Question
Question: The straight line passing through the point of intersection of the straight lines \( x - 3y + 1 = 0 ...
The straight line passing through the point of intersection of the straight lines x−3y+1=0 and 2x+5y−9=0 and having infinite slope and at a distance of 2 units from the origin, has the equation
(1) x=2
(2) 3x+y−1=0
(3) y=1
(4) None of these
Solution
Hint : A straight line is an endless one dimensional figure that has no width. The straight line is a combination of endless points joined on both sides of a point. A straight line does not have any curve in it. The straight line can be horizontal, vertical or slanted. When two or more straight lines cross each other in a plane, they are called intersecting lines.
Complete step-by-step answer :
First we find the intersection point of the straight lines
Take x−3y+1=0 ……………………………..(1)
2x+5y−9=0 …………………………………..(2)
From equation (1) , we get
x=3y−1 ………………………………(3)
Use equation (3) in equation (2) , we get
2(3y−1)+5y−9=0
Multiplying and we get
⇒6y−2+5y−9=0
⇒11y−11=0
⇒11y=11
We divide both sides by 11 , we get
⇒1111y=1111
⇒y=1
Put the value of y=1 in equation (3) , we get
x=3×1−1
⇒x=3−1
⇒x=2
Therefore the intersection point is (2,1) .
From the given data we have infinite slope , means θ=2π
Therefore the straight line parallel to Y-axis i.e., x=a , where a is the distance from origin.
And also given that the distance from origin is 2 units.
Then we have the straight line x=2 .
Therefore the option (1) is correct.
So, the correct answer is “Option 1”.
Note : The slope of a line is a number measured as its “steepness”, usually denoted by the letter m . It is the change in y for a unit change in x for a change in x along the line. The formula of slope of a straight line is m=tanθ , where θ is the angle between the X-axis and the straight line.