The value of the `lambda` if the lines `(2x+3y+4)+lambda(6x-y+12)=0` are

Answer

N/a

Explanation

(i) Given equation of the line is
`(2x+3y+4)+lambda(6x-y+12)=0" (i)"`
If line is parallel to Y-axis i.e., it is perpendicular to X-axis.
Slope = m = `tan90^(@) = infty`
From line (i), `x(2+6lambda)+y(3-lambda)+4+12lambda=0`
and slope `=(-2+6lambda)/(3-lambda)`
`rArr (-2-6lambda)/(3-lambda)=infty`
`rArr (-2-6lambda)/(3-lambda)=(1)/(0)rArrlambda=3`
(ii) if the line (i) is perpendicular to the line `7x+y-4=0` or `y=-7x+4`
`:' (-(2+6lambda))/(3-lambda)(-7)=-1`
`rArr 14+42lambda=-3+lambda`
`rArr 41lambda=-17`
`rArr lambda=-(17)/(41)`
(iii) If the line (i) passes through the point (1,2).
Then, `(2+6+4)+lambda(6-2+12)=0`
`rArr 12+16lambda=0rArrlambda=-(3)/(4)`
(iv) If the line is parallel to X-axis the slope =0
Then, `(-(2+6lambda))/(3-lambda)=0`
`rArr -(2+6lambda)=0rArrlambda=-(1)/(3)`
So, the correct matches are (i) `to` (d), (ii) `to` (c ), (iii) `to` (a), (iv) `to` (b).