To find the ratio in which the line segment joining points \( P(-4, -2) \) and \( Q(10, 4) \) is divided by the y-axis, we follow these steps:
- First, understand that the y-axis will intersect the line segment at a point where the x-coordinate is zero. We use this property to find the x-intercept of the line.
- The equation of the line through points \( P(x_1, y_1) = (-4, -2) \) and \( Q(x_2, y_2) = (10, 4) \) is given by the formula for the line connecting two points:
\(y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)\)
- Substituting the values:
\(y + 2 = \frac{4 + 2}{10 + 4} (x + 4)\)
\(y + 2 = \frac{6}{14}(x + 4)\)
\(y + 2 = \frac{3}{7}(x + 4)\)
- Setting \( x = 0 \) to find the intersection with the y-axis:
\(y + 2 = \frac{3}{7}(4)\)
\(y + 2 = \frac{12}{7}\)
\(y = \frac{12}{7} - \frac{14}{7}\)
\(y = -\frac{2}{7}\)
- The intersection point on the y-axis is \( \left(0, -\frac{2}{7}\right) \).
- Now, calculate the distance of this intersection point from the points \( P \) and \( Q \) using the distance formula. The distance is given by:
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
- Distance from \( P(-4, -2) \) to intersection \( \left(0, -\frac{2}{7}\right) \):
\(d = \sqrt{(0 + 4)^2 + \left(-\frac{2}{7} + 2\right)^2}\)
\(= \sqrt{16 + \left(\frac{14-2}{7}\right)^2}\)
\(= \sqrt{16 + \left(\frac{12}{7}\right)^2}\)
\(= \sqrt{16 + \frac{144}{49}}\)
\(= \sqrt{\frac{784}{49} + \frac{144}{49}}\)
\(= \sqrt{\frac{928}{49}}\)
- Distance from \( Q(10, 4) \) to intersection \( \left(0, -\frac{2}{7}\right) \):
\(d = \sqrt{(10 - 0)^2 + \left(4 + \frac{2}{7}\right)^2}\)
\(= \sqrt{100 + \left(\frac{30}{7}\right)^2}\)
\(= \sqrt{100 + \frac{900}{49}}\)
\(= \sqrt{\frac{4900}{49} + \frac{900}{49}}\)
\(= \sqrt{\frac{5800}{49}}\)
- The ratio of the distances will be: \(\frac{\sqrt{\frac{928}{49}}}{\sqrt{\frac{5800}{49}}} = \frac{\sqrt{928}}{\sqrt{5800}}\).
- Calculate this ratio to get the required ratio of division: \( \frac{2}{5} \).
- The line is divided by the y-axis in the ratio \(2:5\).
Therefore, the correct answer is \(2:5\).