Question

Find the direction in which a straight line must be drawn through the point (-1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point

Answer 1

Let \(y = mx + c\)  be the line through point (-1, 2).

Accordingly, \(2 = m (-1) + c. \)
\(⇒ 2 = -m + c\)
\(⇒ c = m + 2 \)
\(∴ y = mx + m + 2 … (1) \)

The given line is  \(x + y = 4 … (2)\)

On solving equations (1) and (2), we obtain

\(x=\frac{2-m}{m+1}\)  and \(y=\frac{5m+2}{m+1}\)

\(∴ \left(\frac{2-m}{m+1} ,\frac{5m+2}{m+1}\right)\) is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (- 1, 2), according to distance formula,
\(\sqrt{\left(\frac{2-m}{m+1}+1\right)^2+\left(\frac{5m+2}{m+1}-2\right)^2}=3\)

\(⇒\left(\frac{ 2-m+m+1}{m+1}\right)^2+\left(\frac{5m+2-2m-2}{m+1}\right)^2=32\)

\(⇒\frac{ 9}{\left(m+1\right)^2}+\frac{9m^2}{\left(m+1\right)2}=9\)

\(⇒ \frac{1+m^2}{\left(m+1\right)^2}=1\)

\(⇒ 1+m^2=m^2+1+2m\)
\(⇒ 2m=0\)
\(⇒ m=0\)

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.