Q.
Two straight lines 3x+4y=5 and 4x−3y=15 intersect at the point A. The equations of the lines passing through (1,2) and intersecting the given lines at B and C such that AB=AC are
Given line 3x+4y=5 and 4x−3y=15
Slope of line 3x+4y=5 is
i.e., m1=4−3
Slope of line 4x−3y=15
i.e., m2=34
Let the slope of line BC=m
Given, AB=AC ∵tanα=tanβ 1+34mm−34=1−43m−43−m 3+4m3m−4=4−3m−3−4m 12m2−9m2−16+12m =−9−12m−12m−16m2 ⇒7m2+48m−7=0 &(7m−1)(m+7)=0 &m=71,m=−7
Equation of line passing through (1,2) and slope 71 is, =y−2=71(x−1)=x−7y+13=0
Equation of line passing through (1,2) and slope =−7 is y−2=−7(x−1) 7x+y=9