If (x, y) be on the line joining the two points (1, -3) and (-4, 2), prove that x+ y+ 2 = 0.

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Solution

If 3 points are collinear, then the area of the triangle formed by them is zero.

Area of triangle = $\frac{1}{2} | (x_{1} y_{2} - y_{1} x_{2}) + (x_{2} y_{3} - y_{2} x_{3}) + (x_{3} y_{1} - y_{3} x_{1}) | = 0$

$\Rightarrow (x_{1} y_{2} - y_{1} x_{2}) + (x_{2} y_{3} - y_{2} x_{3}) + (x_{3} y_{1} - y_{3} x_{1}) = 0$

$\Rightarrow (2 - 12) + (- 4 y - 2 x) + (- 3 x - y) = 0$

$\Rightarrow - 10 - 5 x - 5 y = 0$

$∴ x + y + 2 = 0$

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If (x, y) be on the line joining the two points (1, -3) and (-4, 2), prove that x+ y+ 2 = 0.

If 3 points are collinear, then the area of the triangle formed by them is zero. Area of triangle = 12|(x1y2−y1x2)+(x2y3−y2x3)+(x3y1−y3x1)|=0 ⇒(x1y2−y1x2)+(x2y3−y2x3)+(x3y1−y3x1)=0 ⇒(2−12)+(−4y−2x)+(−3x−y)=0 ⇒−10−5x−5y=0 ∴x+y+2=0