Coordinate Geometry: CAT previous year questions

Solution

In parallelogram PQRS, $S = P + R - Q = (-3 + 9 - 1, -2 + 1 + 5) = (5, 4)$.

Line through $Q(1, -5)$ and $S(5, 4)$ has slope $\dfrac{9}{4}$.

At $y = 0$: $x = 1 + \dfrac{5 \times 4}{9} = \dfrac{29}{9}$.

Solution

$x^2 + y^2 + 4x - 4y = (x + 2)^2 + (y - 2)^2 - 8$, so the condition is $4 \leq (x + 2)^2 + (y - 2)^2 \leq 8$: a ring with radii 2 and $2\sqrt{2}$ centred at $(-2, 2)$.

The line $y = x + 4$ passes through the centre, so it cuts the ring in half.

Area $= \dfrac{1}{2}\pi(8 - 4) = 2\pi$.

Solution

The triangle is right-angled at $(1, 2)$ with legs 6 and 8, so the hypotenuse is 10.

Inradius of a right triangle $= \dfrac{6 + 8 - 10}{2} = 2$.

Solution

C: centre $(-2, 3)$, radius $\sqrt{4 + 9 + 3} = 4$.

Tangents meeting at $60^\circ$ come from a point at distance $\dfrac{r}{\sin 30^\circ} = 8$ from the centre.

L is the circle with centre $(-2, 3)$ and radius 8; it touches $x = 6$ at $(6, 3)$.

Solution

Between $x = 0$ and $x = 5$, $|x - 5| = 5 - x$, so $|x - y| = 7 - x$.

This gives the lines $y = 7$ and $y = 2x - 7$.

At $x = 0$ the region runs from $y = -7$ to $7$ (length 14); at $x = 5$ from $y = 3$ to $7$ (length 4).

Area $= \dfrac{1}{2}(14 + 4) \times 5 = 45$.

Solution

In parallelogram ABCD, the diagonals bisect each other, so $D = A + C - B$.

$D = (1 - 2 - 3, 1 + 8 - 4) = (-4, 5)$.

Solution

The region is $|x| \leq 1 + y$ for $0 \leq y \leq 1$: a trapezium.

Width at $y = 0$ is 2 and at $y = 1$ is 4.

Area $= \dfrac{1}{2}(2 + 4) \times 1 = 3$.

Solution

The diagonals of a parallelogram bisect each other, so the midpoint of $(2, 1)$ and $(-3, -4)$, which is $(-0.5, -1.5)$, lies on the other diagonal.

$-0.5 + 9(-1.5) + c = 0 \Rightarrow c = 14$.

Solution

For $0 \leq x \leq 2$: $y = |x - 2| + 4 = 6 - x$.

Area $= \int_0^2 (6 - x)\,dx = 12 - 2 = 10$ (a trapezium with parallel sides 6 and 4, width 2).

Solution

The centre lies on $x = 2$ (perpendicular bisector of $(0, 0)$ and $(4, 0)$). Let it be $(2, k)$.

$4 + k^2 = 1 + (k - 9)^2 \Rightarrow 18k = 78 \Rightarrow k = \dfrac{13}{3}$.

$R^2 = 4 + \dfrac{169}{9} = \dfrac{205}{9}$, so area $= \dfrac{205\pi}{9}$.