Question

If the equation of the line joining the points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is \[ ax+by=c \] and the distance between \(A\) and \(B\) is \(\sqrt{a^2+b^2}\), then \(c^2=\)

• A. \(\dfrac{x_1^2+y_1^2}{x_2y_2}\)
• B. \(\dfrac{x_1+x_2}{y_1+y_2}\)
• C. \((x_1y_2-x_2y_1)^2\)
• D. \((x_1x_2-y_1y_2)^2\)

Hint:

The equation of the line through \[ (x_1,y_1)\ \text{and}\ (x_2,y_2) \] can be written as \[ \boxed{ (y_2-y_1)x-(x_2-x_1)y=x_1y_2-x_2y_1. } \] The constant term directly gives the required value of \(c\).

Answer 1

The Correct Option is C

Step 1: Write the equation of the line through the two points. The equation of the line joining \[ A(x_1,y_1) \quad\text{and}\quad B(x_2,y_2) \] is \[ (y_2-y_1)x-(x_2-x_1)y = x_1y_2-x_2y_1. \] Comparing with \[ ax+by=c, \] we obtain \[ a=y_2-y_1,\qquad b=x_1-x_2,\qquad c=x_1y_2-x_2y_1. \]

Step 2: Verify the given condition. Now, \[ a^2+b^2 = (y_2-y_1)^2+(x_2-x_1)^2 = AB^2. \] Since \[ AB=\sqrt{a^2+b^2}, \] the given condition is satisfied.

Step 3: Find \(c^2\). Therefore, \[ c^2 = (x_1y_2-x_2y_1)^2. \] Hence, \[ \boxed{(x_1y_2-x_2y_1)^2}. \] Thus, \[ \boxed{(C)} \] is the correct answer.