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The line l₁ passes through the point (9, −4) and has gradient $ rac{1}{3}$ - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 1
Question 10
The line l₁ passes through the point (9, −4) and has gradient $ rac{1}{3}$. (a) Find an equation for l₁ in the form $ax + by + c = 0$, where a, b and c are integers... show full transcript
Worked Solution & Example Answer:The line l₁ passes through the point (9, −4) and has gradient $ rac{1}{3}$ - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 1
Step 1
Find an equation for l₁ in the form $ax + by + c = 0$
Answer
The gradient of l₁ is rac{1}{3}. We can use the point-slope form of the equation of a line:
Substituting the given point (9, -4):
y + 4 = rac{1}{3}(x - 9)
Rearranging yields:
y + 4 = rac{1}{3}x - 3
rac{1}{3}x - y - 7 = 0
Multiplying through by 3 to eliminate the fraction gives:
Thus, the equation of l₁ is .
Step 2
Calculate the coordinates of P
Answer
For line l₂, which passes through the origin (0, 0) and has gradient -2, the equation can be written as:
To find the intersection point P of l₁ and l₂, we substitute into the equation of l₁:
This simplifies to:
7x - 21 = 0\ x = 3$$ Then substituting $x$ back into the equation of l₂ to find $y$: $$y = -2(3) = -6$$ Thus, the coordinates of P are (3, -6).Step 3
calculate the exact area of ΔOCP
Answer
The coordinates of points O, C, and P are:
- O(0, 0)
- C(0, -7)
- P(3, -6)
To find the area of triangle OCP, we use the formula:
ext{Area} = rac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|
where , , are the vertices of the triangle.
Substituting the coordinates:
= rac{1}{2} |0(-7 + 6) + 0(-6 - 0) + 3(0 + 7)| = rac{1}{2} |0 + 0 + 21| = rac{21}{2}
Therefore, the exact area of ΔOCP is rac{21}{2}.