Question

How do you write an equation in standard form when given slope and point on line $\left( 1,5 \right)$ , $m=-3$ ?

Answer 1

The given problem is a very simple example of coordinate geometry. It is a sub topic of the equation of straight lines. For this problem we need to remember some of the formulae and basic equations of coordinate geometry. In this very problem, we are going to use the formula for the equation of a straight line of slope-point form, where the slope of the line is given and a point on that line is given.

Complete step by step answer:
Now starting off with our solution, we need to recall that in our problem the slope given is $m=-3$ and the point on the required line is $\left( 1,5 \right)$ . Now, according to the given formula we know that, if $\left( {{x}_{1}},{{y}_{1}} \right)$ be a given point on the line with slope $m$ , then we can write the equation of the straight line as $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ .
Now plugging into the values of the given point in the place of ${{x}_{1}}$ , ${{y}_{1}}$ and $m$ we get,
$y-5=\left( -3 \right)\left( x-1 \right)$
Now multiplying the terms in the right hand side, we get,
$y-5=-3x+3$
Now, we know that the general form of the equation of a straight line is $y=mx+c$ , where $m$ is the slope and $c$ is the y-intercept. Transforming our intermediate equation into the form of the general equation, we get,
$y=-3x+8$
The above equation is thus the equation of the straight line of slope $m$ and passes through the point $\left( 1,5 \right)$ .

Note:
From the answer that we have got, of the equation of straight line, we can clearly say that the slope is $m=-3$ and the y-intercept is $8$ , as in the general equation $y=mx+c$ , $c$ is the y-intercept. The y-intercept can also be calculated pretty easily using the known formula ${{y}_{1}}-m{{x}_{1}}$ . This problem can also be solved graphically, where we have to plot the required line.