Question

if point (2,0) (0,5) and (x,y) are collinear show that x/2+y/5=1 using determinant

Answer 1

Let us consider a general case first. Consider three points:

$A\, (x_1,y_1) \\ \\ B \, (x_2,y_2) \\ \\ C \, (x_3,y_3)$

For points A, B and C to be collinear, the condition is:


$\boxed{\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right| = 0 }$


Here, the three points are (2,0), (0,5) and (x,y). 

Let us apply the condition of collinearity:

[tex]\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right| = 0 \\ \\ \\ \implies \left| \begin{array}{ccc} 2 & 0 & 1 \\ 0 & 5 & 1 \\ x & y & 1 \end{array} \right| = 0 \\ \\ \\ \implies 2 \left| \begin{array}{cc} 5 & 1 \\ y & 1 \end{array}\right| - 0\left| \begin{array}{cc} 0 & 1 \\ x & 1 \end{array}\right| + 1\left| \begin{array}{cc} 0 & 5 \\ x & y \end{array}\right| = 0 \\ \\ \\ \implies 2(5 - y) - 0 + 1(0-5x) = 0 \\ \\ \\ \implies 10 -2y - 5x = 0[/tex]


[tex]\implies 10 = 2y + 5x \\ \\ \\ \implies 1 = \frac{2}{10}y + \frac{5}{10}x \\ \\ \\ \implies 1 = \frac{y}{5} + \frac{x}{2} \\ \\ \\ \implies \boxed{\frac{x}{2}+\frac{y}{5}=1} \\ \\ \\ Hence Proved [/tex]


Hope it helps
Purva
Brainly Community

Answer 2

Step-by-step explanation:

I hope you get your answer