The line perpendicular to \(4x-5y+1=0\) and passing through the point of intersection of the straight line \(x+2y-10=0\) and \(2x+y+5=0\) is ?
\(y+\dfrac{5x}{4}=\dfrac{50}{3}\)
\(5x+4y=1\)
\(5x+4y=0\)
\(y+\dfrac{5x}{4}=\dfrac{-50}{3}\)
The Correct Option is C
Solution and Explanation
Given that
\(x+2y-10=0\)……………………(1)
\(2x+y+5=0\)………………………(2)
Now, multiply \(2\) in equation (1)
⇒\(2x+4y-20=0\)………………...(3)
By subtracting (1) from (3) we get ;
⇒3y-25=0
by solving this ,
\(y=\dfrac{25}{3},\)
putting this value in the place of \(y\) in equation (1), we get;
\(x=\dfrac{-20}{3}\)
Hence, checking the options which satisfy the value of intersection point, is 5x+4y=0..(Ans.)
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c