Upon drawing a rough sketch of the coordinates given, we realise that this is a right-angled triangle.<img 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"> The three side lengths are 6, 8 and 10 unitsThe inradius of a circle can be calculated using the formula:&nbsp;$rs=Area$, where s is the semi-perimeter of the triangle&nbsp;The Area would be&nbsp;$\frac{1}{2}\times\ 6\times\ 8=24$and the semi-perimeter would be&nbsp;$\frac{10+6+8}{2}=12$Giving the inradius to be&nbsp;$\frac{24}{12}=2$ unitsTherefore, 2 is the correct answer.&nbsp;

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)=&gt; $ \sin(30) = \frac{4}{\sqrt{( h+2)^2+(k-3)^2}} $Squaring on both sides:$ \frac{1}{4} = \frac{16}{(h+2)^2+(k-3)^2} $=&gt; $ (h+2)^2+(k-3)^2 = 64 $When x = 6 =&gt; h = 6 =&gt; $ 64 + (k-3)^2 = 64 $ =&gt; k = 3.=&gt; 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 =&nbsp;($\frac{1-2}{2}, \frac{1+8}{2}$)Let the coordinates of vertex D be (x, y)($\frac{x+3}{2}, \frac{y+4}{2}$) = ($\frac{1-2}{2}, \frac{1+8}{2}$)x = -4 and y = 5The answer is option D.

Equation of circle $x^2 + y^2 + 2gx + 2fy + c = 0 $It passes through $ (0,0),(4,0),(3,9). $$ \Rightarrow \text{Substitute } (0,0):\ c = 0 $$ \Rightarrow \text{Substitute } (4,0):\ 16 + 0 + 8g + 0 = 0 \ ;\ g = -2 $$ \Rightarrow \text{Substitute } (3,9):\ 9 + 81 - 12 + 18f = 0 \ ;\ f = -13/3 $$ \text{Radius of circle } r = \sqrt{g^2 + f^2 - c} $$ \Rightarrow r^2 = \frac{205}{9} $Therefore, Area = $\pi r^2 = \frac{205\pi}{9} $

The midpoints of two diagonals of a parallelogram are the sameHence the midpoint of (2,1) and (-3,-4) lie on x + 9y + c = 0midpoint of (2,1) and (-3,-4) = $\left( \frac{2-3}{2}, \frac{1-4}{2} \right) = \left( -\frac12, -\frac32 \right) $Keeping this point in the above line equation, we get c = 14

<img src="https://lightyellow-stork-291267.hostingersite.com/wp-content/uploads/2025/11/34.CATQuant.jpg">In any right triangle, the circumradius is half of the hypotenuse.Here,&nbsp;&nbsp;$L = \frac{1}{2} \times \text{(length of the hypotenuse)}$$L = \frac{1}{2}\left(\sqrt{15^{2} + 9^{2}}\right)$= $\frac{1}{2}\sqrt{306}$= $\frac{1}{2} \times 17.49$= $8.74$Hence, the integer close to $L$ is $9$.

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$_{1}$. <figure><img class="img-responsive" data-image="blob" src="https://cracku.in/media/uploads/blob_cEQDD2I"></figure> 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&nbsp; Sum of the area of all triangles =&nbsp;&nbsp;T$_{1}$ +&nbsp;T$_{2}$ + T$_{3}$ + ...&nbsp; $\Rightarrow$&nbsp;T$_{1}$ +&nbsp;T$_{1}$/$4$ +&nbsp;T$_{1}$/$4^2$ + ...&nbsp; $\Rightarrow$ $\dfrac{T_{1}}{1 - 0.25}$ $\Rightarrow$ $\dfrac{4}{3}*T_{1}$ $\Rightarrow$ $\dfrac{4}{3}*144\sqrt{3}$ $\Rightarrow$ $192\sqrt{3}$ Hence, option D is the correct answer.&nbsp;

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.&nbsp; <figure><img class="img-responsive" data-image="blob" src="https://cracku.in/media/uploads/blob_0bQpIFH"></figure> 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.&nbsp;

The following equation will form a square of side $2\sqrt{2}$. The area of the square = $(2\sqrt{2})^2$ = 8 units. <img alt="2" class="img-responsive" data-image="2.png" src="https://cracku.in/media/uploads/2_GFmFx4i.png">

The graph of the given function is as shown below. <figure><img class="img-responsive" data-image="6.PNG" height="220" src="https://cracku.in/media/uploads/6_EoGfon7.PNG" width="246"></figure> We can see that the shortest distance of the point (1/2, 1) will be 1 unit.

<figure><img class="img-responsive" data-image="Screenshot_10.png" src="https://cracku.in/media/uploads/Screenshot_10_cwTKby1.png"></figure> 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 =&gt; c = -8

The number of points on x = 1 is 39. The number of points on x = 2 is 38&nbsp;and so on till x = 39, which has one point. So, the total is 1+2+3+...+39&nbsp;= $\frac{39*40}{2}$ = 780.

The triangle we have is : <img alt="image" class="img-responsive" data-image="image.png" src="https://cracku.in/media/uploads/image_N5kgqHD.png"> 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 =&nbsp;$\frac{1}{2}\times\ base\times\ height$ So we get&nbsp;$\frac{1}{2}\times2\times\ 1$=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 = $\sqrt {(3-(-2))^2 + (8-(-7))^2}$ = $5\sqrt10$

CAT Coordinate Geometry

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

Question 16.

Instructions

Question 17.

Free CAT Quant Questions

CAT Coordinate Geometry

Question 1.

<div><p>The coordinates of the three vertices of a triangle are: (1, 2), (7, 2), and (1, 10). Then the radius of the incircle of the triangle is</p></div>

Question 2.

<div><p>Let C be the circle $x^2+y^2+4 x-6 y-3=0$ and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to $60^{\circ}$. Then, the point at which L touches the line x=6 is</p></div>

Question 3.

<div><p>Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (&minus;2, 8), respectively. Then, the coordinates of the vertex D are</p></div>

Question 4.

<div><p>The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is</p></div>

Question 5.

<div><p>The points (2 , 1) and (-3 , -4) are opposite vertices of a parellelogram. If the other two vertices lie on the line x + 9y + c = 0, then c is</p></div>

Question 6.

<div><p>Let T be the triangle formed by the straight line 3x + 5y - 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is</p></div>

Question 7.

<div><p>The $(x, y)$ coordinates of vertices $P$, $Q$ and $R$ of a parallelogram $PQRS$ are $(-3, -2)$, $(1, -5)$ and $(9, 1)$, respectively. If the diagonal $SQ$ intersects the $x$-axis at $(a, 0)$, then the value of $a$ is:</p></div>

Question 8.

Question 9.

<div>A triangle ABC has area 32 sq units and its side BC, of length 8 units, lies on the line x = 4. Then the shortest possible distance between A and the point (0,0) is</div>

Question 10.

<div>The area of the closed region bounded by the equation&nbsp;<br />I x I + I y I = 2 in the two-dimensional plane is</div>

Question 11.

<div>The shortest distance of the point $(\frac{1}{2},1)$ from the curve y = I x -1I + I x + 1I is</div>

Question 12.

<div>The points (2, 5) and (6, 3) are two end points of a diagonal of a rectangle. If the other diagonal has the equation y =3x+c,then c is</div>

Question 13.

<div>Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X,Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is</div>

Question 14.

<div>The area of the triangle whose vertices are (a,a), (a + 1, a + 1) and (a + 2, a) is</div>

Question 15.

<div>ABCD is a rhombus with the diagonals AC and BD intersection at the origin on the x-y plane. The equation of the straight line AD is x + y = 1. What is the equation of BC?</div>

Question 16.

<div>The points of intersection of three lines $2x+3y-5=0, 5x-7y+2=0$ and $9x-5y-4=0$</div>

Instructions

<div>Instructions <br />For the following questions answer them individually</div>

Question 17.

<div>What is the distance between the points A(3, 8) and B(-2,-7)?</div>

Free CAT Quant Questions