Question

Find the equations of the sides of the triangle, the coordinates of whose vertices are (1, 4), (2, -3) and (-1, -2).

• A. 3x-y+1=0
• B. x+3y+7=0
• C. 7x- y-11=0
• D. 7x+y-11=0

Answer 1

Answer: (A), (B), (D)

The correct options are
A

3x-y+1=0

B

x+3y+7=0

D

7x+y-11=0

If $A(x_1,y_1)$ and $B(x_2,y_2)$ be two points, then the equation of line will be,

$(y-y_1)=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)$

Equation of AB,

$y-4=\frac{(-3-4)}{(2-1)}(x-1)$
$\Rightarrow y-4=-7(x-1)$
$\Rightarrow y-4=-7x+7$
$\Rightarrow 7x+y-11=0$

Equation of BC,

$y+3=\frac{(-2+3)}{(-1-2)}(x-2)$
$\Rightarrow y+3=\frac{-1}{3}(x-2)$
$\Rightarrow 3y+9=-x+2$
$\Rightarrow x+3y+7=0$

Equation of AC,

$y-4=\frac{(-2-4)}{(-1-1)}(x-1)\Rightarrow y-4=3(x-1)\Rightarrow y-4=3x-3\Rightarrow 3x-y+1=0$