The points A(1, 7), B(20, 7) and C(p, q) form the vertices of a triangle ABC, as shown in Figure 2. The point D(8, 2) is the mid-point of AC. (a) Find the value of p and the value of q. (b) The line l, which passes through D and is perpendicular to AC, intersects AB at E. (c) Find an equation for l, in the form ax + by + c = 0, where a, b and c are integers. (d) Find the exact x-coordinate of E.

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The points A(1, 7), B(20, 7) and C(p, q) form the vertices of a triangle ABC, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2

Question 9

The points A(1, 7), B(20, 7) and C(p, q) form the vertices of a triangle ABC, as shown in Figure 2. The point D(8, 2) is the mid-point of AC. (a) Find the value of ... show full transcript

Worked Solution & Example Answer:The points A(1, 7), B(20, 7) and C(p, q) form the vertices of a triangle ABC, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2

Step 1

Find the value of p and the value of q.

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Answer

To find the values of p and q, we first calculate the slope of the line AC. Using the coordinates of A(1, 7) and D(8, 2), we can find the slope (m) as follows:

$m_{AC} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 7}{8 - 1} = \frac{-5}{7}.$

Since D is the midpoint of AC, we can derive the coordinates of C(p, q) by using the midpoint formula, which states:

$D(x_D, y_D) = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2} \right).$

Thus,

$8 = \frac{1 + p}{2} \Rightarrow p = 15$

and

$2 = \frac{7 + q}{2} \Rightarrow q = -3.$

Therefore, the values are ( p = 15 ) and ( q = -3 ).

Step 2

The line l, which passes through D and is perpendicular to AC, intersects AB at E.

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Answer

To find the slope of line l, which is perpendicular to AC, we take the negative reciprocal of the slope of AC:

$m_{l} = -\frac{1}{m_{AC}} = -\frac{7}{5}.$

Using point-slope form, the equation of line l through point D(8, 2) is:

$y - 2 = -\frac{7}{5}(x - 8)$

Rearranging this gives:

$y = -\frac{7}{5}x + \frac{56}{5} + 2$

$y = -\frac{7}{5}x + \frac{66}{5}.$

Multiplying through by 5 to eliminate the fraction:

$5y = -7x + 66 \Rightarrow 7x + 5y - 66 = 0.$

Step 3

Find the exact x-coordinate of E.

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Answer

To find the intersection E of lines AB and l, we first note that AB is horizontal (since both A and B have the same y-coordinate of 7). Therefore, the equation of line AB is:

$y = 7.$

Substituting y = 7 into the equation of line l:

$7 = -\frac{7}{5}x + \frac{66}{5}.$

Multiplying through by 5:

$35 = -7x + 66 \Rightarrow 7x = 66 - 35 \Rightarrow 7x = 31 \Rightarrow x = \frac{31}{7}.$

Therefore, the exact x-coordinate of E is ( \frac{31}{7} ).

The points A(1, 7), B(20, 7) and C(p, q) form the vertices of a triangle ABC, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2

Worked Solution & Example Answer:The points A(1, 7), B(20, 7) and C(p, q) form the vertices of a triangle ABC, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2005 - Paper 2

Find the value of p and the value of q.

The line l, which passes through D and is perpendicular to AC, intersects AB at E.

Find the exact x-coordinate of E.

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