# Given distance 2 from origin to line 5x+12y-a=0 , calculate a?

### 2 Answers | Add Yours

You need to use the following formula that helps you to evaluate the distance from a point `A(x_A,y_A)` to a line `ax + by + c = 0` , such that:

`d = |ax_A + by_A + c|/(sqrt(a^2 + b^2))`

The problem provides the distance `d = 2` from origin point `(0,0)` , to the line `5x + 12y - a = 0` , such that:

`2 = |5*0 + 12*0 - a|/(sqrt(5^2 + 12^2))`

`2 = |-a|/sqrt(25+144) => 2 = |a|/sqrt169 => |a| = 2*13 = 26`

Using the absolute value definition yields:

`|x| = t => x = +-t`

By definition, yields:

`|a| = 26 => a = +-26`

**Hence, evaluating the constant term `a` , under the given conditions, yields `a = +-26` .**

**Sources:**

Let `P(x_1,y_1)` be point and `ax+by+c=0` be the given line.Then `+-d=(ax_1+by_1+c)/sqrt(a^2+b^2)` be the distace from P to the line.

Given line

5x+12y-a=0 and point be (0,0) ,distace is 2 unit.

Thus

`+-2=(5xx0+12xx0-a)/sqrt(5^2+12^2)`

`+-2=(-a)/13`

`a=+-26` unit.