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JEE Main 2021
Straight Lines
Straight Lines and Pair of Straight Lines
Hard

Question

If the line segment joining the points (5,2)(5,2) and (2,a)(2, a) subtends an angle π4\frac{\pi}{4} at the origin, then the absolute value of the product of all possible values of aa is :

Options

Solution

Key Concepts and Formulas

  • Slope of a line: The slope of a line passing through points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
  • Angle between two lines: If θ\theta is the angle between two lines with slopes m1m_1 and m2m_2, then tanθ=m2m11+m1m2\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|. This formula holds when 1+m1m201 + m_1m_2 \neq 0.
  • Absolute value property: If x=a|x| = a, where a>0a > 0, then x=ax = a or x=ax = -a.

Step-by-Step Solution

Step 1: Define the points and identify the required slopes. We are given points P(5,2)P(5, 2) and Q(2,a)Q(2, a). We need to find the angle between the lines OPOP and OQOQ, where OO is the origin (0,0)(0, 0). Let m1m_1 be the slope of OPOP and m2m_2 be the slope of OQOQ. Explanation: This step sets up the problem by defining the points and identifying the key slopes we need to calculate.

Step 2: Calculate the slope of line OPOP. Using the slope formula with points O(0,0)O(0, 0) and P(5,2)P(5, 2): m1=2050=25m_1 = \frac{2 - 0}{5 - 0} = \frac{2}{5} Explanation: We apply the slope formula to find the slope of the line segment connecting the origin and point P.

Step 3: Calculate the slope of line OQOQ. Using the slope formula with points O(0,0)O(0, 0) and Q(2,a)Q(2, a): m2=a020=a2m_2 = \frac{a - 0}{2 - 0} = \frac{a}{2} Explanation: We apply the slope formula to find the slope of the line segment connecting the origin and point Q.

Step 4: Apply the angle formula. We are given that the angle between the lines is π4\frac{\pi}{4}, so tan(π4)=1\tan\left(\frac{\pi}{4}\right) = 1. Plugging the slopes m1m_1 and m2m_2 into the angle formula: 1=a2251+a2251 = \left| \frac{\frac{a}{2} - \frac{2}{5}}{1 + \frac{a}{2} \cdot \frac{2}{5}} \right| Explanation: We substitute the known slopes and angle into the formula for the tangent of the angle between two lines.

Step 5: Simplify the expression inside the absolute value. 1=5a4101+a5=5a4105+a5=5a41055+a=5a42(5+a)=5a410+2a1 = \left| \frac{\frac{5a - 4}{10}}{1 + \frac{a}{5}} \right| = \left| \frac{\frac{5a - 4}{10}}{\frac{5 + a}{5}} \right| = \left| \frac{5a - 4}{10} \cdot \frac{5}{5 + a} \right| = \left| \frac{5a - 4}{2(5 + a)} \right| = \left| \frac{5a - 4}{10 + 2a} \right| Explanation: We simplify the complex fraction within the absolute value to make it easier to solve.

Step 6: Solve the absolute value equation. Since x=1|x| = 1, we have two cases: x=1x = 1 or x=1x = -1.

Case 1: 5a410+2a=1\frac{5a - 4}{10 + 2a} = 1 5a4=10+2a5a - 4 = 10 + 2a 3a=143a = 14 a=143a = \frac{14}{3}

Case 2: 5a410+2a=1\frac{5a - 4}{10 + 2a} = -1 5a4=102a5a - 4 = -10 - 2a 7a=67a = -6 a=67a = -\frac{6}{7} Explanation: We consider both possible cases arising from the absolute value and solve for aa in each case.

Step 7: Calculate the product of the possible values of aa and its absolute value. The two possible values for aa are 143\frac{14}{3} and 67-\frac{6}{7}. Their product is: P=143(67)=14637=272337=4P = \frac{14}{3} \cdot \left(-\frac{6}{7}\right) = -\frac{14 \cdot 6}{3 \cdot 7} = -\frac{2 \cdot 7 \cdot 2 \cdot 3}{3 \cdot 7} = -4 The absolute value of the product is: P=4=4|P| = |-4| = 4 Explanation: We calculate the product of the two possible values of aa and then find the absolute value of that product.

Common Mistakes & Tips

  • Forgetting the negative case in absolute value equations: Remember to consider both positive and negative possibilities when solving an absolute value equation.
  • Arithmetic errors: Be careful with fractions and simplification steps to avoid arithmetic mistakes.
  • Checking for extraneous solutions: While not strictly necessary here, always check if your solutions make the denominator zero in any of the intermediate steps.

Summary

We used the formula for the angle between two lines to set up an equation involving the unknown aa. By considering both positive and negative cases of the absolute value, we found two possible values of aa. The absolute value of the product of these values is 4.

The final answer is 4\boxed{4}, which corresponds to option (A).

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