Question

The distance of the lines \[2x-3y=4\]from the point \[(1,1)\] measured parallel to the line \[x+y=1\] is
1) \[\sqrt{2}\]
2) \[\dfrac{5}{\sqrt{2}}\]
3) \[\dfrac{1}{\sqrt{2}}\]
4) \[6\]

Answer 1

Hint: For solving these types of questions firstly we have to learn the distance formula and after that we will observe the equations given in the question then we will substitute the given point in the equations and find the intersection points of the two lines and when we will get two points we will simply apply the distance formula and we will get our required answer.

Complete step by step answer:
If we have two points then there is exactly one line segment connecting both points together. The distance between the two points is known as the length of that line segment connecting them. And the most important thing is that the distance is always positive (because distance can never be negative). And it is always a scalar quantity i.e. it has only magnitude and it does not depend on the direction. Minimum distance between two points will be zero.
It’s very easy to calculate the distance between two points (when the coordinates are given) by using the distance formula. We calculate the distance by subtracting the \[x\] and \[y\] coordinates of the points and then squaring those terms and after that add the resultant of \[x\] and \[y\] coordinates and then perform the square root operation. Let’s understand distance formula with the help of an example:
Let say there are two points \[P\] and \[Q\] in an \[XY\] plane. The coordinates of point \[P\] are \[({{x}_{1}},{{y}_{1}})\] and of point \[Q\] are \[({{x}_{2}},{{y}_{2}})\]. Then the distance formula between two points \[PQ\] is given by:
\[PQ=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
As we have given that \[x+y=1\] \[..........(i)\]
Let the equation of line parallel to the \[(i)\] is:
\[\Rightarrow x+y+k=0\] \[..........(ii)\]
Let \[(ii)\] passes through the point \[(1,1)\] so the line is along direction of \[x+y=1\].
Therefore, \[x+y+k=0\]
\[\begin{align}
  & \Rightarrow 1+1+k=0 \\
 & \Rightarrow 2+k=0 \\
 & \Rightarrow k=-2 \\
\end{align}\]
As we get the value of \[k\] , so \[(ii)\] becomes as:
\[\Rightarrow x+y-2=0\] \[..........(iii)\]
And we have given \[2x-3y=4\] \[..........(iv)\]
Now we will solve \[(iii)\] and \[(iv)\]
For solving these both, firstly we have to multiply \[(iii)\] with number \[2\] and then solve it with \[(iv)\]
After solving, we will get:
\[\Rightarrow x=2,y=0\]
So, we can say that the point of intersection is \[(2,0)\]
Let the distance between point \[(1,1)\] and \[(2,0)\] be \[d\]
So, \[d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
\[\Rightarrow d=\sqrt{{{(2-1)}^{2}}+{{(1-0)}^{2}}}\]
\[\Rightarrow d=\sqrt{2}\].

So, the correct answer is “Option 1”.

Note: Parallel lines are the lines that never met with each other and extended till the infinity. If there are two parallel lines, then the value of their slopes will always be equal. We can see parallel lines in our day to day life like railway tracks, roads, sidewalks, lines on a writing pad etc.