Explanation
x+2y−9=0 ---(1)
3x+5y−5=0 ---(2)
ax+by−1=0 ---(3)
Solving (1) and (2) simultaneously we get
y=22,x=−35
Now, equation (1), (2) and (3) will be concurrent, that is they will pass through one point if y=22,x=−35 satisfy the third equation ax+by−1=0.
Substituting the values of 'x' and 'y' in this equation we get −35a+22b−1=0 ---(4)
And another equation given is 22x−35y−1=0 ---(5)
x=b & y=a
That is 22x−35y−1=0 passes through (b,a)
We know that concurrent lines have same intersection points.Then
∴3x+2y−5=0 --(1)
2x−5y+3=0 ---(2)
5x+by+c=0 ---(3)
From 1 & 2,
6x+4y−10=0
6x−15y+9=0
⇒19y−19=0
y=1 & x=1
So, substituting x=1 & y=1, in equation (3),
5x+by+c=0
5+b+c=0
∴b+c=−5
Step -1: Find equation
Since A,O,C are linear they will lie on same line.
Given points are A(1,2),O(0,0)and C(a,b)
Therefore,
area ΔAOC = 0
∴12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=0
Here, x1=1,x2=0and x3=a
and y1=2,y2=0and y3=b
Step -2: Find the relation
12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=0
⇒12[1(0−b)+0(b−2)+a(2−0)]=0
⇒−b + 2a=0
⇒2a = b
Hence, the correct answer is option C.
For the points AB=(x,y),BC=(1,2) and CD=(7,0) to be collinear they must have area =0.
Thus, area of a triangle ABC = 12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))=0
=12(2x×2+6y−14)=0
=(2x+6y−14)=0
Thus, we have the relation x+3y=7 for the points to be collinear.
Given that PQR are three collinear points andPQ=2.5. Let us now find the distance of PR
Recall Distance Formula is =$$ \sqrt { { \left( { x}_{ 2 }{ { x }_{ 1 } } \right) }^{ 2}+{ \left( { y }_{ 2 }{ { y }_{ 1 } } \right) }^{ 2 } } \\$$
Distance PR=√(113)2)+(104)2=√100=10
Let us now consider S is the midpoint of PR,then we have the co-ordinates of S as 11+32,10+42=(7,7).
∴ it is clear that Q is the
midpoint of PS as we have know PQ=2.5 units.
Thus, to get the co-ordinates of Q takemidpoint of PS which is 3+72,4+72=(5,11/2).
Step - 1: Find equation
Let A, B, C are the three points
Since A, B, C are linear they will lie on same line.
Given points are A(0,0),B(1,2)and C(x,y)
area ΔABC = 0
⇒12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=0
Here, x1=0,x2=1and x3=x
and y1=0,y2=2and y3=y
Step - 2: Find the relation
⇒12[0(2−y)+1(y−0)+x(0−2)]=0
⇒y−2x=0
⇒2x = y
Hence, the correct answer is option B.
Please disable the adBlock and continue. Thank you.