The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6). Find equa

1.	The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6). Find equations of the sides.
Answer» Equation of the line in two point form is \(\frac {y-y_1}{y_2-y_1} = \frac {x-x_1}{x_2-x_1}\) Equation of side AB is \(\frac {y-4}{3-4} = \frac {x-1}{2-1}\) y – 4 = -1(x – 1) y – 4 = -x + 1 x + y = 5 Equation of side BC is \(\frac {y-3}{6-3} = \frac {x-2}{1-2}\) -1(y – 3) = 3(x – 2) -y + 3 = 3x – 6 ∴ 3x + y = 9 Since both the points A and C have same x co-ordinates i.e. 1 the points A and C lie on a line parallel to Y-axis. ∴ The equation of side AC is x = 1.

The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6). Find equations of the sides.

Answer»

Equation of the line in two point form is \(\frac {y-y_1}{y_2-y_1} = \frac {x-x_1}{x_2-x_1}\)

Equation of side AB is \(\frac {y-4}{3-4} = \frac {x-1}{2-1}\)

y – 4 = -1(x – 1)

y – 4 = -x + 1

x + y = 5

Equation of side BC is \(\frac {y-3}{6-3} = \frac {x-2}{1-2}\)

-1(y – 3) = 3(x – 2)

-y + 3 = 3x – 6

∴ 3x + y = 9

Since both the points A and C have same x co-ordinates i.e. 1

the points A and C lie on a line parallel to Y-axis.

∴ The equation of side AC is x = 1.