Question

The point with integral coordinates on the line \(x+y=1\), that lie at a distance \(2\) units from the line \(5x+12y=0\), is

• A. \((-7,8)\)
• B. \((-1,2)\)
• C. \((-12,13)\)
• D. \((-2,3)\)
• E. \((-3,4)\)

Hint:

For questions involving a point on one line and fixed distance from another line, first use the line equation to reduce variables, then apply the point-to-line distance formula.

Answer 1

The Correct Option is D

Step 1: Use the condition that the point lies on \(x+y=1\).
If a point lies on the line \(x+y=1\), then its coordinates must satisfy:
\[ x+y=1 \] Now we will test the given options quickly.

Step 2: Check which options lie on the line.
For each option:
\[ (-7)+8=1 \] \[ (-1)+2=1 \] \[ (-12)+13=1 \] \[ (-2)+3=1 \] \[ (-3)+4=1 \] So all the given options lie on the line \(x+y=1\). Hence, now we use the distance condition.

Step 3: Recall the distance formula from a point to a line.
The distance of a point \((x_1,y_1)\) from the line \(ax+by+c=0\) is:
\[ \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}} \] Here the line is:
\[ 5x+12y=0 \] So the distance of \((x,y)\) from this line is:
\[ \frac{|5x+12y|}{\sqrt{5^2+12^2}}=\frac{|5x+12y|}{13} \]

Step 4: Apply the given distance condition.
The point must be at distance \(2\) units from the line. Therefore:
\[ \frac{|5x+12y|}{13}=2 \] \[ |5x+12y|=26 \]

Step 5: Use the line equation \(x+y=1\).
From \[ x+y=1 \] we get: \[ x=1-y \] Substitute into \(|5x+12y|=26\):
\[ |5(1-y)+12y|=26 \] \[ |5-5y+12y|=26 \] \[ |5+7y|=26 \]

Step 6: Solve for \(y\).
So we have two cases:
\[ 5+7y=26 \quad \text{or} \quad 5+7y=-26 \] From the first:
\[ 7y=21 \Rightarrow y=3 \] Then \[ x=1-3=-2 \] From the second:
\[ 7y=-31 \Rightarrow y=-\frac{31}{7} \] which is not an integer. Since the point must have integral coordinates, this case is rejected.

Step 7: State the final answer.
Thus the required integral point is:
\[ (-2,3) \] This matches option \((4)\). Hence, the correct answer is:
\[ \boxed{(-2,3)} \]