Find the equations of line joining points (−2,3) and (1,4)?
Answer 1:
The equation for line joining two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is given by \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\) and for given points it is \(y=\frac{1}{3} x+\frac{11}{3}\)
Explanation:
Let the slope intercept form of equation be \(y=m x+c\)
here we do not know the slope \(m\) and \(y-\) intercept \(C\)
What we know is that this passes through the two coordinate pairs, say \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\).
As such we have three equations
\(y=m x+c \ldots \ldots(1)\)
\(y_{1}=m x_{1}+c \ldots \ldots(2) \text { and }\)
\(y_{2}=m x_{2}+c \ldots \ldots(3)\)
Now using these let us eliminate \(m\) and \(c\)
subtracting (2) from (1), we get \(\left(y-y_{1}\right)=m\left(x-x_{1}\right) \ldots \ldots(4)\)
and subtracting (2) from (3), we get \(\left(y-2-y_{1}\right)=m\left(x_{2}-x_{1}\right) \ldots \ldots(5)\)
Dividing (4) by (5)
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)As the two points are \((-2,3)\) and \((1,4)\), the equation is
Answer 2:
For any point on the line, the coordinate pair in slope-intercept form is
Explanation:
For slope m and intercept c, the equation is y = m x +c.
The slope intercept form for coordinates is (x, m x +c ).
The slope of the line through the given points is \(m=\frac{4-3}{1-(-2)}=\frac{1}{3}\)
Also, from (1, 4). 4 = i/3(1) + c. So, c = 11/3.
So, the answer is \(\left(x, \frac{1}{3} x+\frac{11}{3}\right)\).
Answer 3:
The equation of the line is:
\(y=\frac{1}{3} x+\frac{11}{3}\) which can be written as \(y=\frac{1}{3} x+3 \frac{2}{3}\)
Explanation:
If you are given the coordinates of 2 points on a line, substituting them into the formula below allows you to find the equation immediately. In the process you also calculate the slope.
