Straight Lines
Distance from a point to a line
Grade 11
Question:
<p>Two roads are represented by the equation \(y - x = 6\) and \(x + y = 8\). An inspection bungalow has to be so constructed that it is at a distance of 100 from each of the roads. Possible location of the bungalow is given by</p>
<p>\(\left(100\sqrt{2} + 1,\, 7\right)\)</p>
<p>\(\left(1 - 100\sqrt{2},\, 7\right)\)</p>
<p>\(\left(1,\, 7 + 100\sqrt{2}\right)\)</p>
<p>\(\left(1,\, 7 - 100\sqrt{2}\right)\)</p>
Step-by-Step Solution
Key Concept: The locus of points equidistant from two lines is found using the angle bisector formula: the point must satisfy equal perpendicular distances from both lines simultaneously.
<p><strong>Step 1:</strong> Rewrite the lines in standard form:</p><p>Line 1: x - y + 6 = 0</p><p>Line 2: x + y - 8 = 0</p><p><strong>Step 2:</strong> Use the angle bisector formula. For point (x,y) equidistant from both lines:</p><p>$$\frac{x - y + 6}{\sqrt{1^2 + (-1)^2}} = ± \frac{x + y - 8}{\sqrt{1^2 + 1^2}}$$</p><p><strong>Step 3:</strong> Simplify:</p><p>$$\frac{x - y + 6}{\sqrt{2}} = ± \frac{x + y - 8}{\sqrt{2}}$$</p><p>$$x - y + 6 = ±(x + y - 8)$$</p><p><strong>Step 4:</strong> Taking the positive sign:</p><p>x - y + 6 = x + y - 8 → 2y = 14 → y = 7</p><p><strong>Step 5:</strong> Taking the negative sign:</p><p>x - y + 6 = -(x + y - 8) → x - y + 6 = -x - y + 8 → 2x = 2 → x = 1</p><p><strong>Step 6:</strong> Verify: Both lines intersect at (1, 7). The two angle bisectors are y = 7 and x = 1, which pass through the intersection point and are perpendicular to each other.</p><p>∴ Answer: A (The locus is the pair of lines x = 1 and y = 7)</p>
Correct Answer: A