Question

find the equation of the straight line passing through the points (-1,2)and(5,-1)and also find the area of the triangle formed by it with the axes of coordinates​

Answer 1

Therefore the required equation of line is  $x+2y=3$

Therefore the area of the triangle is $= \frac{9}{4}$ square units

Step-by-step explanation:

Given the coordinate of two points are (-1,2) and (5,-1).

The equation of line which passes through the points $(x_1,y_1)$ and $(x_2,y_2)$ is

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

Here $x_1 =-1 , y_1 = 2$ ,$x_2 =5$ and $y_2=-1$

Therefore the required equation of line is

$y-2=\frac{-1-2}{5-(-1)} [x-(-1)]$

$\Leftrightarrow y-2=\frac{-3}{6} (x+1)$

$\Leftrightarrow 2(y-2)= -1(x+1)$

$\Leftrightarrow x+2y=3$

$\Leftrightarrow \frac{x}{3} +\frac{y}{\frac{3}{2} } =1$

Therefore the line intercepts the x-axis at (3,0) and the y-axis at $(0,\frac{3}{2} )$.

Therefore the area of a triangle whose vertices are $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is

$=\frac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here $x_1 = 3, y_1=0 , x_2=0,y_2=\frac{3}{2}$ and $x_3 =0 ,y_3 =0$

Therefore the area of the triangle is

$=\frac{1}{2}[3(\frac{3}{2}-0)+0(0-0)+0(3-0)]$ square units

$= \frac{9}{4}$ square units