Question

Show that points p(2,-2) Q(7,3) R(11,-1) and s(6,-6) are vertices of a parallelogram
p(2,-2) Q(7,3) R(11,-1) and s(6,-6) are the vertices of a parallelogram

Answer 1

Answer:

Step-by-step explanation:

P(2,-2),Q(7,3), R(11,-1) and S(6,-6)

By distance formula ,

PQ=(7−2)2+[3−(−2)]2−−−−−−−−−−−−−−−−−−√

∴=52+52−−−−−−√

∴PQ=25+25−−−−−−√

∴PQ=50−−√

∴PQ=5×5×2−−−−−−−−√

∴PQ52–√...(1)

QR=(11−7)2+(−1−3)2−−−−−−−−−−−−−−−−−−√

∴QR=42+(−4)2−−−−−−−−−−√

∴QR=16+16−−−−−−√

∴QR=32−−√

∴QR=2×2×2×2×2×−−−−−−−−−−−−−−−−√

∴QR=42–√ ...(2)

RS=6−11−−−−−√2+[−6−(−1)]2)

∴RS=−5−−−√2+(−5)2)

∴RS=25+25−−−−−−√

∴RS=50−−√

∴RS=5×5×2−−−−−−−−√

∴RS=52–√ ...(3)

PS=(6−2)2+[−6−(−2)]2−−−−−−−−−−−−−−−−−−−−−√

∴PS=42+(−4)2−−−−−−−−−−√

∴16+16−−−−−−√

∴PS=32−−√

∴PS=2×2×2×2×2−−−−−−−−−−−−−−√

∴PS2×2×2–√

∴PS=42–√ ....(4)

In □ PQRS

PQ=RS  ...[From (1) and (3)]

QR=PS  ...[From (2) and (4)]

A quadrilateral is a parallelogram , if both the pairs of its opposite

sides are congruent

∴□ PQRS is parallelogram.

∴P,Q,R and S are vertices of a parallelogram