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The line l_1 passes through the point (9, −4) and has gradient \\frac{1}{3}; (c) Find an equation for l_1 in the form ax + by + c = 0, where a, b and c are integers - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 1
Question 8
The line l_1 passes through the point (9, −4) and has gradient \\frac{1}{3}; (c) Find an equation for l_1 in the form ax + by + c = 0, where a, b and c are integers... show full transcript
Worked Solution & Example Answer:The line l_1 passes through the point (9, −4) and has gradient \\frac{1}{3}; (c) Find an equation for l_1 in the form ax + by + c = 0, where a, b and c are integers - Edexcel - A-Level Maths Pure - Question 8 - 2005 - Paper 1
Step 1
Find an equation for l_1 in the form ax + by + c = 0
Answer
To find the equation of the line l_1 that passes through the point (9, -4) with a gradient of \frac{1}{3}, we can use the point-slope form of the line:
[y - y_1 = m(x - x_1)]
Substituting the given values:
[y - (-4) = \frac{1}{3}(x - 9)]
[y + 4 = \frac{1}{3}x - 3]
[y = \frac{1}{3}x - 7]
Multiplying through by 3 to eliminate the fraction:
[3y = x - 21]
Rearranging gives us:
[x - 3y - 21 = 0]
Thus, the equation of l_1 is (x - 3y - 21 = 0).
Step 2
Calculate the coordinates of P
Answer
The equation for l_2, which has gradient -2 and passes through the origin, is given by:
[y = -2x]
To find the coordinates of the intersection point P of l_1 and l_2, we set both equations equal to each other:
From l_1:
[y = \frac{1}{3}x - 7]
From l_2:
[y = -2x]
Setting them equal:
[\frac{1}{3}x - 7 = -2x]
Multiplying the entire equation by 3 to eliminate the fraction:
[x - 21 = -6x]
Combining terms gives us:
[7x = 21]
[x = 3]
Substituting back to find y using the equation of l_2:
[y = -2(3) = -6]
Thus, the coordinates of P are (3, -6).
Step 3
calculate the exact area of \Delta OCP
Answer
To find the area of triangle OCP, we will use the coordinates O(0,0), C(0,-7), and P(3,-6).
The formula for the area of a triangle given by vertices ((x_1, y_1), (x_2, y_2), (x_3, y_3)) is:
[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| ]
Substituting the coordinates: [ \text{Area} = \frac{1}{2} \left| 0(-7 + 6) + 0(-6 - 0) + 3(0 + 7) \right| ] [ = \frac{1}{2} \left| 0 + 0 + 21 \right| = \frac{21}{2} ] Thus, the exact area of \Delta OCP is \frac{21}{2}.