Question

If the slope of the line \((\sqrt{3}+\sqrt{2})y+(\sqrt{8}-\sqrt{12})x=0\) is \(a-\sqrt{b}\), then \(a^2+b=\)

• A. 186
• B. 196
• C. 206
• D. 216

Hint:

When surds appear in the slope, rationalizing the denominator usually converts the answer into the form \(a-\sqrt{b}\).

Answer 1

The Correct Option is B

Concept: For a straight line of the form \[ Ax+By+C=0, \] the slope is given by \[ m=-\frac{A}{B}. \] We first simplify the given equation and then determine its slope.

Step 1: Simplify the coefficients. Given \[ (\sqrt{3}+\sqrt{2})y+(\sqrt{8}-\sqrt{12})x=0 \] Using \[ \sqrt{8}=2\sqrt{2}, \qquad \sqrt{12}=2\sqrt{3}, \] the equation becomes \[ (\sqrt{3}+\sqrt{2})y+(2\sqrt{2}-2\sqrt{3})x=0. \] Factorizing the coefficient of \(x\), \[ (\sqrt{3}+\sqrt{2})y+2(\sqrt{2}-\sqrt{3})x=0. \]

Step 2: Find the slope. Comparing with \[ Ax+By+C=0, \] we get \[ A=2(\sqrt{2}-\sqrt{3}), \qquad B=\sqrt{3}+\sqrt{2}. \] Hence, \[ m=-\frac{2(\sqrt{2}-\sqrt{3})}{\sqrt{3}+\sqrt{2}} =\frac{2(\sqrt{3}-\sqrt{2})}{\sqrt{3}+\sqrt{2}}. \] Rationalizing the denominator, \[ m= \frac{2(\sqrt{3}-\sqrt{2})^2} {(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}. \] Since \[ (\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1, \] \[ m=2(3+2-2\sqrt6) \] \[ m=10-4\sqrt6. \] Thus \[ m=10-\sqrt{96}. \] Therefore, \[ a=10, \qquad b=96. \]

Step 3: Calculate \(a^2+b\). \[ a^2+b=10^2+96 \] \[ =100+96 \] \[ =196. \]