Question

Find the point on the line x + y = 4 which is at a unit distance from the line 4x + 3y – 11 = 0.
(a). (4, 0)
(b). (0, 4)
(c). (- 6, 10)
(d). (10, - 8)

Answer 1

Hint: The distance of a point \[({x_0},{y_0})\] from a line \[ax + by + c = 0\] is given as \[\dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}\]. Use this formula to find the coordinates of the point that is on the line x + y =4 and is at a unit distance from 4x + 3y – 11 = 0.

Complete step-by-step answer:
We need to find a point that lies on the line x + y = 4 which is at a unit distance from the line 4x + 3y – 11 = 0.
Let us assume that the point has the coordinates (h, k).

The point lies on the line x + y = 4, then it should satisfy this equation, hence, we have:
\[h + k = 4...........(1)\]
The distance of a point \[({x_0},{y_0})\] from a line \[ax + by + c = 0\] is given as follows:
\[d = \dfrac{{|a{x_0} + b{y_0} + c|}}{{\sqrt {{a^2} + {b^2}} }}\]
The point (h, k) is at unit distance from the line 4x +3y – 11=0. Then, we have:
\[1 = \dfrac{{|4h + 3k - 11|}}{{\sqrt {{4^2} + {3^2}} }}\]
Simplifying, we have:
\[1 = \dfrac{{|4h + 3k - 11|}}{{\sqrt {16 + 9} }}\]
\[1 = \dfrac{{|4h + 3k - 11|}}{{\sqrt {25} }}\]
\[1 = \dfrac{{|4h + 3k - 11|}}{5}\]
Multiplying both sides of the equation with 5, we have:
\[5 = |4h + 3k - 11|..............(2)\]
From equation (1), we have:
\[k = 4 - h\]
Substituting the above equation in equation (2), we have:
\[5 = |4h + 3(4 - h) - 11|\]
Simplifying, we have:
\[5 = |4h + 12 - 3h - 11|\]
\[5 = |h + 1|\]
Removing the modulus, we have:
\[h + 1 = \pm 5\]
\[h = - 6;h = 4\]
For h = - 6, from equation (1), we have:
\[k = 4 - ( - 6)\]
\[k = 10\]
Hence, the point is (- 6, 10)
For h = 4, from equation (1), we have:
\[k = 4 - 4\]
\[k = 0\]
Hence, another point is (4, 0).
Hence, the correct answers are options (a) and (c).

Note: For modulus, you have two equations and you get two solutions. You might make a mistake by directly taking only the positive value but you need to consider both the values.