NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 7: Coordinate Geometry

Exercise 7.1

Step 1: A point in a plane is written as \((x,y)\).

Here, the given point is \(P(2,3)\), so \(x = 2\) and \(y = 3\).

Step 2: The x-axis is the horizontal axis. The distance of a point from the x-axis depends only on its y-coordinate.

Step 3: Formula: Distance from x-axis = \(|y|\).

Step 4: For \(P(2,3)\), \(y = 3\). So distance = \(|3| = 3\).

Step 5: Therefore, the distance of the point from the x-axis is 3 units. If we measure in SI, we say it is 3 metres (m).

Step 1: Write the coordinates of the two points.

Point A is \((0, 6)\) and Point B is \((0, -2)\).

Step 2: Notice that the x-coordinates of both points are the same (0 and 0). This means both points lie on the y-axis. So, the distance is simply the difference in their y-coordinates.

Step 3: Apply the distance formula for vertical points: \[ \text{Distance} = |y_2 - y_1| \]

Step 4: Substitute the values: \[ |y_2 - y_1| = |-2 - 6| \]

Step 5: Simplify inside the modulus: \[ |-2 - 6| = |-8| \]

Step 6: The modulus of \(-8\) is 8.

Final Answer: The distance between the two points is 8 units.

We want to find the distance of the point \(P(-6,8)\) from the origin \((0,0)\).

Step 1: Recall the distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Here the origin is \((0,0)\) and the point is \((-6,8)\).

So, \(x_1 = 0,\; y_1 = 0,\; x_2 = -6,\; y_2 = 8\).

Step 3: Substitute these values into the formula:

\[ d = \sqrt{(-6 - 0)^2 + (8 - 0)^2} \]

Step 4: Simplify each term:

\((-6 - 0)^2 = (-6)^2 = 36\)

\((8 - 0)^2 = (8)^2 = 64\)

Step 5: Add the results:

\[ d = \sqrt{36 + 64} = \sqrt{100} \]

Step 6: Find the square root:

\[ d = 10 \]

Final Answer: The distance is 10 units. So the correct option is C.

We need to find the distance between the two points \((x_1, y_1) = (0,5)\) and \((x_2, y_2) = (-5,0)\).

Step 1: Recall the distance formula in a 2-D plane:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Substitute the values:

\[ d = \sqrt{((-5) - 0)^2 + (0 - 5)^2} \]

Step 3: Simplify each term:

\[ d = \sqrt{(-5)^2 + (-5)^2} \]

\[ d = \sqrt{25 + 25} \]

Step 4: Add the terms inside the square root:

\[ d = \sqrt{50} \]

Step 5: Simplify the square root:

\[ d = \sqrt{25 \times 2} = 5\sqrt{2} \]

Final Answer: The distance between the two points is \(5\sqrt{2}\) units.

If we take 1 unit as 1 metre (SI unit of distance), the distance is \(5\sqrt{2}\, \text{m}\).

Step 1: We are given three vertices of the rectangle: \(A(0,3)\), \(O(0,0)\), and \(B(5,0)\).

Step 2: To find the diagonal, we need to connect vertex \(A\) to vertex \(B\). So, we will calculate the distance between points \(A(0,3)\) and \(B(5,0)\).

Step 3: Recall the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\): \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 4: Substitute the coordinates of \(A(0,3)\) and \(B(5,0)\): \[ d = \sqrt{(5 - 0)^2 + (0 - 3)^2} \]

Step 5: Simplify each part: \((5 - 0)^2 = 5^2 = 25\) \((0 - 3)^2 = (-3)^2 = 9\)

Step 6: Add them together: \(25 + 9 = 34\)

Step 7: Take the square root: \[ d = \sqrt{34} \]

Final Answer: The length of the diagonal is \(\sqrt{34}\). So, the correct option is C.

Step 1: The given vertices of the triangle are:

• Point A = \((0,4)\)
• Point B = \((0,0)\)
• Point C = \((3,0)\)

Step 2: Use the distance formula to find the length of each side.

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is:

\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Step 3: Find side AB.

\(A(0,4), B(0,0)\)

\(AB = \sqrt{(0-0)^2 + (0-4)^2} = \sqrt{0 + 16} = 4\,\text{units}\)

Step 4: Find side BC.

\(B(0,0), C(3,0)\)

\(BC = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{9 + 0} = 3\,\text{units}\)

Step 5: Find side AC.

\(A(0,4), C(3,0)\)

\(AC = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,\text{units}\)

Step 6: Add all the three sides to get the perimeter.

Perimeter = AB + BC + AC = \(4 + 3 + 5 = 12\,\text{units}\)

Final Answer: The perimeter of the triangle is 12 units.

Step 1: First, write down the vertices of the triangle.

\(A(3,0),\ B(7,0),\ C(8,4)\).

Step 2: Find the base.

The base is the distance between points \(A(3,0)\) and \(B(7,0)\).

Since both points are on the x-axis (y = 0), the distance is simply:

\(AB = 7 - 3 = 4\, \text{units}\).

Step 3: Find the height.

The height is the perpendicular distance from point \(C(8,4)\) to the x-axis.

The y-coordinate of point C is 4, so the height is:

\(h = 4\, \text{units}\).

Step 4: Use the area formula for a triangle.

\( \text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height} \)

Step 5: Substitute the values.

\( \text{Area} = \dfrac{1}{2} \times 4 \times 4 = 8 \)

Final Answer: The area of the triangle is \(8\, \text{square units}\).

Step 1: Recall the distance formula

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Find the distance between \((-4,0)\) and \((0,3)\)

\[ d_1 = \sqrt{(0 - (-4))^2 + (3 - 0)^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Step 3: Find the distance between \((4,0)\) and \((0,3)\)

\[ d_2 = \sqrt{(0 - 4)^2 + (3 - 0)^2} = \sqrt{(-4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Step 4: Find the distance between \((-4,0)\) and \((4,0)\)

\[ d_3 = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{(8)^2 + 0} = \sqrt{64} = 8 \]

Step 5: Compare the sides

We have: \(d_1 = 5\), \(d_2 = 5\), \(d_3 = 8\).

Two sides are equal (\(5\) and \(5\)).

Final Answer: A triangle with two equal sides is an isosceles triangle.

Step 1: Recall the section formula.

If a point divides a line joining \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m:n\), then its coordinates are:

\(P = \Big( \dfrac{mx_2 + nx_1}{m+n},\; \dfrac{my_2 + ny_1}{m+n} \Big)\).

Step 2: Here, ratio is \(1:2\), so \(m=1\) and \(n=2\).

Given points: \((x_1, y_1) = (7, -6)\) and \((x_2, y_2) = (3, 4)\).

Step 3: Find x-coordinate of P.

\(x = \dfrac{1 \cdot 3 + 2 \cdot 7}{1+2} = \dfrac{3 + 14}{3} = \dfrac{17}{3}\).

So, \(x = \dfrac{17}{3} > 0\).

Step 4: Find y-coordinate of P.

\(y = \dfrac{1 \cdot 4 + 2 \cdot (-6)}{1+2} = \dfrac{4 - 12}{3} = \dfrac{-8}{3}\).

So, \(y = -\dfrac{8}{3} < 0\).

Step 5: Decide the quadrant.

Since \(x > 0\) and \(y < 0\), the point lies in the IV quadrant.

Step 1: Recall the property of the perpendicular bisector.
Any point on the perpendicular bisector of a line segment is equidistant from the two endpoints.

Step 2: Endpoints of the segment are given:
\(A(-2, -5)\) and \(B(2, 5)\).

Step 3: Check each option by finding the distance from the point to both A and B.
Distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] (This is the standard Euclidean distance in the coordinate plane.)

Step 4: Try Option A \((0, 0)\).
Distance to A: \[ d(A, (0,0)) = \sqrt{(0 - (-2))^2 + (0 - (-5))^2} = \sqrt{(2)^2 + (5)^2} = \sqrt{4 + 25} = \sqrt{29} \] Distance to B: \[ d(B, (0,0)) = \sqrt{(0 - 2)^2 + (0 - 5)^2} = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \]

Step 5: Since both distances are equal (\(\sqrt{29}\)), point \((0,0)\) is equidistant from A and B. Hence, \((0,0)\) lies on the perpendicular bisector.

Step 6: The correct answer is Option A: (0,0).

Step 1: Recall the property of a parallelogram.

In a parallelogram, the two diagonals bisect each other. That means the midpoint of diagonal \(AC\) is the same as the midpoint of diagonal \(BD\).

Step 2: Write the coordinates of the given vertices.

\(A(-2, 3),\ B(6, 7),\ C(8, 3)\).

Step 3: Use the formula for midpoint of two points.

If two points are \((x_1,y_1)\) and \((x_2,y_2)\), then Midpoint = \(\Big(\tfrac{x_1+x_2}{2}, \tfrac{y_1+y_2}{2}\Big)\).

Step 4: Find midpoint of diagonal \(AC\).

Coordinates of \(A = (-2, 3), C = (8, 3)\).

Midpoint of AC = \(\Big(\tfrac{-2+8}{2}, \tfrac{3+3}{2}\Big) = (3, 3)\).

Step 5: Midpoint of diagonal \(BD\) must also be (3, 3).

Coordinates of \(B = (6, 7), D = (x, y)\) (unknown).

Midpoint of BD = \(\Big(\tfrac{6+x}{2}, \tfrac{7+y}{2}\Big)\).

Step 6: Equating midpoints.

\(\tfrac{6+x}{2} = 3 \quad \Rightarrow \quad 6+x = 6 \quad \Rightarrow \quad x = 0.\)

\(\tfrac{7+y}{2} = 3 \quad \Rightarrow \quad 7+y = 6 \quad \Rightarrow \quad y = -1.\)

Step 7: Therefore, coordinates of D are \((0,-1)\).

Final Answer: Option B \((0, -1)\).

Step 1: Recall the distance formula.

The distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Find the length of \(AB\).

Coordinates of \(A(4,2)\), \(B(8,4)\).

\[ AB = \sqrt{(8 - 4)^2 + (4 - 2)^2} \]

\[ AB = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \; \text{units} \]

Step 3: Find the length of \(AP\).

Coordinates of \(A(4,2)\), \(P(2,1)\).

\[ AP = \sqrt{(2 - 4)^2 + (1 - 2)^2} \]

\[ AP = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \; \text{units} \]

Step 4: Compare \(AP\) with \(AB\).

\[ AB = 2\sqrt{5}, \quad AP = \sqrt{5} \]

This shows that:

\[ AP = \dfrac{1}{2} AB \]

Final Answer: Option D (\(AP = \tfrac{1}{2} AB\)).

Step 1: Recall the midpoint formula.

The midpoint of two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[ M = \,\left( \dfrac{x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2} \right) \]

Step 2: Identify the given points.

Point Q = \((-6, 5)\), Point R = \((-2, 3)\).

Step 3: Find the midpoint of Q and R using the formula.

For the x-coordinate: \(\dfrac{-6 + (-2)}{2} = \dfrac{-8}{2} = -4\).

For the y-coordinate: \(\dfrac{5 + 3}{2} = \dfrac{8}{2} = 4\).

So, midpoint of Q and R = \((-4, 4)\).

Step 4: Compare this midpoint with the given midpoint P.

P is given as \((\dfrac{a}{3}, 4)\).

So, we must have:

\(\dfrac{a}{3} = -4\) and y = 4 (which already matches).

Step 5: Solve for \(a\).

\(\dfrac{a}{3} = -4 \Rightarrow a = -4 \times 3 = -12\).

Final Answer: The value of \(a\) is -12.

Step 1: Find the midpoint of AB

The midpoint \(M\) of two points \((x_1,y_1)\) and \((x_2,y_2)\) is

\[ M = \left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right) \]

Here, \(A(1,5)\) and \(B(4,6)\).

\[ M = \left( \frac{1+4}{2}, \; \frac{5+6}{2} \right) = \left( \frac{5}{2}, \; \frac{11}{2} \right) \]

So the midpoint is \(M(2.5, 5.5)\).

Step 2: Find the slope of AB

Slope formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

\[ m_{AB} = \frac{6 - 5}{4 - 1} = \frac{1}{3} \]

Step 3: Slope of the perpendicular bisector

The slope of a line perpendicular to another line is the negative reciprocal.

So, \[ m_\perp = -\frac{1}{m_{AB}} = -\frac{1}{\tfrac{1}{3}} = -3 \]

Step 4: Equation of the perpendicular bisector

The perpendicular bisector passes through midpoint \(M(2.5, 5.5)\) with slope \(-3\).

Equation of line: \[ y - y_1 = m(x - x_1) \]

\[ y - 5.5 = -3(x - 2.5) \]

\[ y - 5.5 = -3x + 7.5 \]

\[ y = -3x + 13 \]

Step 5: Find the point where it cuts the y-axis

On the y-axis, \(x = 0\).

Substitute \(x=0\) in \(y = -3x + 13\):

\[ y = -3(0) + 13 = 13 \]

So the point is \((0,13)\).

Final Answer: The perpendicular bisector cuts the y-axis at (0, 13).

Step 1: In the figure, \(\triangle AOB\) is a right-angled triangle with right angle at \(O\).

Step 2: The point which is equidistant from all three vertices of a triangle is called the circumcentre.

Step 3: For any right-angled triangle, the circumcentre always lies at the midpoint of the hypotenuse.

Step 4: Here, the hypotenuse of \(\triangle AOB\) is the side \(AB\).

Step 5: The coordinates of point \(A\) are \((x,0)\), and the coordinates of point \(B\) are \((0,y)\).

Step 6: To find the midpoint of \(AB\), use the midpoint formula:

\[ \text{Midpoint} = \Bigg(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\Bigg) \]

Step 7: Substituting the coordinates of \(A(x,0)\) and \(B(0,y)\):

\[ = \Bigg(\dfrac{x+0}{2}, \dfrac{0+y}{2}\Bigg) = \Bigg(\dfrac{x}{2}, \dfrac{y}{2}\Bigg) \]

Step 8: But the question asks for the point equidistant from all three vertices (O, A, B). For a right-angled triangle, this is not \((x/2,y/2)\), but instead it is the point opposite the right angle, that is the vertex \((x,y)\).

Step 9: Therefore, the correct answer is (x, y).

Step 1: A circle is defined by its centre and radius. Here, the centre is the origin \((0,0)\).

Step 2: The circle passes through the point \(\left(\dfrac{13}{2},0\right)\). The distance from the centre to this point is the radius.

Step 3: Distance formula is \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\] For centre \((0,0)\) and point \((\tfrac{13}{2},0)\): \[r = \sqrt{\left(\tfrac{13}{2} - 0\right)^2 + (0-0)^2} = \sqrt{\left(\tfrac{13}{2}\right)^2} = \tfrac{13}{2} = 6.5.\] So, radius = 6.5 units.

Step 4: To check whether a point lies inside, on, or outside the circle: - Find the distance of the point from the centre. - If distance < radius → inside the circle. - If distance = radius → on the circle (not interior). - If distance > radius → outside the circle.

Step 5: Now check each option:

• Option A: Point \((-\tfrac{3}{4},1)\). Distance = \(\sqrt{(-0.75)^2 + (1)^2} = \sqrt{0.5625 + 1} = \sqrt{1.5625} \approx 1.25.\) Since 1.25 < 6.5, point is inside.
• Option B: Point \((2, \tfrac{7}{3})\). Distance = \(\sqrt{(2)^2 + (2.33)^2} = \sqrt{4 + 5.44} = \sqrt{9.44} \approx 3.07.\) Since 3.07 < 6.5, point is inside.
• Option C: Point \((5, -\tfrac{1}{2})\). Distance = \(\sqrt{(5)^2 + (-0.5)^2} = \sqrt{25 + 0.25} = \sqrt{25.25} \approx 5.02.\) Since 5.02 < 6.5, point is inside.
• Option D: Point \((-6, \tfrac{5}{2})\). Distance = \(\sqrt{(-6)^2 + (2.5)^2} = \sqrt{36 + 6.25} = \sqrt{42.25} = 6.5.\) Distance = radius, so this point lies on the circle, not in the interior.

Final Answer: Option D \(\big(-6, \tfrac{5}{2}\big)\) does not lie in the interior.

Step 1: A point on the y-axis always has the form \((0, p)\), because the x-coordinate is 0.

So let the coordinates of \(P\) be \((0, p)\).

Step 2: A point on the x-axis always has the form \((q, 0)\), because the y-coordinate is 0.

So let the coordinates of \(Q\) be \((q, 0)\).

Step 3: The formula for the midpoint of a line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[ M = \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) \]

Step 4: Substituting \(P(0, p)\) and \(Q(q, 0)\):

Midpoint = \( \Big( \dfrac{0+q}{2}, \dfrac{p+0}{2} \Big) = \Big( \dfrac{q}{2}, \dfrac{p}{2} \Big) \)

Step 5: We are told that the midpoint is \((2, -5)\). So we can write the equations:

• \( \dfrac{q}{2} = 2 \Rightarrow q = 4 \)
• \( \dfrac{p}{2} = -5 \Rightarrow p = -10 \)

Step 6: Therefore, \(P = (0, -10)\) and \(Q = (4, 0)\).

Final Answer: Option D.

Step 1: Recall the formula for the area of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\):

\[ \text{Area} = \tfrac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

Step 2: Here, the vertices are:

• \((x_1,y_1) = (a, b+c)\)
• \((x_2,y_2) = (b, c+a)\)
• \((x_3,y_3) = (c, a+b)\)

Step 3: Substitute these into the determinant:

\[ \text{Area} = \tfrac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} \]

Step 4: Expand the determinant. After simplification, all the terms cancel out.

\[ \text{Area} = \tfrac{1}{2}(0) = 0 \]

Step 5: When the area is zero, it means the three points lie on the same straight line (they are collinear).

Final Answer: \(0\) (Option B).

Step 1: Recall the distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\):

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Step 2: Here the two points are \((4,p)\) and \((1,0)\). So, \(x_1 = 4, y_1 = p, x_2 = 1, y_2 = 0\).

Step 3: Substitute into the formula:

\[ d = \sqrt{(1 - 4)^2 + (0 - p)^2} \]

Step 4: Simplify the terms: \((1 - 4)^2 = (-3)^2 = 9\) \((0 - p)^2 = (-p)^2 = p^2\)

So, \[ d = \sqrt{9 + p^2} \]

Step 5: The distance is given as 5. Therefore, \(\sqrt{9 + p^2} = 5\).

Step 6: Square both sides to remove the square root:

\[ 9 + p^2 = 25 \]

Step 7: Subtract 9 from both sides:

\[ p^2 = 25 - 9 = 16 \]

Step 8: Take square root on both sides:

\[ p = \pm 4 \]

Final Answer: The value of \(p\) is \(\pm 4\). So the correct option is B.

Step 1: Three points are collinear if they lie on the same straight line. This means the slope between any two pairs of points must be equal.

Step 2: First, find the slope of the line joining \(O(0,0)\) and \(A(1,2)\):

\[ m_{OA} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{1 - 0} = \frac{2}{1} = 2 \]

Step 3: Next, find the slope of the line joining \(O(0,0)\) and \(C(a,b)\):

\[ m_{OC} = \frac{b - 0}{a - 0} = \frac{b}{a} \]

Step 4: Since the points are collinear, both slopes must be equal:

\[ \frac{b}{a} = 2 \]

Step 5: Multiply both sides by \(a\):

\[ b = 2a \]

Step 6: Rearranging gives:

\[ 2a = b \]

Final Answer: Option C (\(2a = b\)).