If the line's 2x+3y−5=0 and (t2+1)x+(2t−3)y+5=0 are perpendicular to each other, then absolute value of sum of all possible real values of t is
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Solution
Given lines l1:2x+3y−5=0l2:(t2+1)x+(2t−3)y+5=0 Slope of both the lines is m1=−23,m2=−t2+12t−3 So, m1×m2=−1⇒23×(t2+12t−3)=−1⇒2t2+2=9−6t ⇒2t2+6t−7=0 ⇒t1+t2=−3 ∴|t1+t2|=3