Question

The point of intersection of the straight lines 2x+3y+4=0,6x-y+12=0 is​

Answer 1

Step-by-step explanation:

The two lines are 

x(2+6k)+y(3−k)+4+12k=0     ......(1)

and 7x+5y−4=0       ......(2)

Let m1 and m2 be the slopes of 1 and 2 respectively. Then,

m1=−3−k2+6k,m2=−57

If lines 1 and 2 are perpendicular. Then,

m1m2=−1 

(−3−k2+6k)(−57)=−1

14+42k=−15+5k

k=−3729