Question

3 and 5 are intercepts of a line $L = 0$, then the distance of $L = 0$ from $(3, 7)$ is

• A. \( \sqrt{31} \)
• B. \( \sqrt{34} \)
• C. \( \frac{21}{\sqrt{34}} \)
• D. \( \frac{\sqrt{34}}{31} \)

Hint:

To calculate the distance from a point to a line, use the formula: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] where \(Ax + By + C = 0\) is the equation of the line.

Answer 1

The Correct Option is C

To find the distance of the line \(L = 0\) from the point \((3, 7)\), we first need the equation of the line. Since the intercepts of the line are 3 on the x-axis and 5 on the y-axis, the equation of the line in intercept form is \(\frac{x}{3} + \frac{y}{5} = 1\).

Rewriting this equation into the standard form \(Ax + By + C = 0\), we obtain:

\(5x + 3y - 15 = 0\)

This is the equation of the line \(L = 0\).

Now, we calculate the distance \(d\) from the point \((3, 7)\) to the line \(5x + 3y - 15 = 0\) using the distance formula:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Substituting the given values, \(A = 5\), \(B = 3\), \(C = -15\), \(x_1 = 3\), \(y_1 = 7\), we get:

\[ d = \frac{|5(3) + 3(7) - 15|}{\sqrt{5^2 + 3^2}} \]

This simplifies to:

\[ d = \frac{|15 + 21 - 15|}{\sqrt{25 + 9}} \]

\[ d = \frac{|21|}{\sqrt{34}} \]

\[ d = \frac{21}{\sqrt{34}} \]

Thus, the distance of the line from the point \((3, 7)\) is \(\frac{21}{\sqrt{34}}\).

Answer 2

The equation of the line with intercepts \(3\) and \(5\) is given by: \[ \frac{x}{3} + \frac{y}{5} = 1 \] The formula for the distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is: \[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \] For the equation \( \frac{x}{3} + \frac{y}{5} = 1 \), we rewrite it as: \[ 5x + 3y - 15 = 0 \] So, \(A = 5\), \(B = 3\), and \(C = -15\). Substituting the point \((3, 7)\) into the distance formula: \[ d = \frac{|5(3) + 3(7) - 15|}{\sqrt{5^2 + 3^2}} = \frac{|15 + 21 - 15|}{\sqrt{25 + 9}} = \frac{21}{\sqrt{34}} \] Thus, the distance is \( \frac{21}{\sqrt{34}} \), which is option (C).