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The line RS cuts the x-axis at the point R and the y-axis at the point S(0, 10), as shown - Leaving Cert Mathematics - Question 5 - 2014
Question 5
The line RS cuts the x-axis at the point R and the y-axis at the point S(0, 10), as shown. The area of the triangle ROS, where O is the origin, is \(\frac{125}{3}\... show full transcript
Worked Solution & Example Answer:The line RS cuts the x-axis at the point R and the y-axis at the point S(0, 10), as shown - Leaving Cert Mathematics - Question 5 - 2014
Step 1
Find the co-ordinates of R.
Answer
To find the coordinates of point R, we can use the area formula for triangle ROS:
We know that the area is (\frac{125}{3}), and we have (OS = 10). So, the equation becomes:
Solving for |RO|:
|RO| = \frac{125}{15} = \frac{25}{3}$$ Now, since point S is at (0, 10) and R lies on the x-axis, we can express R as \(\left( -\frac{25}{3}, 0 \right)\). Therefore, the coordinates of R are \(R \left( -\frac{25}{3}, 0 \right)\).Step 2
Show that the point E(-5, 4) is on the line RS.
Answer
First, we find the slope of line RS using the coordinates of points R and S:
The coordinates of S(0, 10) and R(-\frac{25}{3}, 0) give us:
Now, we can use the point-slope form of the line equation, given point S (0, 10):
y - 10 = \frac{6}{5}x\ y = \frac{6}{5}x + 10$$ Now let's substitute the point E(-5, 4) into this equation: $$4 = \frac{6}{5}(-5) + 10 \ 4 = -6 + 10 \ 4 = 4 \Thus, E is indeed on the line RS.
Step 3
A second line y = mx + c, where m and c are positive constants, passes through the point E and again makes a triangle of area 125/3 with the axes. Find the value of m and the value of c.
Answer
Let's set the equation of the second line:
This line passes through E(-5, 4), so we can substitute these values:
c = 4 + 5m$$ Next, this line also forms a triangle with the axes, which means we need to find the x-intercept and y-intercept to compute the area: - The x-intercept occurs when y = 0: $$0 = mx + c \ \Rightarrow x = -\frac{c}{m}$$ - The y-intercept occurs when x = 0: $$y = c$$ The area A of the triangle formed is given by: $$A = \frac{1}{2} \times |x\text{-intercept}| \times |y\text{-intercept}| = \frac{1}{2} \times \left(-\frac{c}{m}\right) \times c$$ This must equal \(\frac{125}{3}\), so: $$\frac{1}{2} \times \left(-\frac{c^2}{m}\right) = \frac{125}{3}\ -c^2 = \frac{250}{3}m \Substituting for c:
\text{Expanding and simplifying gives a quadratic equation in m:}$$ This can be solved to find the required values for m and c.