Question

How do you write the equation of the line which has slope – 3 and passes through the point (-2, 4).

Answer 1

Now we know that the equation of the line in slope point form is given by the formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ where m is the slope of the equation and $\left( {{x}_{1}},{{y}_{1}} \right)$ is the point lying on the equation. Hence substituting the values in the equation we will get the equation of the line.

Complete step-by-step answer:
Now we know that the equation of a line is a linear equation in two variables of the form $ax+by+c=0$ .
Now let us first understand the meaning of slope.
Slope is nothing by the ratio of y coordinate and x coordinate. Hence slope off line is given by $\dfrac{y}{x}$ .
If $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ are the points on the line then the slope of the line is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ .
For any straight line slope is constant.
Now if we have a point $\left( {{x}_{1}},{{y}_{1}} \right)$ and m is the slope of the line then the equation of the line is given by $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ .
Now consider the given conditions. The slope of the line is given by – 3 and the point (-2, 4) is the point lying on the line. Hence we have m = -3 and ${{x}_{1}}=-2,{{y}_{1}}=4$
Then substituting the values in the equation we get, $y-\left( 4 \right)=\left( -3 \right)\left( x-\left( -2 \right) \right)$ .
$\begin{aligned} & \Rightarrow y-4=-3\left( x+2 \right) \\\ & \Rightarrow y-4=-3x-6 \\\ & \Rightarrow y+3x-4+6=0 \\\ & \Rightarrow 3x+y+2=0 \\\ \end{aligned}$
Hence the equation of the line is 3x + y + 2 = 0.

Note: Note that the equation of line can be written in various forms. The form used above is called the slope point form. Two point form can easily be written replacing m by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ . Similarly slope intercept can be derived by replacing $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 0,c \right)$ in slope point form.