Line y = mx + c passes through the points A(2,1) and B(3,2). Determine

1.	Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c
Answer» 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

Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c

Answer»

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