Angle Between Lines: Examples


Hello again. This lesson will cover a few examples involving angle between lines.

As usual, lets start with the simple ones.

Example 1 Find the angle between
(i) the lines 2x + y – 4 = 0 and 3x – y + 1 = 0.
(ii) the lines joining A(1,2) to B(3,4) and C(6,7) to D(8,5) (i.e. angle between AB and CD)

Solution (i) The slope of the first line is -2 and that of the second is 3. (details)

We can straight away apply the formula now. tanθ = \(|\frac{-2-3}{1+(-2)(3)}|\) = 1. Therefore θ = 45°.

Note that in case we didn’t use the absolute value, we could have obtained tanθ = -1 or θ = 135°. It doesn’t make any difference though. The two different angles correspond to the same situation geometrically (as I’ve explained previously)

(ii) This problem is similar to the previous one. Except that the equations aren’t given (which we do not require anyway). Instead we’ve been provided with points, through which we can calculate the slopes of the lines joining them.

Now, slope of AB = (4 – 2)/(3 – 1) = 1. And, slope of CD = (5 – 7)/(8 – 6) = -1.

We can observe that the product of the two slopes is -1. Hence the two lines are perpendicular (explanation), or the angle between them is 90°

 

Example 2 Find the equation of the line which passes through the point (1, 2) and is
(i) parallel to x + 2y – 3 = 0
(ii) perpendicular to 3x + y = 4

Solution There are atleast two different ways to solve both the parts. I’ll illustrate them both.

(i) Method I

Since the two lines are parallel, the slope of the required line would the same as that of the given line, which equals -1/2.

We have the slope, and a point (1, 2). We can use the point-slope form of the equation now.

The required equation is y – 2 = (-1/2)(x – 1), or x + 2y – 5 = 0. Easy.

Method II

Look at the answer obtained. Both the equations are identical, except for the constants. (I’ve discussed this before)
We can therefore assume the required line as x + 2y + k = 0. The value of the unknown constant ‘k’ will be obtained by the additional condition, i.e. the line passes through the point (1, 2).

Substituting the coordinates, we get (1) + 2(2) + k = 0, or k = -5. The required equation is x + 2y – 5 = 0.

Note that the line could also have been assumed as x + 2y = k, which would have given a different value of k (5), but the equation obtained would have been the same. (x + 2y = 5)

To generalize: Any line parallel to a given line ax + by + c = 0, will be of the form ax + by + k = 0. The value of k will be obtained by some additional given condition.

(ii) Method I
Since the two lines are perpendicular, the product of the slopes of the required line and the given line must equal -1.
We therefore have m.(-3) = -1, or m = 1/3. Using the point-slope form again, we get the required equation as y – 2 = 1/3(x -1), or x – 3y + 5 = 0. Easy again.

Method II

Look again at the answer obtained. If we compare the answer with the given line (i.e. compare the equations of two perpendicular lines), the coefficients of x and y have interchanged with a negative sign on one of them. (3x + y = 4 and x – 3y + 5 = 0)

We can therefore assume the required line as x – 3y + k = 0. (because this line already has the required slope of 1/3, which makes it perpendicular to the given line)

The value of the unknown constant ‘k’ will be obtained by the additional condition, i.e. the line passes through the point (1, 2).

Substituting the coordinates, we get (1) – 3(2) + k = 0, or k = 5. The required equation is thus x – 3y + 5 = 0.

To generalize: Any line parallel to a given line ax + by + c = 0, will be of the form bx – ay + k = 0. The value of k will be obtained by some additional given condition.

To be continued

Leave a comment