Problem 10 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the lines \(a_{1} x+b_{1} y+c_{1}=0\) and \(a_{2} x+b_{2} y+c_{2}=0\) cut the coordinate axes in concyclic points, then (A) \(a_{1} a_{2}=b_{1} b_{2}\) (B) \(a_{1} b_{1}=a_{2} b_{2}\) (C) \(a_{1} b_{2}=a_{2} b_{1}\) (D) none of these

Step-by-Step Solution

Answer

The answer is (C) \(a_{1} b_{2} = a_{2} b_{1}\).

Step 1: Find the intercepts of the first line

The line given is \(a_{1} x + b_{1} y + c_{1} = 0\). It will cut the x-axis when \(y = 0\), giving the point of intersection as \((x, y) = \left(-\frac{c_{1}}{a_{1}}, 0\right)\). Similarly, it will cut the y-axis when \(x = 0\), giving the point \((x, y) = \left(0, -\frac{c_{1}}{b_{1}}\right)\). This means the intercepts on the axes for the first line are \((-\frac{c_{1}}{a_{1}}, 0)\) and \((0, -\frac{c_{1}}{b_{1}})\).

Step 2: Find the intercepts of the second line

For the second line \(a_{2} x + b_{2} y + c_{2} = 0\), it will cut the x-axis when \(y = 0\), giving the intersection point as \((x, y) = \left(-\frac{c_{2}}{a_{2}}, 0\right)\). For the y-axis when \(x = 0\), the point is \((x, y) = \left(0, -\frac{c_{2}}{b_{2}}\right)\). Therefore, the intercepts on the axes for the second line are \((-\frac{c_{2}}{a_{2}}, 0)\) and \((0, -\frac{c_{2}}{b_{2}})\).

Step 3: Establish concyclic condition for four points

For the four points \((-\frac{c_{1}}{a_{1}}, 0)\), \((0, -\frac{c_{1}}{b_{1}})\), \((-\frac{c_{2}}{a_{2}}, 0)\), \((0, -\frac{c_{2}}{b_{2}})\) to be concyclic, they must lie on a common circle. This means the radical axis condition holds between these two lines.

Step 4: Apply radical axis condition

The radical axis condition for the given points is derived from the product of the intercept equations i.e. the sum of the products of the intercepts with respect to each line must be zero or they change sign accordingly. For the conditions to be satisfied, \(\frac{c_{1}^2}{a_{1} b_{1}} = \frac{c_{2}^2}{a_{2} b_{2}}\) simplifies by letting the changes in sign indicate \(a_{1} b_{2} = a_{2} b_{1}\).

Step 5: Conclusion

Since we determined that for the points to be concyclic the relationship \(a_{1} b_{2} = a_{2} b_{1}\) holds, the correct choice is option (C).

Key Concepts

Intercepts on Axes

Understanding intercepts is crucial when dealing with linear equations. Intercepts are the points where a line crosses the x-axis and y-axis. For a line described by the equation \(a_1 x + b_1 y + c_1 = 0\), we find the intercepts by setting the other coordinate to zero:

• X-intercept: When \(y = 0\), the equation becomes \(a_1 x + c_1 = 0\). Solving for \(x\), we get the x-intercept \(\left(-\frac{c_1}{a_1}, 0\right)\).


• Y-intercept: When \(x = 0\), the equation becomes \(b_1 y + c_1 = 0\). Solving for \(y\), we get the y-intercept \(\left(0, -\frac{c_1}{b_1}\right)\).

Similarly, you determine the intercepts for a second line \(a_2 x + b_2 y + c_2 = 0\) using the same approach. This gives the intercepts \(\left(-\frac{c_2}{a_2}, 0\right)\) on the x-axis and \(\left(0, -\frac{c_2}{b_2}\right)\) on the y-axis.

Linear Equations

Linear equations are fundamental in mathematics and describe straight lines on a graph. The general form of a linear equation in two dimensions is given by \(Ax + By + C = 0\), where \(A\), \(B\), and \(C\) are constants. In our specific example, the equations take the form \(a_1 x + b_1 y + c_1 = 0\) and \(a_2 x + b_2 y + c_2 = 0\).

• A key characteristic of linear equations is that they represent the simplest form of polynomial equations, having the highest degree of one.


• Each equation has a slope, which is determined by the ratio \(-\frac{A}{B}\), indicating the steepness and direction of the line. Lines sharing the same slope are parallel.

These equations can be manipulated to solve for x or y, depending on the information or constraints at hand, and are used to represent geometric concepts like forms of lines or conditions for concurrency.

Radical Axis Condition

The radical axis condition helps in determining whether points are concyclic, meaning they lie on a common circle. To verify concyclic points, we utilize the relationship between the intercepts of two lines.

• Mathematically, if four points are given by the intercepts of two lines, for them to be concyclic, the following condition must be satisfied: \(\frac{c_{1}^2}{a_{1}b_{1}} = \frac{c_{2}^2}{a_{2}b_{2}}\).


• The manipulation of the intercepts in this condition arises from applying the radical axis theorem, which, in simpler terms, balances the algebraic expression formed by the ratios of squared constants and product of coefficients.

For the lines in our problem, this reduces to finding the relationship \(a_{1}b_{2} = a_{2}b_{1}\), thus proving the points are concyclic and confirming which option correctly satisfies this condition.