What is the equation of the straight line passing through the point [tex]\((0,5)\)[/tex] and having a gradient of [tex]\(-2\)[/tex]?

[tex]\[
\begin{array}{l}
\text{Gradient (}m\text{)} = -2 \\
\text{y-intercept} = 5
\end{array}
\][/tex]

Equation: [tex]\( y = -2x + 5 \)[/tex]



Answer :

To determine the equation of a straight line passing through a given point and having a specific gradient, we can use the slope-intercept form of the line equation, which is given by:

[tex]\[ y = mx + c \][/tex]

where:
- [tex]\( m \)[/tex] is the gradient (slope) of the line,
- [tex]\( c \)[/tex] is the y-intercept (the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is 0).

Given:
- The line passes through the point [tex]\((0, 5)\)[/tex],
- The gradient [tex]\( m \)[/tex] of the line is [tex]\(-2\)[/tex].

First, let's identify the gradient and the y-intercept.

1. Gradient ( [tex]\( m \)[/tex] ):
[tex]\[ m = -2 \][/tex]

2. Y-intercept ( [tex]\( c \)[/tex] ):
Since the line passes through the point [tex]\((0, 5)\)[/tex], the y-intercept [tex]\( c \)[/tex] is the y-coordinate of the point when [tex]\( x \)[/tex] is 0. Therefore:
[tex]\[ y\text{-intercept} = 5 \][/tex]

Now, substitute the gradient and the y-intercept into the slope-intercept form:

[tex]\[ y = -2x + 5 \][/tex]

Therefore, the equation of the line is:
[tex]\[ y = -2x + 5 \][/tex]

So, summarizing:

- Gradient ([tex]\( m \)[/tex]) = -2,
- Y-intercept = 5,
- Equation: [tex]\( y = -2x + 5 \)[/tex].