CAT Coordinate Geometry Questions & Solutions

A sample of real CAT Coordinate Geometry past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 17 Coordinate Geometry questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   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

   • $\dfrac{27}{7}$

   • $\dfrac{10}{3}$

   • $\dfrac{13}{4}$

   • $\dfrac{29}{9}$

   Show solution

   Given (P(-3,-2)), (Q(1,-5)), and (R(9,1)).

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

   Diagonal SQ passes through S(5,4) and Q(1,-5).

   Slope m = $\dfrac{-5-4}{1-5} = \dfrac{9}{4}$
   

   Equation of SQ is   $y - 4 = \dfrac{9}{4}(x - 5)$
   

   At the x-axis, y = 0: $-4 = \dfrac{9}{4}(a - 5)$

   $-16 = 9(a - 5)$

   $9a = 29$

   $a = \dfrac{29}{9}$
   

   So, $a = {\dfrac{29}{9}}$

2. Q2.CAT 2024

   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

   Answer: 2

   Show solution

   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. 

3. Q3.CAT 2023

   Let C be the circle $x^{2} + y^{2} + 4x - 6y - 3 = 0$ 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

   • (6, 6)

   • (6, 3)

   • (6, 8)

   • (6, 4)

   Show solution

   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)

4. Q4.CAT 2020

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

   • 12

   • 13

   • 15

   • 14

   Show solution

   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

5. Q5.CAT 2020

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

   • $\frac{14\pi}{3}$

   • $\frac{123\pi}{7}$

   • $\frac{12\pi}{5}$

   • $\frac{205\pi}{9}$

   Show solution

   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 =  $\pi\ r^2=\frac{205\pi\ }{9}$

6. Q6.CAT 2018

   Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be

   • $188\sqrt{3}$

   • $248\sqrt{3}$

   • $164\sqrt{3}$

   • $192\sqrt{3}$

   Show solution

   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. 

7. Q7.CAT 2018

   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

   • $8$ units

   • $4$ units

   • $2\sqrt{2}$ units

   • $4\sqrt{2}$ units

   Show solution

   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. 

8. Q8.CAT 2017

   The area of the closed region bounded by the equation 

   I x I + I y I = 2 in the two-dimensional plane is

   • $4\pi$ sq. units

   • 4 sq. units

   • 8 sq. units

   • $2\pi$ sq. units

   Show solution

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

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

   
9. Q9.CAT 2017

   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

   • -5

   • -6

   • -7

   • -8

   Show solution

   

   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

10. Q10.CAT 2005

    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

    • 780

    • 800

    • 820

    • 741

    Show solution

    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.

11. Q11.CAT 2000

    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?

    • x+y=-1

    • x-y=-1

    • x+y=1

    • None of the above

    Show solution

    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.

12. Q12.CAT 1996

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

    • form a triangle

    • are on lines perpendicular to each other

    • are on lines parallel to each other

    • are coincident

    Show solution

    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.

CAT Coordinate Geometry previous year questions with solutions — AthenaPrep