Question 1:
Solution:
Area of △ PQR = 3112 |6 5 m 1 6−1 61|=31|(6)(6)+(5)(−1)+(m)(1)−(1)(5)−(6)(m)−(−1)(6)|=62|36−5+m−5−6m+6|=62|32−5m|=6232−5m=±62−5m=62−32 or −5m=−62−32m=−6 or m=945
The vertices of a triangle are P (6, 1), Q (5, 6) and R (m, -1). Given that the area of the triangle is 31 unit2, find the values of m.
Area of △ PQR = 3112 |6 5 m 1 6−1 61|=31|(6)(6)+(5)(−1)+(m)(1)−(1)(5)−(6)(m)−(−1)(6)|=62|36−5+m−5−6m+6|=62|32−5m|=6232−5m=±62−5m=62−32 or −5m=−62−32m=−6 or m=945
Question 2:
Solution:
((3m)(2)+(3t)(3)3+2,(m)(2)+(2u)(3)3+2)=(t,u)6m+9t5=t6m+9t=5t6m=−4tm=−23t→(1)2m+6u5=u2m+6u=5u2m=−uFrom (1),2(−2t3)=−ut=34u
The points P (3m, m), Q (t, u) and R (3t, 2u) lie on a straight line. Q divides PR in the ratio 3 : 2. Express t in terms of u.
((3m)(2)+(3t)(3)3+2,(m)(2)+(2u)(3)3+2)=(t,u)6m+9t5=t6m+9t=5t6m=−4tm=−23t→(1)2m+6u5=u2m+6u=5u2m=−uFrom (1),2(−2t3)=−ut=34u