Question

Show that the origin is equidistant from the lines 4x + 3y + 10 = 0; 5x − 12y + 26 = 0 and 7x + 24y = 50.

Answer 1

Let us write down the normal forms of the given lines.

First line: 4x + 3y + 10 = 0

$$\begin{gathered} \Rightarrow -4x-3y=10 \\ \Rightarrow -\frac{4}{\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}}x-\frac{3}{\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}}y=\frac{10}{\sqrt{{\left(-4\right)}^2+{\left(-3\right)}^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow -\frac{4}{5}x-\frac{3}{5}y=2 \\ \therefore p=2 \end{gathered}$$

Second line: 5x − 12y + 26 = 0

$$\begin{gathered} \Rightarrow -5x+12y=26 \\ \Rightarrow -\frac{5}{\sqrt{{\left(-5\right)}^2+{12}^2}}x+\frac{12}{\sqrt{{\left(-5\right)}^2+{12}^2}}y=\frac{26}{\sqrt{{\left(-5\right)}^2+{12}^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow -\frac{5}{13}x+\frac{12}{13}y=2 \\ \therefore p=2 \end{gathered}$$

Third line: 7x + 24y = 50

$$\begin{gathered} \Rightarrow \frac{7}{\sqrt{7^2+{24}^2}}x+\frac{24}{\sqrt{7^2+{24}^2}}y=\frac{50}{\sqrt{7^2+{24}^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow \frac{7}{25}x+\frac{24}{25}y=2 \\ \therefore p=2 \end{gathered}$$

Hence, the origin is equidistant from the given lines.