If each of the points (x1,4),(−2,y1) lies on the line joining the points (2,−1) and (5,−3), then the point P(x1,y1) lies on the line
A
6(x+y)−25=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2x+6y+1=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2x+3y−6=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6(x+y)+25=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B2x+6y+1=0 The equation of the line joining the points (2,−1) and (5,−3) is given by y+1=−1+32−5(x−2) ⇒2x+3y−1=0....(i)
Since (x1,4) and (−2,y1) lie on the line 2x+3y−1=0, therefore 2x1+12−1=0⇒x1=−112
and −4+3y1−1=0⇒y1=53
Thus, by checking the options (x1,y1) satisfies 2x+6y+1=0