CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that OP.OQ + OR.OS = OR.OP + OQ.OS = OQ.OR + OP.OS Then the triangle PQR has S as its



A

centroid

loader
B

orthocenter

loader
C

incentre

loader
D

circumcentre

loader

Solution

The correct option is B

orthocenter


OP.OQ + OR.OS = OR.OP + OQ.OS
OP(OQ-OR)+OS(OR-OQ) = 0
⇒(OP-OS)(OQ-OR)=0
SP.RQ = 0
Similarly SR.PQ = 0 and SQ.PR = 0
∴S is orthocentre.


Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image