Standard +0.3 This is a straightforward coordinate geometry problem requiring identification of intercepts (A = (a,0), B = (0,b)), application of the midpoint formula, and simultaneous equations with the distance formula. All steps are standard techniques with no novel insight required, making it slightly easier than average.
5 The line \(\frac { x } { a } + \frac { y } { b } = 1\), where \(a\) and \(b\) are positive constants, intersects the \(x\) - and \(y\)-axes at the points \(A\) and \(B\) respectively. The mid-point of \(A B\) lies on the line \(2 x + y = 10\) and the distance \(A B = 10\). Find the values of \(a\) and \(b\).
\(M\) has coordinates \(\left(\frac{a}{2}, \frac{b}{2}\right)\)
M1*
Uses Pythagoras with their \(A\) & \(B\)
B1\(\checkmark\)
\(\checkmark\) on their \(A\) and \(B\)
\(M\) lies on \(2x + y = 10 \rightarrow a + \frac{b}{2} = 10\)
M1*
Subs into given line, using their \(M\), to link \(a\) and \(b\)
Sub \(\rightarrow a^2 + (20-2a)^2 = 100\) or \(\left(10 - \frac{b}{2}\right)^2 + b^2 = 100\)
DM1
Forms quadratic in \(a\) or in \(b\)
\(\rightarrow a = 6, b = 8\)
A1
cao
## Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A(a, 0)$ and $B(0, b)$, $a^2 + b^2 = 100$ | B1 | soi |
| $M$ has coordinates $\left(\frac{a}{2}, \frac{b}{2}\right)$ | M1* | Uses Pythagoras with their $A$ & $B$ |
| | B1$\checkmark$ | $\checkmark$ on their $A$ and $B$ |
| $M$ lies on $2x + y = 10 \rightarrow a + \frac{b}{2} = 10$ | M1* | Subs into given line, using their $M$, to link $a$ and $b$ |
| Sub $\rightarrow a^2 + (20-2a)^2 = 100$ or $\left(10 - \frac{b}{2}\right)^2 + b^2 = 100$ | DM1 | Forms quadratic in $a$ or in $b$ |
| $\rightarrow a = 6, b = 8$ | A1 | cao |
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5 The line $\frac { x } { a } + \frac { y } { b } = 1$, where $a$ and $b$ are positive constants, intersects the $x$ - and $y$-axes at the points $A$ and $B$ respectively. The mid-point of $A B$ lies on the line $2 x + y = 10$ and the distance $A B = 10$. Find the values of $a$ and $b$.
\hfill \mbox{\textit{CAIE P1 2016 Q5 [6]}}