Question

If two distinct points lying at a unit distance from \[ 4x+3y-10=0 \] and passing through \[ x+y=4 \] are separated by a distance \(d\), then the value of \(d\) is

• A. \(10\sqrt{2}\)
• B. \(10\)
• C. \(\sqrt{2}\)
• D. \(200\)

Hint:

Distance of a point \((x_1,y_1)\) from a line \[ ax+by+c=0 \] is \[ \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}. \]

Answer 1

The Correct Option is A

Step 1: Understand the problem setup.
We need points on the line $x + y = 4$ that are at a unit distance from the line $4x + 3y - 10 = 0$. We must find the distance between these two points.

Step 2: Parametrise points on $x + y = 4$.
Any point on $x + y = 4$ can be written as $(t, 4 - t)$ for some real $t$.

Step 3: Apply the distance condition.
The distance from $(t, 4-t)$ to $4x + 3y - 10 = 0$ is \[ \frac{|4t + 3(4-t) - 10|}{\sqrt{16 + 9}} = \frac{|4t + 12 - 3t - 10|}{5} = \frac{|t + 2|}{5}. \] Set this equal to 1: \[ |t + 2| = 5. \]

Step 4: Solve for $t$.
\[ t + 2 = 5 \Rightarrow t = 3, \quad \text{or} \quad t + 2 = -5 \Rightarrow t = -7. \]

Step 5: Find the two points.
For $t = 3$: point is $(3, 1)$. For $t = -7$: point is $(-7, 11)$.

Step 6: Compute the distance between the two points.
\[ d = \sqrt{(3 - (-7))^2 + (1 - 11)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2}. \]
\[ \boxed{10\sqrt{2}} \]