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Coordinate Geometry questions for IPMAT, with solutions

10 free Coordinate Geometry practice questions for IPMAT Indore, Rohtak and JIPMAT, 5 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Coordinate Geometry on Learn.

1. Question 1Multiple choiceWritten by IPMATwayCircles in the Plane

   The equation of the circle with centre $(-1, 2)$ and radius $3$ is

   1. A.$x^2+y^2-2x+4y-4=0$
   2. B.$x^2+y^2+2x-4y-9=0$
   3. C.$x^2+y^2+2x-4y+4=0$
   4. D.$x^2+y^2+2x-4y-4=0$
   Show the answer and solution

   Answer: D. $x^2+y^2+2x-4y-4=0$

   A circle with centre $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. Here $(x+1)^2+(y-2)^2=9$. Expanding: $x^2+2x+1+y^2-4y+4=9$, i.e. $x^2+y^2+2x-4y-4=0$.

2. Question 2Multiple choiceWritten by IPMATwayStraight Lines

   The line through $(2, 5)$ and $(4, 9)$ meets the $y$-axis at

   1. A.$(0, 9)$
   2. B.$(0, 3)$
   3. C.$(0, 1)$
   4. D.$(0, -1)$
   Show the answer and solution

   Answer: C. $(0, 1)$

   Slope $m=\frac{9-5}{4-2}=\frac{4}{2}=2$. Using the point $(2,5)$: $y-5=2(x-2)$, i.e. $y=2x+1$. The line meets the $y$-axis where $x=0$: $y=1$, the point $(0,1)$.

3. Question 3Multiple choiceAsked in IPMAT Indore 2019Regions, Areas and Loci

   The area enclosed by the curve $2|x|+3|y|=6$ is

   1. A.$12$ square units
   2. B.$3$ square units
   3. C.$4$ square units
   4. D.$24$ square units
   Show the answer and solution

   Answer: A. $12$ square units

   Because of $|x|$ and $|y|$ the curve is symmetric in both axes. In the first quadrant it is the segment $2x+3y=6$ from $(3,0)$ to $(0,2)$. So the curve is a rhombus with vertices $(\pm3,0)$ and $(0,\pm2)$, diagonals $6$ and $4$. Area $=\frac{1}{2}\cdot 6\cdot 4=12$ square units.

4. Question 4Type the answerWritten by IPMATwayRegions, Areas and Loci

   The area of the region $|x-2|+|y|\le 4$ is

   Show the answer and solution

   Answer: 32

   Put $X=x-2$, $Y=y$; the shift does not change area. $|X|+|Y|\le 4$ is a square with vertices $(\pm4,0)$, $(0,\pm4)$ in the new coordinates, diagonals $8$ and $8$. Area $=\frac{1}{2}\cdot 8\cdot 8=32$.

5. Question 5Multiple choiceAsked in IPMAT Indore 2024Circles in the Plane

   If the shortest distance of a given point to a given circle is $4$ cm and the longest distance is $9$ cm, then the radius of the circle is

   1. A.$2.5$ cm or $6.5$ cm.
   2. B.$6.5$ cm
   3. C.$5$ cm or $13$ cm
   4. D.$2.5$ cm
   Show the answer and solution

   Answer: A. $2.5$ cm or $6.5$ cm.

   The nearest and farthest points lie on the line through the point and the centre. If the point is outside, the two distances are $d-r$ and $d+r$, giving $r=\frac{9-4}{2}=2.5$. If it is inside, they are $r-d$ and $r+d$, giving $r=\frac{9+4}{2}=6.5$. Both are possible.

6. Question 6Multiple choiceWritten by IPMATwayCircles in the Plane

   The circle concentric with $x^2+y^2-6x+8y+9=0$ and passing through $(-1, 1)$ is

   1. A.$x^2+y^2-6x+8y-7=0$
   2. B.$x^2+y^2-6x+8y-16=0$
   3. C.$x^2+y^2+6x-8y-16=0$
   4. D.$x^2+y^2-6x+8y+16=0$
   Show the answer and solution

   Answer: B. $x^2+y^2-6x+8y-16=0$

   Concentric circles have the same centre, so the required circle is $x^2+y^2-6x+8y+c=0$ with only $c$ changed. Substituting $(-1,1)$: $1+1+6+8+c=0$, so $c=-16$. The circle is $x^2+y^2-6x+8y-16=0$.

7. Question 7Multiple choiceAsked in IPMAT Indore 2021Straight Lines

   The x-intercept of the line that passes through the intersection of the lines $x+2y=4$ and $2x+3y=6$, and is perpendicular to the line $3x-y=2$ is

   1. A.$2$
   2. B.$0.5$
   3. C.$4$
   4. D.$6$
   Show the answer and solution

   Answer: D. $6$

   Solve $x+2y=4$, $2x+3y=6$: $y=2$, $x=0$. The line $3x-y=2$ has slope $3$, so the required slope is $-\frac{1}{3}$: $y-2=-\frac{1}{3}x\Rightarrow x+3y=6$. At $y=0$, $x=6$.

8. Question 8Multiple choiceAsked in IPMAT Indore 2025Straight Lines

   Let $A(1, 3)$ and $B(5, 1)$ be two points. If a line with slope $m$ intersects $AB$ at an angle $45^\circ$, then the possible values of $m$ are:

   1. A.$5, -\frac{1}{5}$
   2. B.$-3, \frac{1}{3}$
   3. C.$7, \frac{1}{7}$
   4. D.$3, \frac{1}{3}$
   Show the answer and solution

   Answer: B. $-3, \frac{1}{3}$

   Slope of $AB=\frac{1-3}{5-1}=-\frac{1}{2}$. Angle $45^\circ$ means $\left|\frac{m-m_{AB}}{1+m\,m_{AB}}\right|=\tan 45^\circ=1$. The two cases give $m=\frac{1}{3}$ and $m=-3$ (perpendicular to each other, as the two $45^\circ$ lines must be).

9. Question 9Multiple choiceAsked in IPMAT Indore 2019Regions, Areas and Loci

   The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $(0, 0)$, $(0, 31)$, and $(31, 0)$ is

   1. A.$435$
   2. B.$465$
   3. C.$450$
   4. D.$464$
   Show the answer and solution

   Answer: A. $435$

   A point is strictly inside when $x\ge1$, $y\ge1$ and $x+y<31$. For $x=1$, $y$ runs $1..29$ ($29$ values); for $x=2$, $28$ values; ... ; for $x=29$, $1$ value. Sum $=435$. (Pick: area $480.5$, boundary $93$ points, $I=480.5-46.5+1=435$.)

10. Question 10Multiple choiceWritten by IPMATwayCircles in the Plane

    Two circles in the first quadrant touch both coordinate axes and pass through $(9,2)$. The distance between their centres is

    1. A.$12$
    2. B.$22$
    3. C.$17\sqrt2$
    4. D.$12\sqrt2$
    Show the answer and solution

    Answer: D. $12\sqrt2$

    A circle touching both axes in the first quadrant has centre $(r,r)$ and radius $r$. Passing through $(9,2)$: $(9-r)^2+(2-r)^2=r^2\Rightarrow81-18r+r^2+4-4r+r^2=r^2\Rightarrow r^2-22r+85=0\Rightarrow(r-5)(r-17)=0$. Centres $(5,5)$ and $(17,17)$ lie on the line $y=x$, at distance $\sqrt{12^2+12^2}=12\sqrt2$.