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CBSE Class 10 Coordinate Geometry — practice questions

Explore 50 free Coordinate Geometry questions in CBSE Class 10 Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Coordinate Geometry questions and exercises with answers

Showing questions 1–30 of 50.

If $A(6,1)$, $B(8,2)$, $C(9,4)$ and $D(p,3)$ are the vertices of a parallelogram taken in order, what is the value of $p$?Read the answer and worked solution →The midpoint of the line segment joining $(-6,10)$ and $(3,-8)$ isRead the answer and worked solution →The point dividing the line segment joining $(4,-3)$ and $(8,5)$ internally in the ratio $3:1$ isRead the answer and worked solution →A point on the $y$-axis is equidistant from $A(6,5)$ and $B(-4,3)$. What is that point?Read the answer and worked solution →If a point $P(x,y)$ is equidistant from $(7,1)$ and $(3,5)$, which relation must $x$ and $y$ satisfy?Read the answer and worked solution →The points $(3,2)$, $(-2,-3)$ and $(2,3)$ form which type of triangle?Read the answer and worked solution →The distance between two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ in the coordinate plane isRead the answer and worked solution →Two circles have radii in the ratio 2:3. The ratio of their CIRCUMFERENCES is:Read the answer and worked solution →CLASS10 · Mathematics · Section FormulaRead the answer and worked solution →SAT · Mathematics · Slope & InterceptRead the answer and worked solution →SAT · Mathematics · Distance Formula — Gap FillRead the answer and worked solution →Match the number type to an exampleRead the answer and worked solution →Match each logarithm to its valueRead the answer and worked solution →Judge these points and axes claimsRead the answer and worked solution →Sort points into their quadrantsRead the answer and worked solution →If a point is reflected in the x-axis, the new co-ordinates of (a, b) becomeRead the answer and worked solution →The distance of point (5, 0) from the origin isRead the answer and worked solution →The mid-point of the segment joining (-2, -3) and (4, 5) isRead the answer and worked solution →A triangle is formed by the vertices (4,1), (1,1) and (3,5). The triangle is:Read the answer and worked solution →The area of the triangle whose vertices are (1, 2), (-4, -3) and (4, 1) is:Read the answer and worked solution →Three vertices of a parallelogram are (0, 0), (3, 5) and (6, 2). The fourth vertex equals:Read the answer and worked solution →The fourth vertex of a parallelogram with three given vertices (1, 3), (5, 4), (2, 5) is (taking the missing vertex opposite (1, 3)):Read the answer and worked solution →The midpoint of the line segment joining (2, 3) and (8, 5) has coordinates:Read the answer and worked solution →A circle is drawn with the segment joining (2, 3) and (6, 5) as a diameter. The centre of the circle has coordinates:Read the answer and worked solution →Two opposite vertices of a parallelogram are at (4, 5) and (1, 3). The intersection of its diagonals lies at:Read the answer and worked solution →The point dividing the segment joining (2, 3) and (8, 6) in the ratio 1 : 2 (counting from the first point) has coordinates:Read the answer and worked solution →The distance of a point from the y-axis is called its:Read the answer and worked solution →The distance of a point from the x-axis is called its:Read the answer and worked solution →A point on the x-axis has coordinates of the form:Read the answer and worked solution →The graph of ax + by + c = 0, with a and b not both zero, is a:Read the answer and worked solution →

Worked practice questions

If $A(6,1)$, $B(8,2)$, $C(9,4)$ and $D(p,3)$ are the vertices of a parallelogram taken in order, what is the value of $p$?

  1. 5
  2. 6
  3. 7
  4. 8
Show answer and solution

Correct answer: C. 7

Principle: diagonals of a parallelogram bisect each other. 1. Midpoint of $AC$ is $\left(\dfrac{6+9}{2},\dfrac{1+4}{2}\right)=\left(\dfrac{15}{2},\dfrac{5}{2}\right)$. 2. Midpoint of $BD$ is $\left(\dfrac{8+p}{2},\dfrac{2+3}{2}\right)=\left(\dfrac{8+p}{2},\dfrac{5}{2}\right)$. 3. Since these midpoints are equal, $\dfrac{15}{2}=\dfrac{8+p}{2}$, so $15=8+p$ and hence $p=7$. Option D is tempting if one adds the last coordinates incorrectly, while A and B do not make the diagonal midpoints match.

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The midpoint of the line segment joining $(-6,10)$ and $(3,-8)$ is

  1. $\left(-\dfrac{3}{2},1\right)$
  2. $\left(-\dfrac{3}{2},2\right)$
  3. $( -3,1)$
  4. $\left(\dfrac{3}{2},1\right)$
Show answer and solution

Correct answer: A. $\left(-\dfrac{3}{2},1\right)$

The midpoint formula gives $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$. **So the midpoint is $\left(\dfrac{-6+3}{2},\dfrac{10+(-8)}{2}\right)=\left(-\dfrac{3}{2},1\right)$.**

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The point dividing the line segment joining $(4,-3)$ and $(8,5)$ internally in the ratio $3:1$ is

  1. $(7,3)$
  2. $(6,1)$
  3. $(5,-1)$
  4. $(8,3)$
Show answer and solution

Correct answer: A. $(7,3)$

Using the section formula, $x=\dfrac{3\cdot 8+1\cdot 4}{3+1}=7$ and $y=\dfrac{3\cdot 5+1\cdot(-3)}{3+1}=3$. **Hence the required point is $(7,3)$.**

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More Mathematics topics

Introduction to TrigonometryMensuration: Surface Areas & VolumesReal Numbers and PolynomialsArithmetic ProgressionStatistics and ProbabilityLinear Equations in Two VariablesQuadratic EquationsTriangles and Circles