Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.
(i) \(x + 7y = 0 \)
(ii) \(6x + 3y - 5 = 0 \)
(iii) \(y = 0\)
(i) \(x + 7y = 0 \)
(ii) \(6x + 3y - 5 = 0 \)
(iii) \(y = 0\)
Solution and Explanation
(i)The given equation is \(x + 7y = 0\)
It can be written as
\(y =\frac{-1}{7}x + 0............(1)\)
This equation is of the form \(y = mx + c\), where \(m = \frac{-1}{7}\) and \(c = 0\)
Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept are \(\frac{-1}{7}\) and 0 respectively.
(ii) The given equation is \( 6x + 3y – 5 = 0\)
It can be written as \(y=\frac{1}{3}(-6x+5)\)
\(y=-2x +\frac{ 5}{3}.........(2)\)
This equation is of the form \(y = mx + c,\) where \(m =-2\) and \(c = \frac{5}{3}.\)
Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are -2 and \(\frac{5}{3}\) respectively.
(iii) The given equation is \( y = 0\)
It can be written as \(y = 0 × x + 0.....(3)\)
This equation is of the form\( y = mx + c\), where \(m = 0\) and \(c = 0\).
Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.
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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