Question

Find the points on the x-axis, whose distances from the line \(\frac{ x}{3}+\frac{y}{4}=1\) are 4 units.

Answer 1

The given equation of line is \(\frac{ x}{3}+\frac{y}{4}=1\)
or\( 4x + 3y – 12 = 0......(1)\)
On comparing equation (1) with general equation of line \(Ax + By + C = 0\), we obtain \(A = 4, B = 3\), and \(C = -12. \)
Let \((a, 0)\) be the point on the x-axis whose distance from the given line is 4 units.
It is known that the perpendicular distance (d) of a line \(Ax + By + C = 0\) from a point \( (x_1, y_1)\) is given by

\(d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\)
Therefore, 

\(4=\frac{|4a+3\times0-12|}{\sqrt{4^2+3^2}}\)

\(⇒4=\frac{|4a-12|}{5}\)

\(⇒|4a-12|=20\)
\(⇒±(4a-12)=20\)
\(⇒(4a-12)=20\) or \(-(4a-12)=20\)
\(⇒4a=20+12\) or \(4a=-20+12\)
\(⇒a=8\space or -2\)
Thus, the required points on the x-axis are \((-2, 0) \)and \((8, 0).\)