# Angle Between Lines: More Examples

Hi. I’ll illustrate a few more examples involving the formula for angle between two lines.

… continued from here.

Q3. Find the orthocenter of the triangle whose vertices are A(0, 0), B(1, 2) and C(2, 1).

Sol. The orthocenter is the point of concurrency of the altitudes of the triangle. What we’ll need is the equation of the altitudes (any two) and find their point of intersection.

Now that we’ve an idea about perpendicular lines, we can easily calculate the slopes of the altitudes, which are perpendicular to the respective sides.

The slope of BC is -1, which will make the slope of the altitude from A, say AD, equal to 1 (I’ve skipped all the explanation. Details here.)

Therefore using the point-slope form, the equation of AD is x – y = 0. Similarly, the equation of the altitude from B is 2x + y – 4 = 0.

Solving these two equations of the altitudes will give us the coordinates of the orthocenter, which are (4/3, 4/3)

As an exercise, try to find the coordinates of the circumcenter of this triangle (i.e. the point of concurrency of the perpendicular bisectors of the sides). The method will be quite similar.

Q4. An isosceles right angled triangle has its hypotenuse along the line 2x + y = 4. If the opposite vertex is (3, 2), then find the equations to its sides.

Sol. Here’s how you can think about the problem:

(i) We need equations of lines, which means we need two conditions for each line. One is given straight away – the lines pass through (3,2). What about the other?

(ii) The lines are sides of an isosceles right angled triangle, which means that both of them make an angle of 45° with the hypotenuse

(iii) Using the formula for angle between lines, we can obtain the slopes of both the sides. Then we can use the point-slope form to obtain their equations.

Here’s a figure to illustrate the situation…

Okay.. to the formulas!

tan45° = $$|\frac{m – (-2)}{1+(-2)m}|$$. This gives two values (due to the modulus sign) of m: 3 and -1/3 : which are the slopes of the two sides.

Finally, the equations of the sides are y-2 = -1/3(x-3) and y-2 = 3(x-3). Phew!

This might have got a little over your head. But don’t worry. Try solving the problem again, on your own. Think sequentially, step by step. You should be able to figure things out.

One more…

Q5. Find the equation of the line which is perpendicular to the line 3x + 4y = 12,  and intersects it at a point on the X axis.

We know a point through which it passes – the same point at which the given line intersects the X axis.

The point is pretty easy to find out. The y-coordinate of any point on the X-axis is 0. Substitute y=0 in the given equation to get x-coordinate as 4. Therefore, the required point is (4,0).

And, we know its slope, which will be equal to negative of the reciprocal of the given line’s slope. (using the formula m1m2= -1).

Thus, the slope of the required line is 4/3. And the required equation is y – 0 = (4/3)(x-4)

We can also use the second method which I discussed previously (Q2 (ii), Method II).

Any line perpendicular to the line 3x + 4y – 12 = 0, will be of the form 4x – 3y + k = 0. Since the line passes through (4, 0), we have 4(4) – 3(0) + k = 0, or k = -16.

Therefore the required equation is 4x – 3y – 16 = 0.

I think these examples should be sufficient to get you going with angles between lines. Next, I’ll talk about calculation of the distance of a point from a given line.