Question

Q)1) Show that line AB is perpendicular to line BC, where A(1, 2), B (2, 4) and
C(0, 5).​

Answer 1

If AB and BC are perpendicular to each other, the product of their slopes must be -1.

     Slope(AB) = (y₂ - y₁)/(x₂ - x₁)

                       = (4 - 2)/(2 - 1)

                       = 2         ...(1)

     Slope(BC) = (5 - 4)/(0 - 2)

                       = 1/(-2)

                       = -1/2       ...(2)

∵ Product of slope of AB and BC = 2(-1/2) = -1,     AB and BC are perpendicular

Answer 2

★Question:-

• Show that line AB is perpendicular to line BC, where A(1, 2), B (2, 4) and C(0, 5).

★Solution:-

As it is given that AB and BC are perpendicular to each other, the product of their slopes must be -1.

Slope(AB):- $\large\sf \frac{y_2-y_1}{x_2-x_1}$

$\large\qquad \quad\quad \sf \frac{4 - 2}{2 - 1}$

$\large\sf \qquad\quad \quad2 \: \: \: \: \: \: \: \qquad \qquad \qquad \sf \color {red}(1)$

Slope(BC):- $\large\sf\frac{5-4}{0-2}$

$\large\sf\qquad\quad \quad \frac{ \: \: 1}{ - 2}$

$\large\qquad\quad \quad \sf \frac{ - 1}{ \: \: \: 2} \: \: \: \: \: \: \: \: \: \qquad \quad \qquad \sf \color{red}(2)$

Multiplying (1) and (2),

➺2 (-1/2)

➺(-1 × 1)

➺ -1

∴It is proved that AB is perpendicular to line BC.

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