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Points of Intersection in Analytic Geometry
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The direction with $+x-axis$ in which a straight line will be drawn through the point $\left(1,2\right)$ so that its point of intersection with the line $x+y=4$ may be at a distance $\sqrt { \dfrac { 2 }{ 3 } }$ from the point $\left(1,2\right)$ can be:
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A
${15}^{o}$
💡 Explanation:
Solving the geometry: The line through (1,2) at angle θ to +x-axis meets x+y=4 at distance √(2/3). Using parametric form and distance formula gives θ = 15° as the valid solution. This involves setting up the line equation, finding intersection point, and applying distance constraint.