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 10 - 2005 - Paper 2

Question 10

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 10 - 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 know that D(8, 2) is the midpoint of AC. Using the midpoint formula,

$D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

where A(1,7) and C(p,q) gives us:

2 = \frac{7 + q}{2}$$ From the first equation, solving for p: $$8 \times 2 = 1 + p\\ 16 = 1 + p \\ \Rightarrow p = 15$$ From the second equation, solving for q: $$2 \times 2 = 7 + q\\ 4 = 7 + q\\ \Rightarrow q = 4 - 7 = -3$$ Thus, we find the values: 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 equation of line l, we need the slope of line AC first:

The coordinates of A and C are (1, 7) and (15, -3) respectively, so the slope (m) is given by:

$m_{AC} = \frac{q_2 - q_1}{p_2 - p_1} = \frac{-3 - 7}{15 - 1} = \frac{-10}{14} = -\frac{5}{7}$

The slope of the perpendicular line l is the negative reciprocal:

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

Using point-slope form, the equation of line l (through point D(8, 2)) can be written as:

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

Rearranging gives:

= -\frac{7}{5}x + \frac{66}{5}$$ Multiplying through by 5 to eliminate fractions, we obtain: $$7x + 5y - 66 = 0$$

Step 3

Find the exact x-coordinate of E.

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Answer

To find the x-coordinate of E, we need the intersection of lines l and AB. The equation for AB (a horizontal line through B and A at y = 7) is:

$y = 7$

Substituting this into the equation of line l:

\Rightarrow 7 \times 5 = -7x + 66\\ 35 = -7x + 66\\ \Rightarrow 7x = 66 - 35\\ \Rightarrow 7x = 31\\ \Rightarrow x = \frac{31}{7}$$ Thus, 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 10 - 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 10 - 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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