. What we could actually use here is the point-slope form of a line, which is: $y-y_1=m(x-x_1)$

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 m1 and that of the missing line be m2. By virtue of the concept of parallel lines, we know that m1 = m2. Also, m1 can be solved by using the formula: $m_1=-\displaystyle\frac{a}{b}$

Thus, $m_2=m_1=-\displaystyle\frac{a}{b}$.  Now, proceeding to the point-slope form, we have the equation of the missing line as: $y-y_1=m_2(x-x_1)$ $y-y_1=-\displaystyle\frac{a}{b}\big(x-x_1\big)$ $b(y-y_1)=-a(x-x_1)$ $by-by_1=-ax+ax_1$ $ax+by-(ax_1+by_1)=0$

Hence, we have the equation $ax+by-(ax_1+by_1)=0$ 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 (x1 , y1). We know that by virtue of perpendicular lines, $m_2=-\displaystyle\frac{1}{m_1}$  . Hence, $m_2=\displaystyle\frac{b}{a}$ . Plugging this into the point-slope form, we have: $y-y_1=m_2(x-x_1)$ $y-y_1=\displaystyle\frac{b}{a}(x-x_1)$ $a(y-y_1)=b(x-x_1)$ $ay-ay_1=bx-bx_1$ $ay-bx+(bx_1-ay_1)=0$ $bx-ay-(bx_1-ay_1)=0$

Hence, for summary, in finding the equations of parallel or perpendicular lines to any line ax + by + c = 0, and passing through the point (x1 , y1), the following conditions hold:

Parallel Line: $ax+by-(ax_1+by_1)=0$

Perpendicular Line: $bx-ay-(bx_1-ay_1)=0$

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 $ax+by-(ax_1+by_1)=0$  Since a = 2, b = -3, we have the equation of the line as: $ax+by-(ax_1+by_1)=0$ $2x-3y-(2(1)-3(4))=0$ $2x-3y-(2-12)=0$ $2x-3y-(-10)=0$ $2x-3y+10=0$

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), $2x-3y+c=0$ $2(1)-3(4)+c=0$ $-10+c=0$ $c=10$

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 $(-3, 4)$.

Solution: We know that the perpendicular line is in the form $bx-ay-(bx_1-ay_1)=0$ . Since a = 5 and b = 12, then: $bx-ay-(bx_1-ay_1)=0$ $12x-5y-(12(-3)-5(4))=0$ $12x-5y-(-36-20)=0$ $12x-5y-(-56)=0$ $12x-5y+56=0$

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: $12x-5y+c=0$ $12(-3)-5(4)+c=0$ $-36-20+c=0$ $c=56$

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