Moderate -0.3 This is a multi-part question covering standard coordinate geometry techniques (gradient, equation of line, distance formula, area of triangle). Part (a) and (b) are routine C1 exercises. Parts (c) and (d) require applying the distance formula with an unknown and basic problem-solving, but follow predictable patterns. Slightly easier than average due to straightforward application of standard methods with no conceptual challenges.
8. (a) Find an equation of the line joining \(A ( 7,4 )\) and \(B ( 2,0 )\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
(b) Find the length of \(A B\), leaving your answer in surd form.
The point \(C\) has coordinates ( \(2 , t\) ), where \(t > 0\), and \(A C = A B\).
(c) Find the value of \(t\).
(d) Find the area of triangle \(A B C\).
\(\_\_\_\_\)
Attempt at equation of \(AB\); follow through their gradient, not e.g. \(-\frac{1}{m}\)
\(4x - 5y - 8 = 0\) (o.e.)
A1
Requires integer form but allow \(5y + 8 = 4x\) etc. Must have "=" or A0
Part (b):
Answer
Marks
Guidance
Answer
Mark
Guidance
\((AB =)\sqrt{(7-2)^2 + (4-0)^2}\)
M1
For an expression for \(AB\) or \(AB^2\); ignore what is "left" of equals sign
\(= \sqrt{41}\)
A1
Part (c):
Answer
Marks
Guidance
Answer
Mark
Guidance
Using isosceles triangle with \(AB = AC\) then \(t = 2 \times y_A = 2 \times 4 = 8\)
B1
For \(t = 8\); may be implied by correct coordinates \((2, 8)\) or value appearing in (d)
Part (d):
Answer
Marks
Guidance
Answer
Mark
Guidance
Area of triangle \(= \frac{1}{2}t \times (7-2)\)
M1
Follow through their \(t\) \((\neq 0)\); must have \((7-2)\) or 5 and the \(\frac{1}{2}\)
\(= 20\)
A1
## Question 8:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $m_{AB} = \frac{4-0}{7-2} = \frac{4}{5}$ | M1 | Attempt at gradient of $AB$; correct substitution in correct formula |
| $y - 0 = \frac{4}{5}(x-2)$ or $y - 4 = \frac{4}{5}(x-7)$ | M1 | Attempt at equation of $AB$; follow through their gradient, not e.g. $-\frac{1}{m}$ |
| $4x - 5y - 8 = 0$ (o.e.) | A1 | Requires integer form but allow $5y + 8 = 4x$ etc. Must have "=" or A0 |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(AB =)\sqrt{(7-2)^2 + (4-0)^2}$ | M1 | For an expression for $AB$ or $AB^2$; ignore what is "left" of equals sign |
| $= \sqrt{41}$ | A1 | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Using isosceles triangle with $AB = AC$ then $t = 2 \times y_A = 2 \times 4 = 8$ | B1 | For $t = 8$; may be implied by correct coordinates $(2, 8)$ or value appearing in (d) |
### Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area of triangle $= \frac{1}{2}t \times (7-2)$ | M1 | Follow through their $t$ $(\neq 0)$; must have $(7-2)$ or 5 and the $\frac{1}{2}$ |
| $= 20$ | A1 | |
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8. (a) Find an equation of the line joining $A ( 7,4 )$ and $B ( 2,0 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
(b) Find the length of $A B$, leaving your answer in surd form.
The point $C$ has coordinates ( $2 , t$ ), where $t > 0$, and $A C = A B$.\\
(c) Find the value of $t$.\\
(d) Find the area of triangle $A B C$.\\
$\_\_\_\_$
\hfill \mbox{\textit{Edexcel C1 2010 Q8 [8]}}