Question

A point equidistant from the lines 4x+3y+10 = 0, 5x-12y+26 = 0 and 7x+24y-50 = 0 is

Answer 1

1) $4x + 3y + 10 = 0$

2) $5x - 12y + 26 = 0$

3) $7x + 24y - 50 = 0$

1) $l_{1} = l_{2}\frac{52 - 25}{65}a + \frac{39 + 60}{65}b = 0 \rightarrow \frac{27}{65}a = -\frac{99}{65}b \rightarrow a = -\frac{11}{3}b$

$$\text{or} \quad \frac{52 + 25}{65}a + \frac{39 - 60}{65}b = 0 \rightarrow \frac{77}{65}a = \frac{21}{65}b \rightarrow a = \frac{3}{11}b$$

2) $l_{1} = l_{3}\frac{20 - 7}{25}a + \frac{15 - 24}{25}b = -4 \rightarrow \frac{13}{25}a = -4 + \frac{11}{25}b \rightarrow a = -\frac{100}{13} + \frac{11}{13}b$

$$\text{or} \quad \frac{20 + 7}{25}a + \frac{15 + 24}{25}b = 0 \rightarrow a = \frac{39}{27}b \rightarrow a = \frac{13}{9}b$$

3) $l_{2} = l_{3}\frac{125 - 91}{325}a + \frac{-300 - 312}{325}b = -4 \rightarrow a = \frac{-325 + 2}{17} + \frac{306}{17}b$

$$\text{or} \quad \frac{125 + 91}{325}a + \frac{-300 + 312}{325}b = 0 \rightarrow a = -\frac{3}{54}b$$

So (a,b) $\in (0,0)$