Distance of a point from a line: Examples

Hi. This lesson will be covering examples related to distance of a point from a line.

As usual, I’ll start with a no-brainer.

Example 1 Find the distance of the point P(2, 3) from the line 4y = 3x + 1.

Solution The given line can be written as 3x – 4y + 1 = 0 (We’ll always have to transform the equation to this form before using the formula)

The required distance is |3(2)-4(3)+1|/\(\sqrt{3^2+(-4)^2}\) = 1


Example The line 12x + 5y + 9 = 0 is tangent to a circle whose center lies at C(1, 1). Find the area of the circle.

Solution Although I haven’t talked about circles in coordinate geometry yet, but you have the basic knowledge which this problem requires — the radius is perpendicular to the tangent at its point of contact.

In other words, the perpendicular distance from the center of the circle to a tangent is equal the circle’s radius. That’s all you need to know! (Oh, and also that the area of a circle equals πr2)

Talking about perpendicular distance, if D be the point of contact, then CD (or radius) = |12(1)+5(1)+9|/\(\sqrt{12^2+5^2}\) = 2

We can now calculate the area of the circle, which is a 4π square units.


Example Show that any point on the line 7x + 9y = 0 is equidistant from the lines 3x – 4y + 5 = 0 and 12x – 5y + 13 = 0.

Solution Let P(x1, y1) be any point on the line 7x + 9y = 0. Then its distance from the first line is |3x1 – 4y1 + 5|/5 and that from the second is |12x1-5y1+13|/13.

They don’t seem to be equal… yet. Because we haven’t used the fact that the point (x1,y1) lies on the line 7x + 9y = 0. That is 7x+ 9y= 0 or x1 = (-9/7)y1.

Now if we substitute the value of x1 in the previous expressions, they’ll both become |11y1-7|, hence equal. (Please do the calculations yourself to verify)

P.S. The line 7x + 9y = 0 bisects the angle between the other two lines. I’ll talk about angle bisectors later.


Example Find the area of the isosceles right angled triangle whose base has the equation 3x – y = 2 and the opposite vertex lies at (3, -3).

Solution The altitude of the triangle is the perpendicular distance of the base from the vertex, which is |3(3)-(-3)-2|/\(\sqrt{10}=\sqrt{10}\)

Since the triangle is isosceles and right angled, the equal angles will measure 45°, and using some trigonometry, we can calculate the length of the base, which will be 2\(\sqrt{10}\).

Therefore, the required area will be (1/2). \(2\sqrt{10}.\sqrt{10}\)=100 square units.


I think these examples should be enough to give a clear idea of the application of the formula.

Next, I’ll talk about finding distance between parallel lines, using this formula. See you there.

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