Class 10 Maths · Chapter 7

Coordinate Geometry

440 Class 10 Maths questions on Coordinate Geometry from the NCERT textbook, NCERT Exemplar and CBSE board papers 2015–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Maths previous year papers, with answers →

Questions · page 21 of 22

1. Q401 CBSE 2017
   If the ​points $A(k+1,2k)$, $B(3k,2k+3)$ and $C(5k-1,5k)$ are ​collinear, ​then find ​the ​value ​of $k$.
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   The points are collinear for $k=\frac{1}{2}$ or $k=2$.

   Step-by-step solution →
2. Q402 CBSE 2017
   If $a\ne b\ne 0$, prove ​that the ​points $(a,a^2),(b,b^2),(0,0)$ will ​not ​be collinear.
   Show answer

   The area of the triangle is $\frac{1}{2}|ab(b-a)|$, which is nonzero since $a\ne 0$, $b\ne 0$ and $a\ne b$. Hence the points $(a,a^2)$, $(b,b^2)$, $(0,0)$ are not collinear.

   Step-by-step solution →
3. Q403 CBSE 2017
   If the points $A(k+1,2k)$, $B(3k,2k+3)$ ​and $C(5k-1,5k)$ ​are collinear, ​then find the ​value ​of $k$.
   Show answer

   The points are collinear for $k=2$ or $k=\frac{1}{2}$.

   Step-by-step solution →
4. Q404 CBSE 2017
   If $A(-7,5)$, $B(-6,-7)$, $C(-3,-8)$ ​and $D(2,3)$ ​are the vertices ​of a quadrilateral $ABCD$, ​then find the area of ​the quadrilateral.
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   Area of quadrilateral $ABCD=72$ square units.

   Step-by-step solution →
5. Q405 CBSE 2016
   Find the ​points on $x$-axis ​which ​are at a distance $2\sqrt{5}$ ​from ​the ​point $(7,-4)$.
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   The required points are $(9,0)$ and $(5,0)$.

   Step-by-step solution →
6. Q406 CBSE 2016
   Find ​that ​value of $k$, ​for ​which ​the ​points $(8,1)$, $(3,k)$ ​and $(2,-5)$ ​are collinear.
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   $k=-4$.

   Step-by-step solution →
7. Q407 CBSE 2016
   Let $P$ and $Q$ be the points of trisection of ​the ​line ​segment joining the points $A(2,-2)$ and $B(-7,4)$ such ​that $P$ ​is nearer to $A$. Find ​the coordinates ​of $P$ and $Q$.
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   $P=(-1,0)$ and $Q=(-4,2)$.

   Step-by-step solution →
8. Q408 CBSE 2016
   Find ​the ​ratio in which the point $(-3,k)$ divides the line-segment joining the ​points $(-5,-4)$ and $(-2,3)$. ​Also find the ​value ​of $k$.
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   The point divides the segment in the ratio $2:1$ and $k=\frac{2}{3}$.

   Step-by-step solution →
9. Q409 CBSE 2016
   Find the ratio in which ​the $y$-axis ​divides the line ​segment joining the ​points $A(5,-6)$ and $B(-1,-4)$. Also ​find the coordinates of the ​point of division.
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   The $y$-axis divides $AB$ in the ratio $5:1$, and the point of division is $\left(0,-\frac{13}{3}\right)$.

   Step-by-step solution →
10. Q410 CBSE 2016
    Prove ​that ​the ​points $(3,0)$, $(6,4)$ ​and $(-1,3)$ are the vertices ​of a right ​angled isosceles triangle.
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    $AB=CA=5$ (isosceles) and $A{B}^2+C{A}^2=B{C}^2=50$ (right angle at $A$), so the triangle is right angled isosceles.

    Step-by-step solution →
11. Q411 CBSE 2016
    Let $P$ and $Q$ be the points of trisection ​of ​the line ​segment joining the points $A(2,-2)$ ​and $B(-7,4)$ such that $P$ ​is ​nearer to $A$. ​Find the coordinates of $P$ ​and $Q$.
    Show answer

    $P(-1,0)$ and $Q(-4,2)$.

    Step-by-step solution →
12. Q412 CBSE 2016
    Prove ​that ​the points $(2,-2)$, $(-2,1)$ ​and $(5,2)$ ​are ​the ​vertices of a right angled triangle. Also find the ​area of this triangle.
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    The points form a right-angled triangle with the right angle at $A(2,-2)$, and its area is $12.5$ square units.

    Step-by-step solution →
13. Q413 CBSE 2016
    The $x$-coordinate ​of ​a ​point $P$ ​is twice ​its $y$-coordinate. If $P$ is equidistant ​from $Q(2,-5)$ ​and $R(-3,6)$, find the coordinates ​of $P$.
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    The point is $P=(16,8)$.

    Step-by-step solution →
14. Q414 CBSE 2016
    If the ​point $P(x,y)$ ​is ​equidistant from ​the points $A(a+b,b-a)$ and $B(a-b,a+b)$, prove ​that $bx=ay$.
    Show answer

    Expanding $P{A}^2=P{B}^2$ and cancelling reduces to $-4bx+4ay=0$, i.e. $bx=ay$, as required.

    Step-by-step solution →
15. Q415 CBSE 2016
    If the point $P(x,y)$ ​is equidistant ​from ​the ​points $A(a+b,b-a)$ ​and $B(a-b,a+b)$. ​Prove that $bx=ay$.
    Show answer

    Hence $bx=ay$, as required (proved).

    Step-by-step solution →
16. Q416 CBSE 2016
    If ​the point $C(-1,2)$ divides internally the line-segment ​joining the points $A(2,5)$ and $B(x,y)$ in the ratio $3:4$, ​find ​the ​value ​of $x^2+y^2$.
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    $B=(-5,-2)$, so $x^2+y^2=29$.

    Step-by-step solution →
17. Q417 CBSE 2016
    In Fig. 6, $ABC$ is a triangle coordinates of ​whose ​vertex $A$ are $(0,-1)$. $D$ and $E$ ​respectively are ​the mid-points ​of the sides $AB$ and $AC$ and their coordinates are $(1,0)$ and $(0,1)$ respectively. If $F$ ​is the mid-point of $BC$, find the ​areas of $△ABC$ and $△DEF$.

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    $\text{Area of }△ABC=4$ square units and $\text{Area of }△DEF=1$ square unit.

    Step-by-step solution →
18. Q418 CBSE 2016
    If the ​point $P(x,y)$ ​is equidistant from ​the points $A(a+b,b-a)$ ​and $B(a-b,a+b)$. ​Prove that $bx=ay$.
    Show answer

    Starting from $PA=PB$ and simplifying gives $bx=ay$, as required.

    Step-by-step solution →
19. Q419 CBSE 2016
    Find ​the area of $△ABC$ whose two vertices $A$ ​and $B$ ​have ​coordinates $(4,7)$ and $(5,2)$ respectively ​and $D(6,-1)$ is the midpoint ​of $BC$.
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    The area of $△ABC$ is $2$ square units.

    Step-by-step solution →
20. Q420 CBSE 2016
    In ​fig. 7, ​the vertices ​of $△ABC$ are $A(4,6)$, $B(1,5)$ and $C(7,2)$. A line-segment $DE$ is drawn to intersect the ​sides $AB$ ​and $AC$ at $D$ and $E$ respectively such that $\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{3}$. Calculate ​the area of $△ADE$ and compare ​it ​with area of $△ABC$.

    Show answer

    Area of $△ADE=\frac{5}{6}$ sq units, and $\frac{[ADE]}{[ABC]}=\frac{1}{9}$ (since $[ABC]=\frac{15}{2}$ sq units).

    Step-by-step solution →