If the line $(2 x+3 y+4)+\lambda(6 x-y+12)=0$ is perpendicular to the line $7 x+5 y=2$, then $\lambda=$

If the line $(2 x+3 y+4)+\lambda(6 x-y+12)=0$ is perpendicular to the line $7 x+5 y=2$, then $\lambda=$
  1. $\frac{-27}{39}$
  2. $\frac{-29}{37}$
  3. $\frac{-27}{37}$
  4. $\frac{-28}{37}$

Solution

No solution. Refer to answer key.

Asked in: AP EAMCET 2017 (24 Apr Shift 2)

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