Given, A(2, 1) and B(3,2) Equation of the line in two point form is
\(\frac {y-y_1}{y_2-y_1} = \frac {x-s_1}{x_2-x_1}\)
∴ The equation of the required line is \(\frac {y-1}{2-1}= \frac{x-2}{3-2}\)
∴\(\frac {y-1}{1}= \frac {x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1 Comparing this equation with y = mx + c,
we get m = 1 and c = – 1
Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1
