Q.A point equidistant from the lines $4x+3y+10=0$, $5x-12y+26=0$ and $7x+24y-50=0$ is
(A) $(1,-1)$
(B) $(1,1)$
(C) $(0,0)$
(D) $(0,1)$

The key idea is that a point equidistant from three lines must satisfy the distance formula from a point to a line, with the distances equal in absolute value.

Step 1: Distance from point $(x_1, y_1)$ to line $Ax + By + C = 0$ is

Step 2: For the given lines, the denominators are:

$\sqrt{4^2+3^2}=5$, $\sqrt{5^2+(-12)^2}=13$, $\sqrt{7^2+24^2}=25$. …

We need a point whose perpendicular distances to all three given lines are equal. Let’s recall the formula for the distance from a point $(x_1, y_1)$ to a line $ax + by + c = 0$:

The absolute value means distance is always non-negative. For a point to be equidistant from two lines, the expressions for these distances must be equal in magnitude — but the signs inside the absolute values can differ. That’s why we consider both the positive and negative angle bisectors.

Now, instead of solving the full system of three equations, we can test each option. That’s faster and avoids messy algebra.

1. Test option (A): $(1, -1)$

   For $4x+3y+10=0$:

   $|4(1) + 3(-1) + 10| = |4 - 3 + 10| = |11| = 11$

   $\sqrt{4^2 + 3^2} = 5$ → distance $= \frac{11}{5} = 2.2$

   For $5x-12y+26=0$:

   $|5(1) -12(-1) + 26| = |5 + 12 + 26| = |43| = 43$

   $\sqrt{5^2 + (-12)^2} = 13$ → distance $= \frac{43}{13} \approx 3.3077$

   Not equal. So (A) fails.

2. Test option (B): $(1, 1)$

   Line 1: $|4 + 3 + 10| = 17$, distance $= 17/5 = 3.4$

   Line 2: $|5 - 12 + 26| = |19| = 19$, distance $= 19/13 \approx 1.4615$

   Not equal. (B) fails.

3. Test option (C): $(0, 0)$

   Line 1: $|0 + 0 + 10| = 10$, distance $= 10/5 = 2$

   Line 2: $|0 + 0 + 26| = 26$, distance $= 26/13 = 2$

   Line 3: $7x+24y-50=0$: $|0 + 0 - 50| = 50$, $\sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$, distance $= 50/25 = 2$

   All three distances are exactly 2. So (C) works. …