Question

Find all the points on x+y=2, which are at a distance of $\frac{4}{5}$ units from 4x+3y=7.

• A. (5, -3)
• B. (-3, 5)
• C. (6, -4)
• D. (-4, 6)

Answer 1

Answer: (A), (B)

The correct options are
A

(5, -3)

B

(-3, 5)

We will first find a general form of a point on the line x+y=2 in terms of a parameter t. We will use the given condition that those points are a distance of $\frac{4}{5}$ units from 4x+3y=7 to find the value of t.
Our line is x+y=2
Let x=t, then y=2-t
(We can also say x=t-1. We will get y=3-t. both are correct. There are other choices also).
So any point on x+y=2 can be written as (t, 2-t). We want to find the points which are at $\frac{4}{5}$ unit distance from 4x+3y=7
$\Rightarrow |\frac{4t+3(2-t)-7}{\sqrt{4^2+3^2}}|=\frac{4}{5}$

$\Rightarrow |4t+6-3t-7|=4$

$\Rightarrow |t-1|=4$

$\Rightarrow t-1=+4ort-1=-4$

$\Rightarrow t=5ort=-3$

$\Rightarrow$ The points are (5,-3) or (-3,5)[The points are (t,2-t)]
We can see that the midpoint of these two lines is the intersection of given two lines.