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

Exercise 7.2

Step 1: Write the coordinates of both triangles.

• For \(\triangle ABC\): A(-2, 0), B(2, 0), C(0, 2)
• For \(\triangle DEF\): D(-4, 0), E(4, 0), F(0, 4)

Step 2: Compare corresponding vertices.

• D(-4, 0) is exactly 2 times A(-2, 0).
• E(4, 0) is exactly 2 times B(2, 0).
• F(0, 4) is exactly 2 times C(0, 2).

Step 3: This means every coordinate of triangle DEF is obtained by multiplying the coordinates of triangle ABC by 2. This is called a dilation (enlargement) with scale factor = 2.

Step 4: In a dilation:

• The lengths of sides become proportional (here ratio = 1:2).
• The angles remain the same (equal in measure).

Step 5: Since corresponding sides are proportional and corresponding angles are equal, by the definition of similarity, we conclude that:

\(\triangle ABC \sim \triangle DEF\).

Step 1: Look at the coordinates of A and B.

A has coordinates \((-4, 6)\). B has coordinates \((-4, -6)\).

Step 2: Since the x-coordinate of both A and B is \(-4\), the line segment AB is vertical, lying on the line \(x = -4\).

Step 3: For any point to lie on AB, its x-coordinate must also be \(-4\), and its y-coordinate must be between -6 and 6 (the y-values of A and B).

Step 4: Point P has coordinates \((-4, 2)\). The x-coordinate is \(-4\), so it is on the same vertical line.

Step 5: Now check the y-coordinate of P. The y-value is 2. This lies between -6 and 6, because \(-6 \leq 2 \leq 6\).

Final Step: Since both conditions are satisfied, point P lies on the line segment AB.

To check if three points are collinear, we find the slope of the line joining any two points, and then compare it with the slope of the line joining another pair of points.

Step 1: Take the first two points: \((0,5)\) and \((0,-9)\).

Since both points have the same x-coordinate (0), the line between them is vertical.

The slope of a vertical line is undefined.

Step 2: Now take the points \((0,5)\) and \((3,6)\).

The slope formula is:

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

Here, \((x_1,y_1) = (0,5)\) and \((x_2,y_2) = (3,6)\).

So,

\[ m = \frac{6 - 5}{3 - 0} = \frac{1}{3} \]

Step 3: Compare the slopes.

- Slope of line through \((0,5)\) and \((0,-9)\) = undefined.

- Slope of line through \((0,5)\) and \((3,6)\) = \(\tfrac{1}{3}\).

Step 4: Since the slopes are not equal, the three points do not lie on the same straight line.

Therefore, the statement is False.

Step 1: Find the midpoint of AB

The coordinates of A are \((-1,1)\) and of B are \((3,3)\).

Midpoint formula: \(M = \left(\tfrac{x_1+x_2}{2}, \tfrac{y_1+y_2}{2}\right)\).

So, \(M = \left(\tfrac{-1+3}{2}, \tfrac{1+3}{2}\right) = (1,2)\).

The perpendicular bisector will pass through this midpoint \((1,2)\).

Step 2: Find the slope of AB

Slope formula: \(m = \tfrac{y_2-y_1}{x_2-x_1}\).

So, slope of AB = \(\tfrac{3-1}{3-(-1)} = \tfrac{2}{4} = \tfrac{1}{2}\).

Step 3: Find the slope of the perpendicular bisector

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

So, slope of perpendicular bisector = \(-\tfrac{1}{\tfrac{1}{2}} = -2\).

Step 4: Equation of the perpendicular bisector

Using point-slope form: \(y - y_1 = m(x - x_1)\).

Here, point is \((1,2)\) and slope is \(-2\).

Equation: \(y - 2 = -2(x - 1)\).

Simplify: \(y - 2 = -2x + 2 \Rightarrow y = -2x + 4\).

Step 5: Find where this line cuts the y-axis

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

Put \(x=0\) in \(y = -2x + 4\): \(y = -2(0) + 4 = 4\).

So the line meets the y-axis at \((0,4)\).

Step 6: Compare with given point

The question says the intersection is at \((0,2)\), but we found it is \((0,4)\).

Conclusion: The statement is False.

Step 1: To check if three points can form a triangle, we must see if they are collinear (i.e., lie on the same straight line). If the points are collinear, they cannot form a triangle.

Step 2: We test collinearity by comparing the slopes of two pairs of points. - Slope of line between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by:

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

Step 3: Calculate slope of line AB (between points A(3,1) and B(12,−2)): \[ m_{AB} = \frac{-2 - 1}{12 - 3} = \frac{-3}{9} = -\tfrac{1}{3} \]

Step 4: Calculate slope of line AC (between points A(3,1) and C(0,2)): \[ m_{AC} = \frac{2 - 1}{0 - 3} = \frac{1}{-3} = -\tfrac{1}{3} \]

Step 5: Since \(m_{AB} = m_{AC} = -\tfrac{1}{3}\), all three points A, B, C lie on the same straight line (they are collinear).

Step 6: If points are collinear, they cannot form the vertices of a triangle. Hence, the statement is True.

Step 1: Recall the property of a parallelogram.

In a parallelogram, the opposite sides are parallel and equal in length. That means, if points A, B, C, D form a parallelogram, then:

• \(\overrightarrow{AB} = \overrightarrow{DC}\)
• \(\overrightarrow{BC} = \overrightarrow{AD}\)

Step 2: Find vector \(\overrightarrow{AB}\).

\(A(4,3), B(6,4)\)

\(\overrightarrow{AB} = (x_2 - x_1, y_2 - y_1) = (6-4, 4-3) = (2,1)\)

Step 3: Find vector \(\overrightarrow{DC}\).

\(D(-3,5), C(5,-6)\)

\(\overrightarrow{DC} = (5 - (-3), -6 - 5) = (8,-11)\)

Check: \(\overrightarrow{AB} = (2,1)\), \(\overrightarrow{DC} = (8,-11)\). They are not equal and not parallel.

Step 4: Find vector \(\overrightarrow{BC}\).

\(B(6,4), C(5,-6)\)

\(\overrightarrow{BC} = (5-6, -6-4) = (-1,-10)\)

Step 5: Find vector \(\overrightarrow{AD}\).

\(A(4,3), D(-3,5)\)

\(\overrightarrow{AD} = (-3-4, 5-3) = (-7,2)\)

Check: \(\overrightarrow{BC} = (-1,-10)\), \(\overrightarrow{AD} = (-7,2)\). They are not equal and not parallel.

Step 6: Since both pairs of opposite sides are neither equal nor parallel, the given points do not form a parallelogram.

Final Answer: False.

Step 1: The circle is centered at the origin \(O(0,0)\). Since the point \(P(5,0)\) lies on the circle, the distance from the center to \(P\) is the radius.

Step 2: Calculate the radius:

\(OP = \sqrt{(5-0)^2 + (0-0)^2} = \sqrt{25} = 5\,\text{units}\).

Step 3: Now calculate the distance from the center \(O(0,0)\) to the point \(Q(6,8)\):

\(OQ = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,\text{units}\).

Step 4: Compare the distances:

The radius = 5 units, but the distance \(OQ = 10\,\text{units}\).

Step 5: Since \(10 > 5\), point \(Q\) lies outside the circle.

To check if a point lies on the perpendicular bisector of a line segment, we must test whether the point is equidistant (same distance) from both end points of the segment.

Step 1: Write the coordinates of all points.

• \(A(2,7)\)
• \(P(6,5)\)
• \(Q(0,-4)\)

Step 2: Use the distance formula:

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

Step 3: Calculate distance \(AP\).

\[ AP = \sqrt{(2 - 6)^2 + (7 - 5)^2} = \sqrt{(-4)^2 + (2)^2} = \sqrt{16 + 4} = \sqrt{20} \]

Step 4: Calculate distance \(AQ\).

\[ AQ = \sqrt{(2 - 0)^2 + (7 - (-4))^2} = \sqrt{(2)^2 + (11)^2} = \sqrt{4 + 121} = \sqrt{125} \]

Step 5: Compare the two distances.

We have \(AP = \sqrt{20}\) and \(AQ = \sqrt{125}\).

Conclusion: Since \(AP \neq AQ\), point \(A\) is not equidistant from \(P\) and \(Q\). Therefore, \(A\) does not lie on the perpendicular bisector of the line segment \(PQ\).

Step 1: Recall what trisection means.

Trisection of a line segment means dividing it into 3 equal parts. So, two points divide the line segment in the ratio \(1:2\) and \(2:1\).

Step 2: Formula for internal division of a line.

If a point \(P(x, y)\) divides the line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m:n\), then

\[ P(x, y) = \Bigg( \dfrac{m x_2 + n x_1}{m+n},\, \dfrac{m y_2 + n y_1}{m+n} \Bigg) \]

Step 3: Apply the formula for ratio \(1:2\).

Here, we take \(A(7, -2) = (x_1, y_1)\) and \(B(1, -5) = (x_2, y_2)\).

Ratio = \(1:2\). That means \(m = 1, n = 2\).

Step 4: Calculate the x-coordinate.

\[ x = \dfrac{1 \cdot 7 + 2 \cdot 1}{1+2} = \dfrac{7 + 2}{3} = \dfrac{9}{3} = 3 \]

Wait! This is not \(5\). Let’s try ratio \(2:1\) instead (because we need both trisection points).

Step 5: Calculate again for ratio \(2:1\).

Now \(m = 2, n = 1\).

\[ x = \dfrac{2 \cdot 7 + 1 \cdot 1}{2+1} = \dfrac{14 + 1}{3} = \dfrac{15}{3} = 5 \]

Step 6: Calculate the y-coordinate for ratio \(2:1\).

\[ y = \dfrac{2 \cdot (-2) + 1 \cdot (-5)}{2+1} = \dfrac{-4 - 5}{3} = \dfrac{-9}{3} = -3 \]

Step 7: Final Result.

So, the point is \((5, -3)\), which matches the given point \(P(5, -3)\).

Conclusion: Yes, the given point is one of the trisection points. Hence, the statement is True.

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 AB.

Points are \(A(-6,10)\) and \(B(-4,6)\).

\[ AB = \sqrt{((-4)-(-6))^2 + (6-10)^2} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

Step 3: Find the distance AC.

Points are \(A(-6,10)\) and \(C(3,-8)\).

\[ AC = \sqrt{((3)-(-6))^2 + (-8-10)^2} = \sqrt{(9)^2 + (-18)^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5} \]

Step 4: Compare the ratio AB : AC.

\[ \frac{AB}{AC} = \frac{2\sqrt{5}}{9\sqrt{5}} = \frac{2}{9} \]

This matches the condition given in the question.

Step 5: Check collinearity by slope method.

Slope of line AB:

\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 10}{-4 - (-6)} = \frac{-4}{2} = -2 \]

Slope of line AC:

\[ m_{AC} = \frac{-8 - 10}{3 - (-6)} = \frac{-18}{9} = -2 \]

Since \(m_{AB} = m_{AC}\), the points are collinear.

Final Conclusion: The three points A, B, C are collinear and \(AB = \tfrac{2}{9} AC\). Hence, the statement is True.

Step 1: A point lies on a circle if the distance between the point and the centre of the circle is equal to the radius.

Step 2: The formula for 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: Here, the centre of the circle is \((3, 5)\) and the point is \((-2, 4)\).

So, \(x_1 = 3,\; y_1 = 5,\; x_2 = -2,\; y_2 = 4\).

Step 4: Substitute values:

\[ d = \sqrt{((-2) - 3)^2 + (4 - 5)^2} \]

\[ d = \sqrt{(-5)^2 + (-1)^2} \]

\[ d = \sqrt{25 + 1} \]

\[ d = \sqrt{26} \]

Step 5: Approximate value:

\( \sqrt{26} \approx 5.1 \; \text{units} \)

Step 6: Compare with the radius:

The radius = 6 units.

The distance = 5.1 units.

Step 7: Since 5.1 units is less than 6 units, the point lies inside the circle, not on it.

Final Answer: The statement is False.

Step 1: Write down the given points.

We have four points: \(A(-1,-2),\ B(4,3),\ C(2,5),\ D(-3,0)\).

Step 2: Find the vector \(\overrightarrow{AB}\).

Formula: \(\overrightarrow{AB} = (x_2 - x_1,\; y_2 - y_1)\).

So, \(\overrightarrow{AB} = (4 - (-1),\; 3 - (-2)) = (5,\; 5)\).

Step 3: Find the vector \(\overrightarrow{BC}\).

\(\overrightarrow{BC} = (2 - 4,\; 5 - 3) = (-2,\; 2)\).

Step 4: Check if \(AB\) is perpendicular to \(BC\).

We use the dot product rule: If \(\overrightarrow{u}\cdot\overrightarrow{v} = 0\), the two vectors are at right angles.

\(\overrightarrow{AB}\cdot\overrightarrow{BC} = (5)(-2) + (5)(2) = -10 + 10 = 0\).

Since the dot product is 0, \(AB \perp BC\). That means angle at \(B\) is a right angle (\(90^\circ\)).

Step 5: Find the vector \(\overrightarrow{CD}\).

\(\overrightarrow{CD} = (-3 - 2,\; 0 - 5) = (-5,\; -5)\).

Step 6: Compare \(\overrightarrow{CD}\) with \(\overrightarrow{AB}\).

We see that \(\overrightarrow{CD} = (-5,-5) = -1 \times (5,5) = -\overrightarrow{AB}\).

This means \(CD\) is parallel and equal in length to \(AB\), but in opposite direction. So, \(AB \parallel CD\).

Step 7: Find the vector \(\overrightarrow{DA}\).

\(\overrightarrow{DA} = (-1 - (-3),\; -2 - 0) = (2,\; -2)\).

Wait! Let's double-check carefully: \(\overrightarrow{DA} = (x_A - x_D,\; y_A - y_D) = (-1 - (-3),\; -2 - 0) = (2, -2)\).

Step 8: Compare \(\overrightarrow{DA}\) with \(\overrightarrow{BC}\).

\(\overrightarrow{BC} = (-2,2)\). Notice \(\overrightarrow{DA} = (2,-2) = -1 \times (-2,2)\).

So, \(DA \parallel BC\).

Step 9: Conclusion.

We have shown that: • \(AB \parallel CD\), • \(BC \parallel DA\), • and angle at \(B\) is a right angle.

Therefore, the quadrilateral formed by points \(A, B, C, D\) is a rectangle.