CAT Coordinate Geometry Questions

Upon drawing a rough sketch of the coordinates given, we realise that this is a right-angled triangle.

The three side lengths are 6, 8 and 10 units

The inradius of a circle can be calculated using the formula: $$rs=Area$$, where s is the semi-perimeter of the triangle 

The Area would be $$\frac{1}{2}\times\ 6\times\ 8=24$$
and the semi-perimeter would be $$\frac{10+6+8}{2}=12$$

Giving the inradius to be $$\frac{24}{12}=2$$ units

Therefore, 2 is the correct answer. 

Given equation of circle = $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$

Center of the circle is (-2,3) and radius of the circle = $$\sqrt{\ g^2+f^2-c}=\sqrt{\ 4+9+3}=4$$

Let us assume the point of the intersection of the tangents is ( h,k)

The angle made by the line joining (h,k) to the centre makes an angle of 30 degrees with the tangent, and sin(30) will be the ratio of the radius and the distance between the center and (h,k)

=> $$\sin\left(30\right)=\dfrac{4}{\sqrt{\left(\ h+2\right)^2+\left(k-3\right)^2}}$$

Squaring on both sides:

$$\dfrac{1}{4}=\dfrac{16}{\left(h+2\right)^2+\left(k-3\right)^2}$$

=> $$\left(h+2\right)^2+\left(k-3\right)^2=64$$

When x = 6 => h = 6 => $$64+\left(k-3\right)^2=64$$ => k = 3.

=> required point is (6,3)

In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.

Midpoint of AC = $$\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

Let the coordinates of vertex D be (x,y)

$$\left(\ \frac{\ x+3}{2},\ \frac{\ y+4}{2}\right)=\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$

x = -4 and y = 5

The answer is option D.

The midpoints of two diagonals of a parallelogram are the same

Hence the midpoint of (2,1) and (-3,-4) lie on $$x+9y+c=0$$

midpoint of (2,1) and (-3,-4) = ($$\frac{2-3}{2},\frac{1-4}{2}$$) = (-1/2 , -3/2)

Keeping this cordinates in the above line equation, we get c = 14

Equation of circle $$x^2+y^2+2gx+2fy+c=0$$

It passes through (0,0), (4,0) and (3,9). Substitute each point in the above equation:

=> On substituting the value (0,0) in the above equation, we obtain: $$c=0$$

=> On substituting the value (4,0) in the above equation, we obtain:  $$16+0+8g+0 = 0$$ ; $$g=-2$$

=> On substituting the value (3,9) in the above equation, we obtain: $$9+81-12+18f = 0$$ ; $$f= -13/3$$

Radius of the circle r = $$\sqrt{\ g^2+f^2-c}$$ => $$r^2=\frac{205}{9}$$

Therefore, Area = $$\pi\ r^2=\frac{205\pi\ }{9}$$

In any right triangle, the circumradius is half of the hypotenuse. Here,L=$$\ \frac{\ 1}{2}\ $$* the length of the hypotenuse = $$\ \frac{\ 1}{2}$$($$\sqrt{\ 15^2+9^2}$$) = $$\ \frac{\ 1}{2}\ $$*$$\sqrt{\ 306}$$ = $$\ \frac{\ 1}{\ 2}\times\ $$17.49 = 8.74

Hence, the integer close to L = 9

We can see that T$$_{2}$$ is formed by using the mid points of T$$_{1}$$. Hence, we can say that area of triangle of T$$_{2}$$ will be (1/4)th of the area of triangle T$$_{1}$$.

Area of triangle T$$_{1}$$ = $$\dfrac{\sqrt{3}}{4}*(24)^2$$ = $$144\sqrt{3}$$ sq. cm

Area of triangle T$$_{2}$$ = $$\dfrac{144\sqrt{3}}{4}$$ = $$36\sqrt{3}$$ sq. cm 

Sum of the area of all triangles =  T$$_{1}$$ + T$$_{2}$$ + T$$_{3}$$ + ... 

$$\Rightarrow$$ T$$_{1}$$ + T$$_{1}$$/$$4$$ + T$$_{1}$$/$$4^2$$ + ... 

$$\Rightarrow$$ $$\dfrac{T_{1}}{1 - 0.25}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*T_{1}$$

$$\Rightarrow$$ $$\dfrac{4}{3}*144\sqrt{3}$$

$$\Rightarrow$$ $$192\sqrt{3}$$

Hence, option D is the correct answer. 

We know that area of the triangle = 32 sq. units, BC = 8 units

Therefore, the height of the perpendicular drawn from point A to BC = 2*32/8 = 8 units.

Let us draw a possible diagram of the given triangle. 

We can see that if A coincide with (-4, 0) then the distance between A and (0, 0) = 4 units.

If we move the triangle up or down keeping the base BC on x = 4, then point A will move away from origin as vertical distance will come into factor whereas horizontal distance will remain as 4 units.

Hence, we can say that minimum distance between A and origin (0, 0) = 4 units. 

The following equation will form a square of side $$ 2\sqrt{2} $$.

The area of the square = $$(2\sqrt{2})^2$$ = 8 units.

The midpoint of one diagonal lies on the other diagonal.

Midpoint is ((2+6)/2, (5+3)/2) = (4,4)
Hence 4 = 3 * 4 + c => c = -8

The graph of the given function is as shown below.

We can see that the shortest distance of the point (1/2, 1) will be 1 unit.

The number of points on x = 1 is 39. The number of points on x = 2 is 38 and so on till x = 39, which has one point.

So, the total is 1+2+3+...+39 = $$\frac{39*40}{2}$$ = 780.

The triangle we have is :

The length of three sides is $$\sqrt 2, \sqrt 2$$ and $$2$$.
This is a right-angled triangle.
Hence, it's area equals $$1/2 * \sqrt 2 * \sqrt 2 = 1$$
So, the correct answer is b)
Alternate Approach :
Area of triangle = $$\frac{1}{2}\times\ base\times\ height$$
So we get $$\frac{1}{2}\times2\times\ 1$$=1 square units

The line should be parallel to AD and should be of equal distance from the origin in the third quadrant. The equation x+y = -1 satisfies all these conditions.

For points to be coincident, value of determinant should not be equal zero, so that they have a unique value of system.

Here value of determinant is not equal to zero, simultaneously not any two lines are parallel or perpendicular.

So system has a unique value

Hence points are coincident.

Distance between two points = $$\sqrt {(3-(-2))^2 + (8-(-7))^2}$$ = $$5\sqrt10$$