O(0,0),B(−3,−1) and C(−1,−3) are vertices of a △OBC.D is a point on OC and E is a point on OB. If the equation of DE is 2x+2y+√2=0, then the ratio in which the line DE divides the altitude of the △OBC is
We have O(0,0),B(−3,−1),C(−1,−3) are vertices of a △OBC and equation of DE is 2x+2y+√2=0
Slope of BC is ‌
−3−(−1)
−1−(−3)
=‌
−2
2
=−1 ∴ Slope of attitude from 0 to BC is 1(⟂ to BC) ∴ Equation of altitude with slope passes through (0,0) ‌y−0‌=1(x−0) ⇒‌y‌=x Equation of BC y+1=(−1)(x+3)⇒y=−x−4 Point of intersection of line BC and altitude ‌x‌=−x−4 ⇒‌2x‌=−4 ⇒‌x‌=−2 ∴‌y‌=−2 ∴ Foot of perpendicular A=(−2,−2) Now, 2x+2y+√2=0 ‌⇒‌‌x+y=−‌