TS EAMCET · Maths · Straight Lines
The line \(\mathrm{L} \equiv 6 x+3 y+\mathrm{k}=0\) divides the line segment joining the points \((3,5)\) and \((4,6)\) in the ratio \(-5: 4\). If the point of intersection of the lines \(\mathrm{L}=0\) and \(x-y+1=0\) is \(\mathrm{P}(\mathrm{g}, \mathrm{h})\) then \(\mathrm{h}=\)
- A 2 g
- B \(2 g-1\)
- C 3 g
- D \(\mathrm{g}+1\)
Answer & Solution
Correct Answer
(D) \(\mathrm{g}+1\)
Step-by-step Solution
Detailed explanation
Point of division: \((x,y) = \left(\frac{(-5)(4) + (4)(3)}{-5+4}, \frac{(-5)(6) + (4)(5)}{-5+4}\right) = (8,10)\) Substitute point into L: \(6(8) + 3(10) + k = 0 \Rightarrow 48 + 30 + k = 0 \Rightarrow k = -78\) Line L: \(6x + 3y - 78 = 0 \Rightarrow 2x + y - 26 = 0\)…
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