Question

The mid-point of the points on $x+y=4$ which are at a unit distance from the line $4x+3y-10=0$ is

• A. (-2,6)
• B. (2,2)
• C. (-7,11)
• D. (1,3)

Answer 1

The correct option is A (-2,6)

Let the required points on the line $x+y=4$ be P$(x_1,y_1)$ and Q$(x_2,y_2)$.

By ASA criterion, the triangles formed by the lines and the perpendiculars are congruent.
If the point of intersection of the lines is O, then OP=OQ

Thus, O is the mid point of PQ.

Hence, the required point is the point of intersection of the lines $x+y=4$ and $4x+3y–10=0$

On solving the equations, we get the point as (-2,6)