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Question: The line passing through the point of intersection of \[x + y = 2\], \[x - y = 0\] and is parallel t...

The line passing through the point of intersection of x+y=2x + y = 2, xy=0x - y = 0 and is parallel to x+2y=5x + 2y = 5 is
A.x+2y=1x + 2y = 1
B.x+2y=2x + 2y = 2
C.x+2y=4x + 2y = 4
D.x+2y=3x + 2y = 3

Explanation

Solution

Hint : Here in this question, we need to find the equation of line which passes through the intersection point of the two lines and also parallel to line x+2y=5x + 2y = 5. For this, first we will calculate the intersection point of the given lines x+y=2x + y = 2 and xy=0x - y = 0 by using an elimination method. Then we will calculate the slope of the line x+2y=5x + 2y = 5 is given by ab - \dfrac{a}{b}. To further simplify by using a slope point formula (yy1)=m(xx1)\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) to get the required solution.

Complete step-by-step answer :
Given,
the line passes through the intersection of x+y=2x + y = 2, xy=0x - y = 0 and parallel to the line x+2y=5x + 2y = 5
Consider the equations
x+y=2x + y = 2 ----(1)
xy=0x - y = 0 ----(2)
Now we have to solve these two equations to find the point of intersection by using an elimination method.
Since the sign of coefficients of yy are different and no need of change the sign by the alternate sign and we simplify to known the unknown value xx
x+y=2 xy=0   x + y = 2\\\ x - y = 0 \;
2x=22x = 2
Divide 2 on both sides, then
x=1\Rightarrow \,\,\,x = 1
We have found the value of xx now we have to find the value of yy . So we will substitute the value xx to any one of the equations (1) or (2) . we will substitute the value of xxto equation (1).
Therefore, we have x+y=2x + y = 2
1+y=2\Rightarrow \,\,\,1 + y = 2
Subtract on both sides, then
y=21\Rightarrow \,\,\,y = 2 - 1
y=1\Rightarrow \,\,\,y = 1
Now, the intersection point of the two lines x+y=2x + y = 2 and xy=0x - y = 0 is given by (1,1)\left( {1,1} \right).
Considering the line x+2y=5x + 2y = 5
We know that the slope of the line ax+by+c=0ax + by + c = 0 is equal to m=abm = - \dfrac{a}{b}.
Now, the slope of the line x+2y=5x + 2y = 5 is given by m=12m = - \dfrac{1}{2}.
From slope point formula (yy1)=m(xx1)\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)
The required equation is
(y1)=12(x1)\Rightarrow \,\,\,\left( {y - 1} \right) = - \dfrac{1}{2}\left( {x - 1} \right)
Multiply both side by 2
2(y1)=(x1)\Rightarrow \,\,\,2\left( {y - 1} \right) = - \left( {x - 1} \right)
2y2=1x\Rightarrow \,\,\,2y - 2 = 1 - x
Add xx and 2 on both sides, then we have
2y2+x+2=1x+x+2\Rightarrow \,\,\,2y - 2 + x + 2 = 1 - x + x + 2
On simplification, we get
x+2y=3\therefore \,\,\,\,\,\,x + 2y = 3
Hence, it’s a required solution.
Therefore, option (4) is the correct answer.
So, the correct answer is “Option B”.

Note : In this type of question while eliminating the term we must be aware of the sign where we change the sign by the alternate sign. We can also use another method to find the equation of the required line. The equation of the line which is parallel to the line ax+by+c=0ax + by + c = 0 is given by . Here substitute the point in the equation ax+by+d=0ax + by + d = 0 and find the value of constant d.