Coordinate geometry: CAT Previous Year Questions

Ans: D

Explanation:Let the coordinates of the vertices be P = (-3, -2), Q = (1, -5), and R = (9, 1). Since PQRS is a parallelogram, the diagonals bisect each other. Let the fourth vertex be S = (x, y).
In a parallelogram, the midpoint of the diagonal PR is the same as the midpoint of the diagonal QS.
Midpoint of PR = $ (\frac{-3+9}{2}, \frac{-2+1}{2}) = (\frac{6}{2}, \frac{-1}{2}) = (3, -\frac{1}{2}) $
Midpoint of QS = $ (\frac{1+x}{2}, \frac{-5+y}{2}) $
Equating the coordinates of the midpoints:
$ \frac{1+x}{2} = 3 \implies 1+x = 6 \implies x = 5 $
$ \frac{-5+y}{2} = -\frac{1}{2} \implies -5+y = -1 \implies y = 4 $
So the coordinates of vertex S are (5, 4).

The diagonal SQ connects points S(5, 4) and Q(1, -5).
The equation of the line passing through S and Q can be found using the two-point form:
$ \frac{y – y_1}{x – x_1} = \frac{y_2 – y_1}{x_2 – x_1} $
Using S(5, 4) as $(x_1, y_1)$ and Q(1, -5) as $(x_2, y_2)$:
$ \frac{y – 4}{x – 5} = \frac{-5 – 4}{1 – 5} $
$ \frac{y – 4}{x – 5} = \frac{-9}{-4} $
$ \frac{y – 4}{x – 5} = \frac{9}{4} $
$ 4(y – 4) = 9(x – 5) $
$ 4y – 16 = 9x – 45 $
$ 4y = 9x – 45 + 16 $
$ 4y = 9x – 29 $

The diagonal SQ intersects the x-axis at (a, 0). This means when y = 0, x = a.
Substitute y = 0 into the equation of the line:
$ 4(0) = 9a – 29 $
$ 0 = 9a – 29 $
$ 9a = 29 $
$ a = \frac{29}{9} $

Correct_Option:D

Ans: 2

Explanation:Let the vertices of the triangle be A = (1, 2), B = (7, 2), and C = (1, 10).
We first calculate the lengths of the sides of the triangle.
Side AB: The y-coordinates are the same (2), so this is a horizontal line. The length is the absolute difference of the x-coordinates: $|7 – 1| = 6$.
Side AC: The x-coordinates are the same (1), so this is a vertical line. The length is the absolute difference of the y-coordinates: $|10 – 2| = 8$.
Side BC: We use the distance formula: $\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
$BC = \sqrt{(7 – 1)^2 + (2 – 10)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

The lengths of the sides are 6, 8, and 10. We notice that $6^2 + 8^2 = 36 + 64 = 100 = 10^2$. This means the triangle is a right-angled triangle, with the right angle at vertex A (where the sides of length 6 and 8 meet).

The area of a right-angled triangle is (1/2) * base * height.
Area = (1/2) * 6 * 8 = 24.

The semi-perimeter (s) of the triangle is half the sum of its side lengths.
$s = (6 + 8 + 10) / 2 = 24 / 2 = 12$.

The radius of the incircle (r) of a triangle can be found using the formula: Area = r * s.
So, $r = Area / s$.
$r = 24 / 12 = 2$.

Alternatively, for a right-angled triangle with legs ‘a’ and ‘b’ and hypotenuse ‘c’, the inradius is given by $r = (a + b – c) / 2$.
Here, $a = 6$, $b = 8$, and $c = 10$.
$r = (6 + 8 – 10) / 2 = (14 – 10) / 2 = 4 / 2 = 2$.

Final_Answer:2

Ans: B

The provided equation for a circle is: $x^{2} + y^{2} + 4x – 6y – 3 = 0$
The coordinates of the circle’s center are (-2,3), and its radius is calculated as $\sqrt{\ g^2+f^2-c} = \sqrt{\ 4+9+3} = 4$.
Let’s denote the point where the tangents intersect as ( h,k).
The line connecting (h,k) to the center of the circle forms a 30-degree angle with each tangent. Consequently, the sine of this angle, sin(30), represents the ratio of the circle’s radius to the distance between the center and the point (h,k).
This gives us: $\sin\left(30\right)=\dfrac{4}{\sqrt{\left(\ h+2\right)^2+\left(k-3\right)^2}}$
Squaring both sides of the equation yields:
$\dfrac{1}{4}=\dfrac{16}{\left(h+2\right)^2+\left(k-3\right)^2}$
Rearranging this, we get: $\left(h+2\right)^2+\left(k-3\right)^2=64$
Given that x = 6, we set h = 6. Substituting this into the equation: $64+\left(k-3\right)^2=64$. This simplifies to $\left(k-3\right)^2=0$, which means k = 3.
Therefore, the required point of intersection is (6,3).

Ans: D

Explanation:In a parallelogram ABCD, the diagonals AC and BD bisect each other. This means that the midpoint of AC is the same as the midpoint of BD.

Let the coordinates of A be $(x_A, y_A) = (1, 1)$.
Let the coordinates of B be $(x_B, y_B) = (3, 4)$.
Let the coordinates of C be $(x_C, y_C) = (-2, 8)$.
Let the coordinates of D be $(x_D, y_D)$.

The midpoint of AC, M_AC, is given by:
$M_{AC} = (\frac{x_A + x_C}{2}, \frac{y_A + y_C}{2})$
$M_{AC} = (\frac{1 + (-2)}{2}, \frac{1 + 8}{2})$
$M_{AC} = (\frac{-1}{2}, \frac{9}{2})$

The midpoint of BD, M_BD, is given by:
$M_{BD} = (\frac{x_B + x_D}{2}, \frac{y_B + y_D}{2})$
$M_{BD} = (\frac{3 + x_D}{2}, \frac{4 + y_D}{2})$

Since M_AC = M_BD, we can equate the corresponding coordinates:
For the x-coordinate:
$\frac{3 + x_D}{2} = \frac{-1}{2}$
$3 + x_D = -1$
$x_D = -1 – 3$
$x_D = -4$

For the y-coordinate:
$\frac{4 + y_D}{2} = \frac{9}{2}$
$4 + y_D = 9$
$y_D = 9 – 4$
$y_D = 5$

So, the coordinates of vertex D are (-4, 5).

Alternatively, in a parallelogram, the vector AB is equal to the vector DC, and the vector AD is equal to the vector BC.

Using $\vec{AB} = \vec{DC}$:
$\vec{AB} = (x_B – x_A, y_B – y_A) = (3 – 1, 4 – 1) = (2, 3)$
$\vec{DC} = (x_C – x_D, y_C – y_D) = (-2 – x_D, 8 – y_D)$

Equating the components:
$2 = -2 – x_D \implies x_D = -2 – 2 = -4$
$3 = 8 – y_D \implies y_D = 8 – 3 = 5$
So, D is (-4, 5).

Using $\vec{AD} = \vec{BC}$:
$\vec{AD} = (x_D – x_A, y_D – y_A) = (x_D – 1, y_D – 1)$
$\vec{BC} = (x_C – x_B, y_C – y_B) = (-2 – 3, 8 – 4) = (-5, 4)$

Equating the components:
$x_D – 1 = -5 \implies x_D = -5 + 1 = -4$
$y_D – 1 = 4 \implies y_D = 4 + 1 = 5$
So, D is (-4, 5).

Correct_Option:D

Ans: D

Explanation:Let the given opposite vertices of the parallelogram be A = (2,1) and C = (-3,-4). Let the other two vertices be B = $(x_1, y_1)$ and D = $(x_2, y_2)$. Since ABCD is a parallelogram, the midpoint of the diagonal AC is the same as the midpoint of the diagonal BD.

Midpoint of AC = $(\frac{2 + (-3)}{2}, \frac{1 + (-4)}{2}) = (\frac{-1}{2}, \frac{-3}{2})$.

Midpoint of BD = $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$.

Equating the midpoints:
$\frac{x_1 + x_2}{2} = \frac{-1}{2} \implies x_1 + x_2 = -1$
$\frac{y_1 + y_2}{2} = \frac{-3}{2} \implies y_1 + y_2 = -3$

The other two vertices B and D lie on the line $x+9y+c=0$.
So, for vertex B: $x_1 + 9y_1 + c = 0$ (Equation 1)
And for vertex D: $x_2 + 9y_2 + c = 0$ (Equation 2)

Add Equation 1 and Equation 2:
$(x_1 + 9y_1 + c) + (x_2 + 9y_2 + c) = 0$
$(x_1 + x_2) + 9(y_1 + y_2) + 2c = 0$

Substitute the values of $x_1 + x_2$ and $y_1 + y_2$ we found earlier:
$(-1) + 9(-3) + 2c = 0$
$-1 – 27 + 2c = 0$
$-28 + 2c = 0$
$2c = 28$
$c = 14$

Correct_Option:D

Ans: D

Explanation:Let the vertices of the triangle be A = (0,0), B = (4,0), and C = (3,9).
We need to find the area of the circle passing through these three points, which is the circumcircle of the triangle. The area of the circle is given by $\pi R^2$, where R is the circumradius.

The circumradius R of a triangle with sides a, b, c and area K is given by the formula $R = \frac{abc}{4K}$.

First, let’s find the lengths of the sides of the triangle.
Side a (opposite to vertex A, i.e., BC):
$a = \sqrt{(3-4)^2 + (9-0)^2} = \sqrt{(-1)^2 + 9^2} = \sqrt{1 + 81} = \sqrt{82}$

Side b (opposite to vertex B, i.e., AC):
$b = \sqrt{(3-0)^2 + (9-0)^2} = \sqrt{3^2 + 9^2} = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10}$

Side c (opposite to vertex C, i.e., AB):
$c = \sqrt{(4-0)^2 + (0-0)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4$

Next, let’s find the area of the triangle K. We can use the determinant formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:
$K = \frac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)|$
Using A=(0,0), B=(4,0), C=(3,9):
$K = \frac{1}{2} |0(0 – 9) + 4(9 – 0) + 3(0 – 0)|$
$K = \frac{1}{2} |0 + 4(9) + 0|$
$K = \frac{1}{2} |36|$
$K = 18$

Now, we can calculate the circumradius R:
$R = \frac{abc}{4K} = \frac{\sqrt{82} \cdot 3\sqrt{10} \cdot 4}{4 \cdot 18}$
$R = \frac{12\sqrt{820}}{72}$
$R = \frac{\sqrt{820}}{6}$
$R = \frac{\sqrt{4 \cdot 205}}{6}$
$R = \frac{2\sqrt{205}}{6}$
$R = \frac{\sqrt{205}}{3}$

The area of the circle is $\pi R^2$:
Area $= \pi \left(\frac{\sqrt{205}}{3}\right)^2$
Area $= \pi \frac{205}{9}$

Let’s check the options:
Option A: $\frac{14\pi}{3}$
Option B: $\frac{123\pi}{7}$
Option C: $\frac{12\pi}{5}$
Option D: $\frac{205\pi}{9}$

Our calculated area matches Option D.

Alternatively, we can find the circumcenter (h, k) and circumradius R by setting the distance from the circumcenter to each vertex equal.
Let the circumcenter be (h, k).
Distance from (h, k) to (0,0): $h^2 + k^2 = R^2$ (1)
Distance from (h, k) to (4,0): $(h-4)^2 + k^2 = R^2$ (2)
Distance from (h, k) to (3,9): $(h-3)^2 + (k-9)^2 = R^2$ (3)

From (1) and (2):
$h^2 + k^2 = (h-4)^2 + k^2$
$h^2 = h^2 – 8h + 16$
$0 = -8h + 16$
$8h = 16$
$h = 2$

Substitute h=2 into (1) and (3):
$2^2 + k^2 = R^2 \implies 4 + k^2 = R^2$ (4)
$(2-3)^2 + (k-9)^2 = R^2$
$(-1)^2 + k^2 – 18k + 81 = R^2$
$1 + k^2 – 18k + 81 = R^2$
$k^2 – 18k + 82 = R^2$ (5)

From (4) and (5):
$4 + k^2 = k^2 – 18k + 82$
$4 = -18k + 82$
$18k = 82 – 4$
$18k = 78$
$k = \frac{78}{18} = \frac{13}{3}$

Now find $R^2$ using equation (4):
$R^2 = 4 + k^2 = 4 + \left(\frac{13}{3}\right)^2 = 4 + \frac{169}{9} = \frac{36 + 169}{9} = \frac{205}{9}$

The area of the circle is $\pi R^2 = \pi \frac{205}{9}$.

Correct_Option:D

Ans: B

Explanation:Let the vertices of the triangle be A = $(x_A, y_A)$, B = $(x_B, y_B)$, and C = $(x_C, y_C)$.
We are given that the side BC lies on the line $x = 4$. This means that the x-coordinates of B and C are both 4.
Let B = $(4, y_B)$ and C = $(4, y_C)$.
The length of the side BC is given as 8 units. Since BC lies on the vertical line $x = 4$, the length of BC is the absolute difference of the y-coordinates:
$|y_C – y_B| = 8$.

The area of triangle ABC is given as 32 sq units. The area of a triangle can be calculated as $\frac{1}{2} \times \text{base} \times \text{height}$.
We can take BC as the base of the triangle. The length of the base BC is 8.
The height of the triangle with respect to the base BC is the perpendicular distance from vertex A to the line containing BC, which is the line $x = 4$.
The x-coordinate of A is $x_A$. The distance from A$(x_A, y_A)$ to the line $x = 4$ is $|x_A – 4|$.
So, the area of triangle ABC is $\frac{1}{2} \times 8 \times |x_A – 4| = 4 \times |x_A – 4|$.
We are given that the area is 32 sq units:
$4 \times |x_A – 4| = 32$
$|x_A – 4| = 8$
This gives two possibilities for $x_A$:
$x_A – 4 = 8$ or $x_A – 4 = -8$
$x_A = 12$ or $x_A = -4$.

We need to find the shortest possible distance between A and the point (0,0). The distance between A = $(x_A, y_A)$ and (0,0) is given by the distance formula:
$d = \sqrt{(x_A – 0)^2 + (y_A – 0)^2} = \sqrt{x_A^2 + y_A^2}$.

To minimize this distance, we need to minimize $x_A^2 + y_A^2$.
The value of $y_A$ can be any real number. To minimize $y_A^2$, we should choose $y_A = 0$.

Now let’s consider the two possible values for $x_A$:
Case 1: $x_A = 12$.
If $y_A = 0$, the coordinates of A are (12, 0).
The distance from A to (0,0) is $\sqrt{12^2 + 0^2} = \sqrt{144} = 12$.

Case 2: $x_A = -4$.
If $y_A = 0$, the coordinates of A are (-4, 0).
The distance from A to (0,0) is $\sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4$.

Comparing the distances from the two cases, the shortest possible distance is 4.

Let’s reconfirm if the y-coordinate of A can be arbitrary. The area calculation only depends on the x-coordinate of A as the height. The y-coordinate of A does not affect the area of the triangle given BC on the line $x=4$. Therefore, to minimize the distance from A to the origin, we need to choose the y-coordinate of A to be 0.

The shortest possible distance between A and the point (0,0) occurs when $y_A = 0$.
If $x_A = 12$, the minimum distance is $\sqrt{12^2 + 0^2} = 12$.
If $x_A = -4$, the minimum distance is $\sqrt{(-4)^2 + 0^2} = 4$.

The shortest of these two distances is 4.

The final answer is $\boxed{4}$ units.

Let’s recheck the options.
Option A: 8 units
Option B: 4 units
Option C: $2\sqrt{2}$ units
Option D: $4\sqrt{2}$ units

Our calculated shortest distance is 4 units, which matches Option B.

The final answer is $\boxed{4}$.

To be more precise in the output format:
Correct_Option:B

Ans: C

Explanation:The equation $|x| + |y| = 2$ represents a square in the two-dimensional plane.
We can analyze this equation by considering the four quadrants:
1. In the first quadrant, $x \ge 0$ and $y \ge 0$, so the equation becomes $x + y = 2$. This is a line segment connecting the points (2, 0) and (0, 2).
2. In the second quadrant, $x \le 0$ and $y \ge 0$, so the equation becomes $-x + y = 2$, or $y = x + 2$. This is a line segment connecting the points (0, 2) and (-2, 0).
3. In the third quadrant, $x \le 0$ and $y \le 0$, so the equation becomes $-x – y = 2$, or $x + y = -2$. This is a line segment connecting the points (-2, 0) and (0, -2).
4. In the fourth quadrant, $x \ge 0$ and $y \le 0$, so the equation becomes $x – y = 2$, or $y = x – 2$. This is a line segment connecting the points (0, -2) and (2, 0).

These four line segments form a square with vertices at (2, 0), (0, 2), (-2, 0), and (0, -2).
The diagonals of this square lie along the x and y axes. The length of the diagonal along the x-axis is from -2 to 2, which is $2 – (-2) = 4$. The length of the diagonal along the y-axis is from -2 to 2, which is $2 – (-2) = 4$.
The area of a square can be calculated as half the product of its diagonals.
Area = $\frac{1}{2} \times d_1 \times d_2$
Here, $d_1 = 4$ and $d_2 = 4$.
Area = $\frac{1}{2} \times 4 \times 4 = \frac{1}{2} \times 16 = 8$ sq. units.

Alternatively, we can consider the square as being composed of four right-angled triangles, each in one quadrant. For the first quadrant, the triangle has vertices at (0, 0), (2, 0), and (0, 2). The base and height of this triangle are 2 and 2 respectively. The area of this triangle is $\frac{1}{2} \times 2 \times 2 = 2$ sq. units. Since there are four such triangles, the total area is $4 \times 2 = 8$ sq. units.

The options provided are:
A: $4\pi$ sq. units
B: 4 sq. units
C: 8 sq. units
D: $2\pi$ sq. units

The calculated area is 8 sq. units, which corresponds to Option C.

Correct_Option:C

Ans: D

Explanation:Let the given endpoints of a diagonal be A = (2, 5) and B = (6, 3).
The midpoint of this diagonal is the center of the rectangle.
Midpoint M = ((2+6)/2, (5+3)/2) = (8/2, 8/2) = (4, 4).
The diagonals of a rectangle bisect each other, so the other diagonal also passes through the midpoint M(4, 4).
The equation of the other diagonal is given as y = 3x + c.
Since the point M(4, 4) lies on this diagonal, it must satisfy the equation.
Substitute the coordinates of M into the equation:
4 = 3(4) + c
4 = 12 + c
c = 4 – 12
c = -8.

The other diagonal has the equation y = 3x – 8.
Therefore, c is -8.

Correct_Option:D