Area of Triangle OPQ: A Geometric Approach Involving Line Equations and Distances

The problem at hand involves finding the area of triangle OPQ where P and Q are points on the x-axis and y-axis, respectively. This problem is approached by analyzing the given line equation 3x 4y 15 and determining the distances and intercepts involved.

Identifying Intercepts

The first step in solving the problem is to identify the intercepts of the line equation with the coordinate axes.

X-Intercept:

When y 0, the equation simplifies to:

3x   4(0)  15 implies 3x  15 implies x  5

This gives us the x-intercept (5, 0), which we denote as point P.

Y-Intercept:

When x 0, the equation simplifies to:

3(0)   4y  15 implies 4y  15 implies y  frac{15}{4}

This gives us the y-intercept (0, frac{15}{4}), which we denote as point Q.

Area Calculation:

The area of triangle OPQ can be calculated using the base and height as follows:

text{Area}  frac{1}{2} times text{base} times text{height}

Here, the base is the distance along the x-axis from O to P, which is P(5, 0). The height is the distance along the y-axis from O to Q, which is Q(0, frac{15}{4}).

text{Base}  OP  5text{Height}  OQ  frac{15}{4}

Substituting these values into the area formula:

text{Area}  frac{1}{2} times 5 times frac{15}{4}  frac{1}{2} times frac{75}{4}  frac{75}{8}

Conclusion:

The area of triangle OPQ is:

boxed{frac{75}{8}}

Alternative Method Involving Distance from Origin

When solving geometric problems, it is also useful to use the distance from a point to a line formula. Given the line 3x 4y 15, the distance from the origin to this line can be calculated as:

text{Distance}  frac{|3(0)   4(0) - 15|}{sqrt{3^2   4^2}}  frac{15}{5}  3

Given that the line intersects the circle centered at the origin with radius 5, we can find the intersection points by solving:

x^2   y^2  25 quad text{with} quad 3x   4y  15

Upon solving, we get the intersection points at (left(-frac{7}{5}, frac{24}{5}right)) and ((5, 0)).

Determining the Area:

The area of triangle OPQ can also be calculated using the distance between the intersection points:

text{Distance}  sqrt{left(-frac{7}{5} - 5right)^2   left(frac{24}{5} - 0right)^2}  sqrt{left(frac{-32}{5}right)^2   left(frac{24}{5}right)^2}  sqrt{frac{1024}{25}   frac{576}{25}}  sqrt{frac{1600}{25}}  8

The area of the triangle can then be calculated as:

text{Area}  frac{1}{2} times 5 times 8  20

Another Approach: Isosceles Triangle Properties

Let's consider the geometric properties of an isosceles triangle OPQ with the origin O at the vertex and OP and OQ as its base sides.

Makes Calculation Simpler:

The slope of the line is:

text{Slope}  -frac{3}{4}

The base angles of the isosceles triangle can be calculated as:

arctanleft(frac{3}{4}right)  36.87^circ

The vertex angle is then:

180^circ - 2 times 36.87^circ  106.26^circ

The area of the isosceles triangle can be calculated with the formula:

text{Area}  frac{1}{2} times text{base} times text{height}text{Area}  frac{1}{2} times 5 times 5 times sin(106.26^circ)text{Area} approx 12 text{ square units}