Question

Find the equation of a straight line passing throught the point listed and having the given gradient. Express your answer in the form 1) ax + by + c = 0 and 11) y = mx + c
Question: (1,2)     gradient 3

Please show all working and explain in detail :)

Answer 1

First let's try to find the equation in this form : y = mx + c 

The gradient is given 3 . In a line's equation, x's coefficient represents the line's gradient.

So equation of a line with the gradient of 3, would look like this ;

$y= 3x + c$

Now a point that the line passes through is given, (1, 2) 

This point's x-coordinate is 1 and y-coordinate is 2.

So we'll plug its x-coordinate value in the equation and also y-coordinate value. So we can solve it.

As you know, $x=1$ and $y=2$

$y = 3x + c$

$2\quad =\quad 3\cdot 1+c\\ \\ 2\quad =\quad 3+c\\ \\ 2-3\quad =\quad c\\ \\ -1\quad =\quad c$

We found c = -1 

Also in a line's equation, c is constant and it represents the line's y-intercept

So let's build the line's equation.

$m=3$ and $c=-1$

$y= mx + c$

$y= 3x -1$

We found the line's equation in this form, $y= mx + c$

Now let's turn it into this form, $ax + by + c = 0$

$y\quad =\quad 3x-1\\ \\ y-3x\quad =\quad -1\\ \\ y-3x+1\quad =\quad 0\\ \\ -3x+y+1\quad =\quad 0$

Final answers,

$\boxed { y\quad =\quad 3x-1 }$

and 

$\boxed { -3x+y+1\quad =\quad 0 }$

I hope this was clear enough :)

Answer 2

$2=3\cdot1+c\\ 2=3+c\\ c=-1\\ \boxed{y=x-1}\\\\ y=x-1\\ \boxed{x-y-1=0}$