Angle Bisectors: Examples


Hello. I’ll cover a few examples related to angle bisectors, before bringing the Straight Line series to a close.

Starting with a simple one…

Example 1 Find the equation of angle bisectors for the pair of lines 3x + 4y – 7 = 0 and 4x + 3y – 7 = 0.
Solution All we have to do is apply the formula (derived here).

The required equations are \(\frac{3x+4y-7}{\sqrt{{3}^2 + {4}^2}}=\pm\frac{4x+3y-7}{\sqrt{{4}^2 + {3}^2}}\)
On rearranging the terms, we get two bisectors as x – y = 0 and x + y = 2.

Note that the two bisectors are perpendicular (and always will be).

We can find a little more information about these bisectors. By finding out the angle between (any one of the) bisectors and one of the lines, we can distinguish between the acute angle bisector and the obtuse angle bisector.

That is, if you find out the angle (θ) between the bisector x – y = 0, and one of the lines, 3x + 4y – 7 = 0, you’ll get |tanθ| = 7, which being greater than 1 indicates that θ is greater than 45°, making x – y = 0 the bisector of the obtuse angle between the given lines. (and the other one of the acute angle)

The figure below will explain the situation better.

straight line angle bisector coordinate geometry

Onto the next..

Example 2 Find the equations of the lines which are equally inclined to the lines 5x + 12y – 17 = 0 and 12x + 5y – 17 = 0, and pass through the point (4, 3).
Solution Now being ‘equally inclined’ has got something to do with the angle bisectors.

Why? Because angle bisectors themselves are equally inclined to the two given lines. But they may or may not pass through the point (3, 2). In case they do, then the equations of the angle bisectors will be the required equations.

Else, we’ll need to find out the equations of lines which are parallel to the angle bisectors and pass through the given point.

The equations to the angle bisectors are \(\frac{5x+12y-17}{\sqrt{{5}^2 + {12}^2}}=\pm\frac{12x+5y-17}{\sqrt{{12}^2 + {5}^2}}\), or x – y = 0 and x + y = 2, which do not pass through the given point (3, 2)

We therefore need to find out the equations of lines which are parallel to these bisectors and pass through (3, 2).
The slopes of the bisectors are 1 and -1. Using the concept of parallel lines, the required equations using point-slope form will be y – 2 = 1(x – 3) and y – 2 = -1 (x – 3), or x – y = 1 and x + y = 5.
Here’s a figure to illustrate:

straight line angle bisector coordinate geometry

The dotted lines indicate the angle bisectors, and the blue lines are the required lines. Notice that the line x + y = 5 is equally inclined with both the lines.

And that’ll be the end.

It was quite a long journey! I’ll return to straight lines sometime later with Pair of Straight Lines and the General Equation of Second Degree, and eventually move on to Conic Sections.

You can now move on to the next topic – Circle.

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