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Question: How do you write an equation of a line with slope \(3\) and the \(y\) intercept is -4 ?...

How do you write an equation of a line with slope 33 and the yy intercept is -4 ?

Explanation

Solution

First of all this is a very simple and a very easy problem. The general equation of a straight line is y=mx+cy = mx + c, where mm is the gradient and y=cy = c is the value where the line cuts the y-axis. The number cc is called the intercept on the y-axis. Based on this provided information we try to find the equation of the straight line.

Complete step-by-step answer:
Given that an equation of a line has the slope equal to 3 and the yyintercept equal to -4.
We know that the equation of the straight line is given by:
y=mx+c\Rightarrow y = mx + c
Where mm is the slope of the straight line and cc is the yyintercept of the straight line.
So given that the slope of the straight line is m=3m = 3
The intercept of the straight line is c=4c = - 4
Substituting the values of the given data, in the general form of the straight line, as shown below:
y=mx+c\Rightarrow y = mx + c
y=(3)x+4\Rightarrow y = \left( 3 \right)x + - 4
So the equation of the straight line is given by:
y=3x4\Rightarrow y = 3x - 4
On further simplifying where, moving all the variable terms and the constants to one side of the equation, as shown below:
3xy4=0\Rightarrow 3x - y - 4 = 0

Final Answer: The equation of the line is 3xy4=03x - y - 4 = 0.

Note:
Please note that while solving such kind of problems, we should understand that if the y-intercept value is zero, then the straight line is passing through the origin, which is in the equation of y=mx+cy = mx + c, if c=0c = 0, then the equation becomes y=mxy = mx, and this line passes through the origin, whether the slope is positive or negative.