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
Approach Solution - 1
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
Approach Solution -2
Step 1: Find the point of intersection of the given lines:
- The equations of the two lines are:
- \( x + 2y - 10 = 0 \quad \text{(1)} \)
- \( 2x + y + 5 = 0 \quad \text{(2)} \)
- To find the point of intersection, solve these equations simultaneously.
- From equation (1):
- \( x = 10 - 2y \).
- Substitute \( x = 10 - 2y \) into equation (2):
- \( 2(10 - 2y) + y + 5 = 0 \).
- \( 20 - 4y + y + 5 = 0 \).
- \( 25 - 3y = 0 \).
- \( y = \frac{25}{3} \).
- Substitute \( y = \frac{25}{3} \) into \( x = 10 - 2y \):
- \( x = 10 - 2\left(\frac{25}{3}\right) \).
- \( x = 10 - \frac{50}{3} \).
- \( x = \frac{30}{3} - \frac{50}{3} = -\frac{20}{3} \).
- The point of intersection is:
- \( \left(-\frac{20}{3}, \frac{25}{3}\right) \).
Step 2: Find the slope of the perpendicular line:
- The given line is \( 4x - 5y + 1 = 0 \).
- Rewriting in slope-intercept form (\( y = mx + c \)):
- \( -5y = -4x - 1 \).
- \( y = \frac{4}{5}x + \frac{1}{5} \).
- The slope of this line is \( \frac{4}{5} \).
- The slope of a line perpendicular to this is the negative reciprocal:
- \( m_{\perp} = -\frac{5}{4} \).
Step 3: Equation of the perpendicular line:
- The perpendicular line passes through \( \left(-\frac{20}{3}, \frac{25}{3}\right) \) and has slope \( -\frac{5}{4} \).
- Using the point-slope form of a line:
- \( y - y_1 = m(x - x_1) \).
- \( y - \frac{25}{3} = -\frac{5}{4}\left(x + \frac{20}{3}\right) \).
- Simplify:
- \( y - \frac{25}{3} = -\frac{5}{4}x - \frac{5}{4} \cdot \frac{20}{3} \).
- \( y - \frac{25}{3} = -\frac{5}{4}x - \frac{100}{12} \).
- \( y - \frac{25}{3} = -\frac{5}{4}x - \frac{25}{3} \).
- \( y = -\frac{5}{4}x - \frac{25}{3} + \frac{25}{3} \).
- \( y = -\frac{5}{4}x \).
- Rewrite in standard form:
- \( y + \frac{5}{4}x = 0 \).
Step 4: Match with the options:
- The equation \( y + \frac{5}{4}x = 0 \) matches Option (C).
Final Answer:
The correct option is (C) \( 5x + 4y = 0 \).
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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