Straight Lines
Intercept form and perpendicular lines
Grade 11
Question:
<p>Let \(L\) be the line passing through the point \(P(1, 2)\) such that its intercepted segment between the co-ordinate axes is bisected at \(P\). If \(L_1\) is the line perpendicular to \(L\) and passing through the point \((-2, 1)\), then the point of intersection of \(L\) and \(L_1\) is</p>
<p>\(\left(\dfrac{4}{5}, \dfrac{12}{5}\right)\)</p>
<p>\(\left(\dfrac{11}{20}, \dfrac{29}{10}\right)\)</p>
<p>\(\left(\dfrac{3}{10}, \dfrac{17}{5}\right)\)</p>
<p>\(\left(\dfrac{3}{5}, \dfrac{23}{10}\right)\)</p>
Step-by-Step Solution
Key Concept: If a line's intercept segment between coordinate axes is bisected at point P(1,2), then P is the midpoint of the segment joining the x-intercept (a,0) and y-intercept (0,b). Use midpoint formula: (a/2, b/2) = (1,2) to find a=2, b=4.
<p><strong>Step 1:</strong> Find line L using the intercept bisection condition.</p><p>Let line L have x-intercept a and y-intercept b. The intercept segment endpoints are (a,0) and (0,b).</p><p>Since P(1,2) bisects this segment: (a/2, b/2) = (1,2)</p><p>Therefore: a = 2 and b = 4</p><p><strong>Step 2:</strong> Write equation of line L in intercept form.</p><p>Line L: x/2 + y/4 = 1, or 2x + y = 4</p><p>Slope of L: m₁ = -2</p><p><strong>Step 3:</strong> Find equation of line L₁ perpendicular to L through (-2,1).</p><p>Slope of L₁: m₂ = 1/2 (negative reciprocal of -2)</p><p>Line L₁: y - 1 = (1/2)(x + 2)</p><p>Simplifying: y = (1/2)x + 2, or x - 2y + 4 = 0</p><p><strong>Step 4:</strong> Find intersection of L and L₁.</p><p>Solve simultaneously:</p><p>2x + y = 4 ... (i)</p><p>x - 2y + 4 = 0 ... (ii)</p><p>From (ii): x = 2y - 4</p><p>Substitute in (i): 2(2y - 4) + y = 4</p><p>4y - 8 + y = 4</p><p>5y = 12 → y = 12/5</p><p>x = 2(12/5) - 4 = 24/5 - 20/5 = 4/5</p><p>∴ Point of intersection: (4/5, 12/5)</p>
Correct Answer: A