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NVAMediumJEE 2023Straight Line Equations

JEE Mathematics 2023 Question with Solution

If the line l1:3y−2x=3l_1 : 3y - 2x = 3 is the angular bisector of the line l2:x−y+1=0l_2 : x - y + 1 = 0 and l3:αx+βy+17=0l_3 : \alpha x + \beta y + 17 = 0, then α2+β2−α−β\alpha^2 + \beta^2 - \alpha - \beta is equal to _____.

Answer

Correct answer:348

Step-by-step solution

Standard Method

Given: l1:3y−2x=3l_1 : 3y - 2x = 3 is the angular bisector of l2:x−y+1=0l_2 : x - y + 1 = 0 and l3:αx+βy+17=0l_3 : \alpha x + \beta y + 17 = 0.

Find: α2+β2−α−β\alpha^2 + \beta^2 - \alpha - \beta.

The solution gives the common intersection point of the bisector and the given line as P=(0,1)P=(0,1).

Since PP lies on l3l_3, substitute P=(0,1)P=(0,1) into the line used in the working:

α(0)−β(1)+17=0\alpha(0) - \beta(1) + 17 = 0

So,

β=−17\beta = -17

Now the solution considers the reflected point

Q′=(−1713,613)Q' = \left(-\frac{17}{13},\frac{6}{13}\right)

and substitutes it into l3l_3 with β=−17\beta=-17:

α(−1713)−(−17)(613)+17=0\alpha\left(-\frac{17}{13}\right) - (-17)\left(\frac{6}{13}\right) + 17 = 0

which simplifies to

−17α13+10213+17=0-\frac{17\alpha}{13} + \frac{102}{13} + 17 = 0

From the given working,

α=7\alpha = 7

Therefore,

α2+β2−α−β=72+(−17)2−7−(−17)\alpha^2 + \beta^2 - \alpha - \beta = 7^2 + (-17)^2 - 7 - (-17) =49+289−7+17= 49 + 289 - 7 + 17 =348= 348

Hence the required numerical value is 348348.

Answer from the extracted working

Given: The extracted solution concludes α=7\alpha=7 and β=−17\beta=-17.

Find: α2+β2−α−β\alpha^2 + \beta^2 - \alpha - \beta.

Substitute these directly:

α2+β2−α−β=72+(−17)2−7−(−17)=49+289−7+17=348\begin{aligned} \alpha^2 + \beta^2 - \alpha - \beta &= 7^2 + (-17)^2 - 7 - (-17) \\ &= 49 + 289 - 7 + 17 \\ &= 348 \end{aligned}

Therefore, the answer is 348348.

Common mistakes

  • Using the sign of β\beta incorrectly. The extracted working uses the form αx−βy+17=0\alpha x - \beta y + 17 = 0 during substitution, which gives β=−17\beta=-17. If the sign convention is mixed up, the final value changes. Follow the exact line form used in the working before substituting coordinates.

  • Substituting the point P=(0,1)P=(0,1) incorrectly into the line equation. Since x=0x=0, the αx\alpha x term becomes 00. Forgetting this leads to an incorrect relation between α\alpha and β\beta. Always evaluate each coordinate carefully.

  • Making an error while squaring the negative value β=−17\beta=-17. The term β2\beta^2 equals (−17)2=289(-17)^2=289, not −289-289. Square the entire signed quantity before combining terms.

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