Let the area of the triangle formed by a straight line L:x+by+c=0 with coordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L makes an angle of 45∘ with the positive x-axis, then the value of b2+c2 is:
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Step-by-Step Solution
Step 1: Find intercepts and area
First, we convert the given equation of the line L:x+by+c=0 into the intercept form ax+dy=1. This allows us to identify the x -intercept as a=−c and the y -intercept as d=−c/b. The area of the triangle formed by the line with the coordinate axes is given by half the product of the absolute values of the intercepts. We are given that this area is 48 square units, so we set up the equation 2∣b∣c2=48.
Step 2: Use perpendicular distance and angle
The perpendicular distance from the origin to the line x+by+c=0 is given by the formula p=12+b2∣c∣. The direction cosines of the normal to the line Ax+By+C=0 are A2+B2A and A2+B2B. Since the perpendicular from the origin makes an angle of 45∘ with the positive x -axis, its direction cosines are cos45∘ and sin45∘. Comparing these, we find that 1+b21=cos45∘=21. Solving this equation gives 1+b2=2, which implies b2=1.
Step 3: Calculate c squared
Now that we have b2=1, we can substitute this value back into the area equation we derived earlier: 2∣b∣c2=48. Since b2=1, ∣b∣=1=1. Substituting this into the equation gives 2(1)c2=48. Multiplying both sides by 2 yields c2=96.
Step 4: Find the value of b squared plus c squared
Finally, we need to find the value of b2+c2. We have already calculated b2=1 and c2=96. Adding these two values together, we get 1+96=97.