TS EAMCET · Maths · Straight Lines

If the lines $L_{1} \equiv 2 x + y + 3 = 0, L_{2} \equiv k x + 2 y - 3 = 0$ and $L_{3} \equiv 3 x - 2 y + 1 = 0$ are concurrent then the cosine of the acute angle between the lines $L_{2} = 0$ and $2 x - 5 y + 7 = 0$ is

A $\frac{1}{\sqrt{2}}$
B $(\frac{15}{2 \sqrt{29}})$
C $(\frac{25}{29})$
D $(\frac{20}{29})$

Verified Solution

Answer & Solution

Correct Answer

(D) $(\frac{20}{29})$

Step-by-step Solution

Detailed explanation

Given, The lines L1≡2x+y+3=0, L2≡kx+2y-3=0 and L3≡3x-2y+1=0 are concurrent, So by using the condition of concurrent we get, 213k2-33-21=0 ⇒22-6-1k+9+3-2k-6=0 ⇒k=-5 Now L2=-5x+2y-3=0 So, slope of L2 will be m1=52 And slope of line 2x-5y+7=0 will…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If the lines L1≡2x+y+3=0,L2≡kx+2y-3=0 and L3≡3x-2y+1=0 are concurrent then the cosine of the acute angle between the lines L2=0 and 2x-5y+7=0 is

Step-by-step Solution

If the lines $L_{1} \equiv 2 x + y + 3 = 0, L_{2} \equiv k x + 2 y - 3 = 0$ and $L_{3} \equiv 3 x - 2 y + 1 = 0$ are concurrent then the cosine of the acute angle between the lines $L_{2} = 0$ and $2 x - 5 y + 7 = 0$ is