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 22 of 22

1. Q421 CBSE 2016
   In ​Fig. 8, 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

   $[ADE]=\frac{5}{6}$ sq units, $[ABC]=\frac{15}{2}$ sq units, and $[ADE]:[ABC]=1:9$.

   Step-by-step solution →
2. Q422 CBSE 2016
   The ​line segment ​joining the ​points $A(3,2)$ and $B(5,1)$ ​is ​divided ​at the point $P$ in the ​ratio $1:2$ and $P$ ​also lies on the line $3x-18y+k=0$. Find the value of $k$.
   Show answer

   The value of $k$ is $19$.

   Step-by-step solution →
3. Q423 CBSE 2016
   Prove that the ​area of a triangle with ​vertices $(t,t-2)$, $(t+2,t+2)$ ​and $(t+3,t)$ ​is ​independent ​of $t$.
   Show answer

   The area is $4$ square units, with no $t$ remaining, so it is independent of $t$. Hence proved.

   Step-by-step solution →
4. Q424 CBSE 2016
   The ​co-ordinates of the points $A$, $B$ ​and $C$ ​are $(6,3)$, $(-3,5)$ ​and $(4,-2)$ respectively. $P(x,y)$ is any ​point in ​the plane. Show that $\frac{\text{ar}(△PBC)}{\text{ar}(△ABC)}=\left|\frac{x+y-2}{7}\right|$.
   Show answer

   Since $\text{ar}(△PBC)=\frac{7}{2}|x+y-2|$ and $\text{ar}(△ABC)=\frac{49}{2}$, their ratio is $\left|\frac{x+y-2}{7}\right|$, as required.

   Step-by-step solution →
5. Q425 CBSE 2016
   In fig. 8, ​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

   Ar$(△ADE)=\frac{5}{6}$ sq units and ar$(△ADE):$ar$(△ABC)=1:9$.

   Step-by-step solution →
6. Q426 CBSE 2015
   If the point $R(5,7)$ is ​the ​mid-point of ​the line-segment joining the ​points $P(3,y)$ ​and $Q(x,9)$, find the ​value ​of $x+y$.
   Show answer

   $x+y=12$

   Step-by-step solution →
7. Q427 CBSE 2015
   Show that ​the points $(a,a)$, $(-a,-a)$ ​and $(-\sqrt{3}a,\sqrt{3}a)$ ​are ​the vertices of an ​equilateral ​triangle.
   Show answer

   Since $A{B}^2=B{C}^2=C{A}^2=8a^2$, the three sides are equal ($AB=BC=CA=2\sqrt{2}|a|$), so the points are the vertices of an equilateral triangle.

   Step-by-step solution →
8. Q428 CBSE 2015
   If $A(5,2)$, $B(2,-2)$ ​and $C(-2,t)$ ​are ​the vertices of a ​right angled ​triangle with $\angle B=90^{\circ }$, then find ​the value of $t$.
   Show answer

   $t=1$.

   Step-by-step solution →
9. Q429 CBSE 2015
   The points $A(4,7)$, $B(p,3)$ and $C(7,3)$ ​are the ​vertices ​of a ​right triangle, right-angled ​at $B$. Find ​the value of $p$.
   Show answer

   $p=4$ (the value $p=7$ is rejected as it makes $B$ coincide with $C$).

   Step-by-step solution →
10. Q430 CBSE 2015
    For ​what values of $k$ are ​the ​points $(8,1)$, $(3,-2k)$ ​and $(k,-5)$ ​collinear?
    Show answer

    $k=2$ or $k=\frac{11}{2}$

    Step-by-step solution →
11. Q431 CBSE 2015
    Find ​the relation between $x$ and $y$ if the ​points $A(x,y)$, $B(-5,7)$ ​and $C(-4,5)$ are ​collinear.
    Show answer

    The required relation is $2x+y+3=0$.

    Step-by-step solution →
12. Q432 CBSE 2015
    Point $A$ lies on the line segment $PQ$ joining $P(6,-6)$ and $Q(-4,-1)$ in such a way ​that $\frac{PA}{PQ}=\frac{2}{5}$. If point $A$ ​also lies ​on ​the line $3x+k(y+1)=0$, find ​the ​value of $k$.
    Show answer

    $k=2$.

    Step-by-step solution →
13. Q433 CBSE 2015
    Find the ​area of ​the ​triangle $ABC$ with $A(1,-4)$ and mid-points of ​sides through $A$ being $(2,-1)$ ​and $(0,-1)$.
    Show answer

    Area of $△ABC=12$ square units.

    Step-by-step solution →
14. Q434 CBSE 2015
    If ​the coordinates ​of points $A$ and $B$ are $(-2,-2)$ ​and $(2,-4)$ respectively, ​find ​the ​coordinates of $P$ such that $AP=\frac{3}{7}AB$, ​where $P$ lies on the line segment $AB$.
    Show answer

    $P=\left(-\frac{2}{7},-\frac{20}{7}\right)$.

    Step-by-step solution →
15. Q435 CBSE 2015
    Find ​the point ​on ​the $x$-axis which is ​equidistant from $(1,-6)$ ​and $(-1,8)$.
    Show answer

    The required point is $(-7,0)$.

    Step-by-step solution →
16. Q436 CBSE 2015
    Find the ​ratio ​in which the line segment joining the ​points $(-5,8)$ and $(4,-10)$ is divided ​by the ​point $(-3,y)$. Also ​find the value ​of $y$.

    Show answer

    The point divides the segment in the ratio $2:7$ and $y=4$.

    Step-by-step solution →
17. Q437 CBSE 2015
    Find ​the ​values of $k$ so ​that the ​area of the triangle ​with vertices $(1,-1)$, $(-4,2k)$ and $(-k,-5)$ is $24$ ​sq. units.
    Show answer

    $k=3$ or $k=-\frac{9}{2}$.

    Step-by-step solution →
18. Q438 CBSE 2015
    The base $BC$ ​of an equilateral ​triangle $ABC$ ​lies on the $y$-axis. The ​coordinates of point $C$ are $(0,-3)$. The origin is the mid-point of the base. Find the coordinates of ​the points $A$ and $B$. ​Also find the coordinates ​of another point $D$ ​such that $BACD$ is a rhombus.
    Show answer

    $A(3\sqrt{3},0)$, $B(0,3)$, and $D(-3\sqrt{3},0)$.

    Step-by-step solution →
19. Q439 CBSE 2015
    If $A(-4,8)$, $B(-3,-4)$, $C(0,-5)$ ​and $D(5,6)$ ​are ​the ​vertices ​of a quadrilateral $ABCD$, ​find ​its area.
    Show answer

    Area of quadrilateral $ABCD=72$ square units.

    Step-by-step solution →
20. Q440 CBSE 2015
    Find ​the ​area of ​the quadrilateral $ABCD$ whose vertices ​are $A(-3,9)$, $B(-2,-3)$, $C(1,-4)$ ​and $D(6,7)$.

    Show answer

    Area of quadrilateral $ABCD=72$ square units

    Step-by-step solution →