Question

The vertices of a triangle are $A(3,-4),B(5,7)$ and $C(-4,5)$. The respective slopes of $AB,BC$ and $AC$ are

• A. $\frac{11}{2},\frac{2}{9},-\frac{9}{7}$
• B. $\frac{11}{2},-\frac{9}{7},-\frac{2}{9}$
• C. $-\frac{9}{7},-\frac{2}{9},\frac{11}{2}$
• D. $-\frac{9}{7},\frac{11}{2}-\frac{2}{9}$

Answer 1

The correct option is A $\frac{11}{2},\frac{2}{9},-\frac{9}{7}$

Let the points be $A(x_1,y_1)=(03,-4),B(x_2,y_2)=(5,7),C(x_3,y_3)=(-4,5)$

We know that the slope of a line passing through the points

$(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$.

$\therefore m_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{7+4}{5-3}=\frac{11}{2}$

$m_{BC}=\frac{y_3-y_2}{x_3-x_2}=\frac{5-7}{-4-5}=\frac{2}{9}$

and $m_{AC}=\frac{y_3-y_1}{x_3-x_1}=\frac{5+4}{-4-3}=-\frac{9}{7}$

$\therefore (m_{AB},m_{BC},m_{AC})=(\frac{11}{2},\frac{2}{9},-\frac{9}{7})$