# IMAT 2014 Q55 [Equation of a Straight Line]

What is the equation of the straight line which passes through (­-6 , 2) and is perpendicular to 4y + 3x = 8?

A. 3y ­- 4x = 18
B. 4y ­- 3x = 26
C. 3y + 4x = ­18
D. 3y ­- 4x = 30
E. 3y + 4x = 30

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The first thing we should solve for is the gradient of this new line. Luckily, we know that if two lines are perpendicular to each other their gradients multiply to give -1 as a product.

First we can rearrange the equation of the given line to say:

y= -3/4x +2

Next, using the equation m_1\times m_2= -1

where -3/4 is m_2, we can construct:

m_1\times -3/4= -1 and therefore m_1= 4/3

Now that we have the gradient of our line, we can use the given points to find the constant.

y= mx+c for the points (-6,2) becomes

2= 4/3(-6) +c

2= -24/3 +c

6= -24 +3c

3c= 30

c= 10

Finally, we can construct our equation to be y= 4/3x+10.

By looking we can see our answer is either D or E.

Notice the questions asks of the equation in a different format, where the constant is the sum.

Rearrange the mentioned equation and it gives us: 3y-4x= 30

Therefore our answer is D.

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