I have a question : This is the Answer of question 3.How did we get -3? The midpoint between (A,B) and (C;D) is Xm= (A+C)/2 and Ym= (B+D)/2 Thus, 6= (2a+(a-2b))/2 12= 3a-2b and
(2a+ a - 2b)/ 2 = 6 That is, 3a - 2b = 12(equation 1) (2a - b+ 4a + 3b)/ 2 = 3 That is, 6a + 2b = 6( equation 2) on solving equation 1 and 2, 3a - 2b = 12 6a + 2b = 6 ____________ 9a = 18 a = 18/9 = 2 substitute this value of a in either of the equation then b = -3
I have a question : This is the Answer of question 3.How did we get -3?
ReplyDeleteThe midpoint between (A,B) and (C;D) is Xm= (A+C)/2 and Ym= (B+D)/2
Thus, 6= (2a+(a-2b))/2
12= 3a-2b and
3= (2a-b+4a+3b)/2
6= 6a+2b adding both equatins ,
18= 9a
a=2 and
b= -3
Its correct
DeleteSubstitute the value of a in either of the equations 6a+2b=6 or 3a- 2b =12
ReplyDeleteThis comment has been removed by the author.
ReplyDeletewhat is the answer for 11?
ReplyDeleteAnswer is uploaded as a slide(above)
Deletecan u post the answer with the steps for question 3 ?
ReplyDelete(2a+ a - 2b)/ 2 = 6
DeleteThat is, 3a - 2b = 12(equation 1)
(2a - b+ 4a + 3b)/ 2 = 3
That is, 6a + 2b = 6( equation 2)
on solving equation 1 and 2,
3a - 2b = 12
6a + 2b = 6
____________
9a = 18
a = 18/9 = 2
substitute this value of a in either of the equation
then b = -3