Question

How to find b in linear equation form $y=mx+b$ if the $2$ coordinates are $(5,6)$ and $(1,0)$?

Answer 1

Hint:
We know the general equation of a straight line is $y=mx+b$ where $m$ is the gradient and $y=b$ is the value where the line cuts the $y-$axis. We know about the cartesian coordinates of points which is $(x,y)$ where $x$ is the abscissa and $y$ is the ordinate.

Complete step by step Solution:
Given that –
Linear equation form $y=mx+b$ and two given coordinates are $(5,6)$ and $(1,0)$
Now we know that if a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ then equation of line is
$(y-y_1)={\displaystyle \frac{y_2-y_1}{x_2-x_1}}(x-x_1)$
Now compare given points $(5,6)$ and $(1,0)$ with $(x_1,y_1)$ and $(x_2,y_2)$
We get $x_1=5,y_1=6$ and $x_2=1,y_2=0$
Now put all value in the equation line which passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ which is $(y-y_1)={\displaystyle \frac{y_2-y_1}{x_2-x_1}}(x-x_1)$
$\Rightarrow (y-6)={\displaystyle \frac{(0-6)}{(1-5)}}(x-5)$
After calculating all values we get
$\Rightarrow (y-6)={\displaystyle \frac{(-6)}{(-4)}}(x-5)$
$\Rightarrow (y-6)={\displaystyle \frac{(-1)\times (6)}{(-1)\times (4)}}(x-5)$
We know that we can multiply and divide any number with $1$ because after multiplication and divide of $1$ we will get same number therefore in above equation we have we have multiply by $1$ with numerator and denomination in left hand side and split negative sign which will cross from each other.
Now after calculating all left hand side numbers we will get $(y-6)={\displaystyle \frac{(3)}{(2)}}(x-5)$
Now we will multiply both side by $2$then we will get
$\Rightarrow (2)\times (y-6)=(2)\times {\displaystyle \frac{(3)}{(2)}}(x-5)$
After calculating all numbers, we will get $(2)\times (y-6)=(3)\times (x-5)$
Now multiplying both side and we will get
$\Rightarrow (2y-12)=(3x-15)$
Now we will transform or convert it in the form of $y=mx+b$ so we will get our required value of b for this we keep $y$ in the left hand side and all values will transfer to the right hand side
$\Rightarrow 2y=3x-15+12$
After calculating all numbers, we will get
$\Rightarrow 2y=3x-3$
For transform or converting it in the form of $y=mx+b$ we have to make coefficient of $y$one for it we will divide both side by $2$then we will get
 $\Rightarrow y={\displaystyle \frac{3}{2}}x-{\displaystyle \frac{3}{2}}$
Now compare above equation with the linear equation form $y=mx+b$ then we will get value of b which is our required value
Therefor after comparing both equation we will get $m={\displaystyle \frac{3}{2}}$ and $b=-{\displaystyle \frac{3}{2}}$

So our required value is $b=-{\displaystyle \frac{3}{2}}$ which is our answer.

Note:
We can solve the above question by directly putting all values in the linear line equation which passes through the two points $(x_1,y_1)$ and $(x_2,y_2)$. Another method is the graph method which we can use to find the length of $y=b$ where the line cuts the $y-$axis.