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

Exercise 7.3

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 length of side AB.

Coordinates: \(A(-5, 6), B(-4, -2)\)

\[ AB^2 = (-4 - (-5))^2 + (-2 - 6)^2 \]

\[ AB^2 = (1)^2 + (-8)^2 = 1 + 64 = 65 \]

So, \(AB = \sqrt{65}\).

Step 3: Find length of side BC.

Coordinates: \(B(-4, -2), C(7, 5)\)

\[ BC^2 = (7 - (-4))^2 + (5 - (-2))^2 \]

\[ BC^2 = (11)^2 + (7)^2 = 121 + 49 = 170 \]

So, \(BC = \sqrt{170}\).

Step 4: Find length of side CA.

Coordinates: \(C(7, 5), A(-5, 6)\)

\[ CA^2 = (-5 - 7)^2 + (6 - 5)^2 \]

\[ CA^2 = (-12)^2 + (1)^2 = 144 + 1 = 145 \]

So, \(CA = \sqrt{145}\).

Step 5: Compare the side lengths.

We have:

• \(AB^2 = 65\)
• \(BC^2 = 170\)
• \(CA^2 = 145\)

Since \(65 \neq 170 \neq 145\), all three sides are of different lengths.

Step 6: Check for right triangle.

If it were a right triangle, one side squared would equal the sum of the other two (Pythagoras theorem). But:

• \(65 + 145 = 210 \neq 170\)
• \(65 + 170 = 235 \neq 145\)
• \(145 + 170 = 315 \neq 65\)

So, it is not a right triangle.

Final Conclusion: The triangle has all unequal sides and is not right-angled. Therefore, it is a scalene triangle.

Step 1: Recall that any point on the x–axis has the form \((x, 0)\), because its y–coordinate is always zero.

Step 2: Use 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 3: Here, one point is \((x, 0)\) and the other is \((7, -4)\). So the distance becomes: \[ d = \sqrt{(x - 7)^2 + (0 - (-4))^2} \]

Step 4: Simplify the second term: \(0 - (-4) = 4\). So, \[ d = \sqrt{(x - 7)^2 + (4)^2} \]

Step 5: The problem says the distance is \(2\sqrt{5}\). That means: \[ \sqrt{(x - 7)^2 + 16} = 2\sqrt{5} \]

Step 6: Square both sides to remove the square root: \[ (x - 7)^2 + 16 = (2\sqrt{5})^2 \]

Step 7: Calculate the right-hand side: \((2\sqrt{5})^2 = 4 \times 5 = 20\). So, \[ (x - 7)^2 + 16 = 20 \]

Step 8: Subtract 16 from both sides: \[ (x - 7)^2 = 20 - 16 = 4 \]

Step 9: Take square root of both sides: \[ x - 7 = \pm 2 \]

Step 10: Solve for \(x\): - If \(x - 7 = 2\), then \(x = 9\). - If \(x - 7 = -2\), then \(x = 5\).

Step 11: So the points are \((9, 0)\) and \((5, 0)\).

Final Answer: There are two points on the x–axis: (5, 0) and (9, 0).

Step 1: Recall the concept.

To identify a quadrilateral, we usually check:

• If opposite sides are parallel (slopes are equal).
• If adjacent sides are perpendicular (product of slopes = -1).
• If lengths of opposite sides are equal.

Step 2: Find slope of each side.

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

• \(AB: m = \dfrac{3 - (-2)}{7 - 2} = \dfrac{5}{5} = 1\).
• \(BC: m = \dfrac{-1 - 3}{11 - 7} = \dfrac{-4}{4} = -1\).
• \(CD: m = \dfrac{-6 - (-1)}{6 - 11} = \dfrac{-5}{-5} = 1\).
• \(DA: m = \dfrac{-2 - (-6)}{2 - 6} = \dfrac{4}{-4} = -1\).

Step 3: Check parallel and perpendicular sides.

• \(AB\) and \(CD\) both have slope 1 ⇒ parallel.
• \(BC\) and \(DA\) both have slope -1 ⇒ parallel.
• \(AB \times BC = 1 \times -1 = -1\) ⇒ perpendicular.

So opposite sides are parallel and adjacent sides are perpendicular.

Step 4: Check lengths of sides (using distance formula).

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

• \(|AB| = \sqrt{(7 - 2)^2 + (3 - (-2))^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}\).
• \(|BC| = \sqrt{(11 - 7)^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\).
• \(|CD| = \sqrt{(6 - 11)^2 + (-6 - (-1))^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\).
• \(|DA| = \sqrt{(2 - 6)^2 + (-2 - (-6))^2} = \sqrt{(-4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\).

Step 5: Conclusion.

We see that:

• Opposite sides are equal (\(|AB| = |CD|\), \(|BC| = |DA|\)).
• Adjacent sides are not equal (so it is not a square).
• Angles are right angles (because slopes are perpendicular).

Therefore, the quadrilateral is a Rectangle.

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 coordinates are:

• Point A: \((-3, -14)\)
• Point B: \((a, -5)\)
• Distance: \(d = 9\)

Step 3: Substitute values into the formula:

\[ 9 = \sqrt{(a - (-3))^2 + (-5 - (-14))^2} \]

Step 4: Simplify the terms:

• \(a - (-3) = a + 3\)
• \(-5 - (-14) = -5 + 14 = 9\)

So:

\[ 9 = \sqrt{(a+3)^2 + (9)^2} \]

Step 5: Square both sides to remove the square root:

\[ 9^2 = (a+3)^2 + 9^2 \]

\[ 81 = (a+3)^2 + 81 \]

Step 6: Subtract 81 from both sides:

\[ 81 - 81 = (a+3)^2 \]

\[ 0 = (a+3)^2 \]

Step 7: Take square root of both sides:

\[ a + 3 = 0 \]

Step 8: Solve for \(a\):

\[ a = -3 \]

Final Answer: \(a = -3\).

Step 1: Recall the idea.

A point is equidistant from two given points if its distance from the first point is the same as its distance from the second point.

All such points lie on the perpendicular bisector of the line segment joining the two points.

Step 2: Find the midpoint of AB.

The coordinates of the midpoint \(M\) are found using the formula:

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

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

\[M = \left( \dfrac{-5 + (-1)}{2}, \dfrac{4 + 6}{2} \right) = \left( \dfrac{-6}{2}, \dfrac{10}{2} \right) = (-3, 5)\]

Step 3: Find the slope of AB.

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

\(m_{AB} = \dfrac{6 - 4}{-1 - (-5)} = \dfrac{2}{4} = \dfrac{1}{2}\)

Step 4: Find the slope of the perpendicular bisector.

The slope of the perpendicular line is the negative reciprocal of \(m_{AB}\).

So, slope of perpendicular bisector = \(-\dfrac{1}{m_{AB}} = -\dfrac{1}{1/2} = -2\)

Step 5: Equation of the perpendicular bisector.

The line passes through midpoint \((-3,5)\) and has slope \(-2\).

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

\(y - 5 = -2(x + 3)\)

\(y - 5 = -2x - 6\)

\(y = -2x - 1\)

Step 6: Choose one point on this line.

If we put \(x = 0\), then \(y = -2(0) - 1 = -1\).

So one point is \((0,-1)\).

Step 7: Conclusion.

Every point on the line \(y = -2x - 1\) is equidistant from \(A\) and \(B\). That means there are infinitely many such points. One example is \((0, -1)\).

Step 1: Write the coordinates of the two given points.

Point \(A = (-5, -2)\), Point \(B = (4, -2)\).

Step 2: Find the midpoint of segment \(AB\).

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

So, \(M = left(\dfrac{-5 + 4}{2}, \dfrac{-2 + (-2)}{2}\right) = left(-\dfrac{1}{2}, -2\right)\).

Step 3: Find the slope of \(AB\).

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

Here, \(y_2-y_1 = -2 - (-2) = 0\). So, slope of \(AB = 0\). This means \(AB\) is a horizontal line.

Step 4: Slope of the perpendicular bisector.

A line perpendicular to a horizontal line is vertical. So the perpendicular bisector is a vertical line passing through the midpoint \(M\).

Equation of perpendicular bisector: \(x = -\dfrac{1}{2}\).

Step 5: Find the point where this line cuts the x–axis.

On the x–axis, \(y = 0\). Substituting in \(x = -\dfrac{1}{2}\), we get point \(Q = left(-\dfrac{1}{2}, 0\right)\).

Step 6: Identify the type of triangle \(QAB\).

Since \(Q\) lies on the perpendicular bisector of \(AB\), the distances \(QA\) and \(QB\) are equal.

Thus, triangle \(QAB\) has two equal sides, so it is an isosceles triangle.

Step 1: Recall the condition for collinear points.
If three points lie on the same straight line, then the slope between the first two points = slope between the second and the third point.

Step 2: Find the slope of the line joining \((5,1)\) and \((-2,-3)\).
Slope formula: \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).

Here, \((x_1, y_1) = (5,1), (x_2, y_2) = (-2,-3)\).

Slope = \(\dfrac{-3 - 1}{-2 - 5} = \dfrac{-4}{-7} = \dfrac{4}{7}\).

Step 3: Now find the slope of the line joining \((-2,-3)\) and \((8,2m)\).

Slope = \(\dfrac{2m - (-3)}{8 - (-2)} = \dfrac{2m + 3}{10}\).

Step 4: For collinearity, equate both slopes.

\(\dfrac{2m + 3}{10} = \dfrac{4}{7}\).

Step 5: Cross multiply to remove fractions.

\(7(2m + 3) = 40\).

\(14m + 21 = 40\).

Step 6: Simplify to solve for \(m\).

\(14m = 40 - 21 = 19\).

\(m = \dfrac{19}{14}\).

Final Answer: \(m = \dfrac{19}{14}\).

Step 1: Since point \(A(2,-4)\) is equidistant from \(P(3,8)\) and \(Q(-10,y)\), it means:

\(AP = AQ\)

Step 2: Use the distance formula:

\(AP = \sqrt{(2-3)^2 + (-4-8)^2}\)

\(AQ = \sqrt{(2-(-10))^2 + (-4-y)^2}\)

Step 3: Square both sides to remove the square root:

\((2-3)^2 + (-4-8)^2 = (2+10)^2 + (-4-y)^2\)

Step 4: Simplify each term:

• \((2-3)^2 = (-1)^2 = 1\)
• \((-4-8)^2 = (-12)^2 = 144\)
• \((2+10)^2 = (12)^2 = 144\)
• \((-4-y)^2 = (y+4)^2\) (since squaring removes the sign)

So the equation becomes:

\(1 + 144 = 144 + (y+4)^2\)

\(145 = 144 + (y+4)^2\)

Step 5: Subtract 144 from both sides:

\((y+4)^2 = 1\)

Step 6: Take square root:

\(y+4 = ±1\)

So, \(y = -4+1 = -3\) or \(y = -4-1 = -5\)

Step 7: Now find \(PQ\) using the distance formula.

For \(y = -3\):

\(PQ = \sqrt{(3 - (-10))^2 + (8 - (-3))^2} = \sqrt{13^2 + 11^2} = \sqrt{169 + 121} = \sqrt{290}\)

For \(y = -5\):

\(PQ = \sqrt{(3 - (-10))^2 + (8 - (-5))^2} = \sqrt{13^2 + 13^2} = \sqrt{169 + 169} = \sqrt{338}\)

Final Answer: \(y = -3\) or \(y = -5\). \(PQ = \sqrt{290}\) when \(y = -3\), and \(PQ = \sqrt{338}\) when \(y = -5\).

Step 1: Write the given vertices:

\(A(-8,4),\ B(-6,6),\ C(-3,9)\)

Step 2: To check if the area is zero, we must see if the three points lie on a straight line (collinear).

Step 3: Find slope of line AB:

\(m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{6 - 4}{-6 - (-8)} = \dfrac{2}{2} = 1\)

Step 4: Find slope of line BC:

\(m_{BC} = \dfrac{9 - 6}{-3 - (-6)} = \dfrac{3}{3} = 1\)

Step 5: Find slope of line AC:

\(m_{AC} = \dfrac{9 - 4}{-3 - (-8)} = \dfrac{5}{5} = 1\)

Step 6: Since all slopes are equal (\(m_{AB} = m_{BC} = m_{AC} = 1\)), all three points lie on the same straight line.

Step 7: If three points are collinear, the "triangle" formed by them has no area (it becomes a straight line).

Final Answer: Area of the triangle \(= 0\ \text{ square units}\).

Step 1: Let the required point on the x-axis be \(P(x,0)\) because every point on the x-axis has \(y = 0\).

Step 2: Suppose this point divides the line joining \(A(-4,-6)\) and \(B(-1,7)\) in the ratio \(m:n\).

Step 3: By the section formula, if a point divides the line joining \((x_1,y_1)\) and \((x_2,y_2)\) in the ratio \(m:n\), then the coordinates are

\[ \left( \dfrac{mx_2 + nx_1}{m+n},\; \dfrac{my_2 + ny_1}{m+n} \right) \]

Step 4: Apply this for \(A(-4,-6)\) and \(B(-1,7)\):

Coordinates of point = \( \Big( \dfrac{m(-1) + n(-4)}{m+n},\; \dfrac{m(7) + n(-6)}{m+n} \Big) \)

Step 5: Since the point lies on the x-axis, its \(y\)-coordinate must be 0.

So, \[ \dfrac{7m - 6n}{m+n} = 0 \]

Step 6: Multiply both sides by \((m+n)\):

\(7m - 6n = 0 \Rightarrow 7m = 6n\)

Therefore, \(m:n = 6:7\).

Step 7: Now find the \(x\)-coordinate:

\[ x = \dfrac{6(-1) + 7(-4)}{6+7} = \dfrac{-6 - 28}{13} = -\dfrac{34}{13} \]

Step 8: Since \(y = 0\), the point is \(\Big(-\dfrac{34}{13}, 0\Big)\).

Final Answer: Ratio is \(6:7\) and the point of division is \(\Big(-\dfrac{34}{13},0\Big)\).

Step 1: Recall the section formula.

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

\[ x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n} \]

Step 2: Write down the given values.

• \(A(x_1,y_1) = \big(\tfrac{1}{2}, \tfrac{3}{2}\big)\)
• \(B(x_2,y_2) = (2, -5)\)
• \(P(x,y) = \big(\tfrac{3}{4}, \tfrac{5}{12}\big)\)

Step 3: Apply the section formula for the \(x\)-coordinate.

\[ \tfrac{3}{4} = \frac{m(2) + n(\tfrac{1}{2})}{m+n} \]

Step 4: Simplify the numerator.

\[ \tfrac{3}{4} = \frac{2m + \tfrac{n}{2}}{m+n} \]

Step 5: Remove the fraction in the numerator.

Multiply numerator and denominator by 2: \[ \tfrac{3}{4} = \frac{4m + n}{2m + 2n} \]

Step 6: Cross multiply.

\[ 3(2m + 2n) = 4(4m + n) \]

\[ 6m + 6n = 16m + 4n \]

Step 7: Rearrange terms.

\[ 6n - 4n = 16m - 6m \] \[ 2n = 10m \]

Step 8: Simplify the ratio.

\[ n = 5m \quad \Rightarrow \quad m:n = 1:5 \]

Step 9: Verify with the \(y\)-coordinate (optional check).

Using \(y = \tfrac{my_2 + ny_1}{m+n}\): Substitute \(m:n = 1:5\), You will get \(y = \tfrac{5}{12}\), which matches the given \(P\).

Final Answer: The point \(P\) divides \(AB\) in the ratio \(1:5\) (that is, \(AP:PB = 1:5\)).

Step 1: Recall the section formula

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

\[ x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n} \]

Step 2: Write the given data

• \(A(3a+1, -3)\)
• \(B(8a, 5)\)
• Point \(P(9a-2, -b)\)
• Ratio \(3:1\) → so \(m=3,\ n=1\)

Step 3: Apply section formula for the x-coordinate

\[ x_P = \frac{3(8a) + 1(3a+1)}{3+1} = \frac{24a + 3a + 1}{4} = \frac{27a + 1}{4} \]

But we are also given that \(x_P = 9a - 2\).

So,

\[ 9a - 2 = \frac{27a + 1}{4} \]

Step 4: Solve for a

Multiply both sides by 4:

\[ 4(9a - 2) = 27a + 1 \] \[ 36a - 8 = 27a + 1 \] \[ 36a - 27a = 1 + 8 \] \[ 9a = 9 \quad \Rightarrow \quad a = 1 \]

Step 5: Apply section formula for the y-coordinate

\[ y_P = \frac{3(5) + 1(-3)}{4} = \frac{15 - 3}{4} = \frac{12}{4} = 3 \]

But we are given \(y_P = -b\).

\[ -b = 3 \quad \Rightarrow \quad b = -3 \]

Final Answer: \(a = 1, \ b = -3\)

Step 1: Recall the midpoint formula

The midpoint of a line joining points \((x_1, y_1)\) and \((x_2, y_2)\) is:

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

Step 2: Apply to given points

Point \(A(10, -6)\), point \(B(k, 4)\).

So, midpoint \((a,b)\) is:

\[ a = \dfrac{10 + k}{2}, \quad b = \dfrac{-6 + 4}{2} \]

Step 3: Simplify the \(y\)-coordinate of midpoint

\[ b = \dfrac{-6 + 4}{2} = \dfrac{-2}{2} = -1 \]

Step 4: Use the condition \(a - 2b = 18\)

Substitute values:

\[ a - 2b = 18 \]

\[ \dfrac{10 + k}{2} - 2(-1) = 18 \]

\[ \dfrac{10 + k}{2} + 2 = 18 \]

Step 5: Solve for \(k\)

Subtract 2 from both sides:

\[ \dfrac{10 + k}{2} = 16 \]

Multiply both sides by 2:

\[ 10 + k = 32 \]

Subtract 10:

\[ k = 22 \]

Step 6: Find length of \(|AB|\)

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

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

Substitute \(A(10,-6)\), \(B(22,4)\):

\[ |AB| = \sqrt{(22 - 10)^2 + (4 - (-6))^2} \]

\[ |AB| = \sqrt{(12)^2 + (10)^2} \]

\[ |AB| = \sqrt{144 + 100} = \sqrt{244} \]

\[ |AB| = 2\sqrt{61} \]

Step 1: The diameter of the circle is given as \(10\sqrt{2}\).

So, the radius is half of the diameter:

\[ r = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \]

Step 2: The formula for 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: The circle passes through the point \((11, -9)\). So, the distance between the centre \((2a, a - 7)\) and the point \((11, -9)\) must be equal to the radius.

\[ (11 - 2a)^2 + (-9 - (a - 7))^2 = (5\sqrt{2})^2 \]

Step 4: Simplify the right-hand side:

\[ (5\sqrt{2})^2 = 25 \times 2 = 50 \]

Step 5: Expand the terms:

\[ (11 - 2a)^2 + (-9 - a + 7)^2 = 50 \]

\[ (11 - 2a)^2 + (-a - 2)^2 = 50 \]

Step 6: Expand each square:

\[ (11 - 2a)^2 = (2a - 11)^2 = 4a^2 - 44a + 121 \]

\[ (-a - 2)^2 = (a + 2)^2 = a^2 + 4a + 4 \]

Step 7: Add them:

\[ 4a^2 - 44a + 121 + a^2 + 4a + 4 = 50 \]

\[ 5a^2 - 40a + 125 = 50 \]

Step 8: Simplify:

\[ 5a^2 - 40a + 75 = 0 \]

Step 9: Divide the whole equation by 5:

\[ a^2 - 8a + 15 = 0 \]

Step 10: Factorize:

\[ (a - 3)(a - 5) = 0 \]

Step 11: Solve for \(a\):

\[ a = 3 \quad \text{or} \quad a = 5 \]

Step 1: Recall the section formula.

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

\[ P = \left( \dfrac{mx_2 + nx_1}{m+n}, \; \dfrac{my_2 + ny_1}{m+n} \right) \]

Step 2: Write down what we know.

• Point \(R = (3, 2)\)
• Point \(B = (5, 1)\)
• Ratio \(RP:PB = 1:2\), so \(m=1, n=2\)

Step 3: Find the coordinates of \(P\).

\[ P = \left( \dfrac{1 \cdot 5 + 2 \cdot 3}{1+2}, \; \dfrac{1 \cdot 1 + 2 \cdot 2}{1+2} \right) \]

\[ P = \left( \dfrac{5 + 6}{3}, \; \dfrac{1 + 4}{3} \right) = \left( \dfrac{11}{3}, \; \dfrac{5}{3} \right) \]

Step 4: Use the line equation.

Point \(P\) lies on the line \(3x - 18y + k = 0\). Substituting \(x = 11/3, y = 5/3\):

\[ 3 \left(\dfrac{11}{3}\right) - 18 \left(\dfrac{5}{3}\right) + k = 0 \]

Step 5: Simplify.

\[ 11 - 30 + k = 0 \]

\[ -19 + k = 0 \]

Step 6: Solve for \(k\).

\[ k = 19 \]

Final Answer: \(k = 19\)

Step 1: We are told that \(D, E, F\) are the midpoints of the sides of \(\triangle ABC\). When we join these midpoints, we get another triangle inside, called the medial triangle (\(\triangle DEF\)).

Step 2: An important property: The area of the medial triangle is always one-fourth of the area of the original triangle. That means, if we can find the area of \(\triangle DEF\), then:

\[ \text{Area}(\triangle ABC) = 4 \times \text{Area}(\triangle DEF) \]

Step 3: To calculate \(\text{Area}(\triangle DEF)\), we use the shoelace formula (a method for finding area from coordinates).

The shoelace formula for points \((x_1,y_1), (x_2,y_2), (x_3,y_3)\) is:

\[ \text{Area} = \tfrac{1}{2}\,\Big|x_1y_2 + x_2y_3 + x_3y_1 - (y_1x_2 + y_2x_3 + y_3x_1)\Big| \]

Step 4: Put the coordinates of \(D, E, F\):

• \(D(-\tfrac{1}{2},\tfrac{5}{2})\)
• \(E(7,3)\)
• \(F(\tfrac{7}{2},\tfrac{7}{2})\)

Now calculate:

\[ = \tfrac{1}{2}\Big|(-\tfrac{1}{2})(3) + (7)(\tfrac{7}{2}) + (\tfrac{7}{2})(\tfrac{5}{2}) - \big[(\tfrac{5}{2})(7) + (3)(\tfrac{7}{2}) + (\tfrac{7}{2})(-\tfrac{1}{2})\big]\Big| \]

Step 5: Simplify step by step:

• First part: \((-\tfrac{1}{2})(3) = -\tfrac{3}{2}\)
• Next: \(7 \times \tfrac{7}{2} = \tfrac{49}{2}\)
• Next: \(\tfrac{7}{2} \times \tfrac{5}{2} = \tfrac{35}{4}\)

Total = \(-\tfrac{3}{2} + \tfrac{49}{2} + \tfrac{35}{4}\)

= \(\tfrac{-6 + 98 + 35}{4} = \tfrac{127}{4}\)

Second part:

• \((\tfrac{5}{2})(7) = \tfrac{35}{2}\)
• \((3)(\tfrac{7}{2}) = \tfrac{21}{2}\)
• \((\tfrac{7}{2})(-\tfrac{1}{2}) = -\tfrac{7}{4}\)

Total = \(\tfrac{35}{2} + \tfrac{21}{2} - \tfrac{7}{4} = \tfrac{70 + 42 - 7}{4} = \tfrac{105}{4}\)

Step 6: Put back into formula:

\[ \text{Area}(DEF) = \tfrac{1}{2} \times \Big|\tfrac{127}{4} - \tfrac{105}{4}\Big| = \tfrac{1}{2} \times \tfrac{22}{4} = \tfrac{11}{4} \]

Step 7: Finally, use the property:

\[ \text{Area}(ABC) = 4 \times \tfrac{11}{4} = 11 \]

Final Answer: The area of \(\triangle ABC\) is 11 square units.

Step 1: Write the coordinates of the points.

\(A(2,9),\ B(a,5),\ C(5,5)\).

Step 2: To check if the triangle is right-angled at \(B\), the lines \(BA\) and \(BC\) must be perpendicular.

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

\(\overrightarrow{BA} = (x_A - x_B,\; y_A - y_B) = (2 - a,\; 9 - 5) = (2 - a,\; 4)\).

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

\(\overrightarrow{BC} = (x_C - x_B,\; y_C - y_B) = (5 - a,\; 5 - 5) = (5 - a,\; 0)\).

Step 5: For perpendicularity, the dot product of the two vectors must be zero:

\(\overrightarrow{BA} \cdot \overrightarrow{BC} = (2 - a)(5 - a) + (4)(0) = (2 - a)(5 - a)\).

Step 6: Solve the equation:

\((2 - a)(5 - a) = 0\).

This gives \(a = 2\) or \(a = 5\).

Step 7: If \(a = 5\), then point B and C become the same point \((5,5)\). That means it will not form a triangle. So, we reject this case.

Therefore, \(a = 2\).

Step 8: Now calculate the area of right-angled triangle.

The legs of the triangle are \(BA\) and \(BC\).

Length of \(BA = \sqrt{(2 - 2)^2 + (9 - 5)^2} = \sqrt{0 + 16} = 4\,\text{units}\).

Length of \(BC = \sqrt{(5 - 2)^2 + (5 - 5)^2} = \sqrt{9 + 0} = 3\,\text{units}\).

Step 9: Area of right-angled triangle = \(\tfrac{1}{2} \times \text{base} \times \text{height}\).

\(= \tfrac{1}{2} \times 4 \times 3 = 6\,\text{square units}\).

Final Answer: \(a = 2\), area of \(\triangle ABC = 6\,\text{square units}\).

Step 1: Recall the idea.

The point \(R\) lies on the line segment from \(P(-1,3)\) to \(Q(2,5)\). We are told that the length \(PR\) is \(\dfrac{3}{5}\) of the total length \(PQ\). This means \(R\) divides the line from \(P\) to \(Q\) in the ratio \(3:2\).

Step 2: Write the coordinates of \(P\) and \(Q\).

\(P(x_1,y_1) = (-1,3), \; Q(x_2,y_2) = (2,5)\)

Step 3: Use the section formula.

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

\[ \Bigg(\dfrac{mx_2+nx_1}{m+n}, \; \dfrac{my_2+ny_1}{m+n}\Bigg) \]

Step 4: Identify the ratio.

Here, \(PR:PQ = \tfrac{3}{5}\). So, the ratio of \(PR:RQ = 3:2\). That means \(m=3\) and \(n=2\).

Step 5: Substitute the values.

\[ x = \dfrac{3(2)+2(-1)}{3+2} = \dfrac{6-2}{5} = \dfrac{4}{5} \]

\[ y = \dfrac{3(5)+2(3)}{3+2} = \dfrac{15+6}{5} = \dfrac{21}{5} \]

Step 6: Write the final coordinates.

So, the point \(R\) is:

\(R\big(\tfrac{4}{5},\tfrac{21}{5}\big)\)

Step 1: Recall that three points are collinear if the slope of AB = slope of BC.

Step 2: Write the coordinates of the points clearly:

• \(A(k+1,\,2k)\)
• \(B(3k,\,2k+3)\)
• \(C(5k-1,\,5k)\)

Step 3: Find slope of line AB:

\[ m_{AB} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{(2k+3) - (2k)}{3k - (k+1)} = \dfrac{3}{2k - 1} \]

Step 4: Find slope of line BC:

\[ m_{BC} = \dfrac{y_3 - y_2}{x_3 - x_2} = \dfrac{5k - (2k+3)}{(5k-1) - 3k} = \dfrac{3k - 3}{2k - 1} \]

Step 5: For collinearity, set \(m_{AB} = m_{BC}\):

\[ \dfrac{3}{2k - 1} = \dfrac{3k - 3}{2k - 1} \]

Step 6: Since denominator \((2k-1)\) is common, equate the numerators:

\[ 3 = 3k - 3 \]

Step 7: Solve for \(k\):

\[ 3k = 3 + 3 = 6 \quad \Rightarrow \quad k = 2 \]

Final Answer: \(k = 2\).

Step 1: We are given line \(2x + 3y - 5 = 0\) and points \(B(8, -9)\) and \(C(2, 1)\).

Step 2: To find the ratio in which the line divides \(BC\), we use the formula:

For a line \(ax + by + c = 0\), the ratio is

\[ |ax_1 + by_1 + c| : |ax_2 + by_2 + c| \]

where \((x_1,y_1)\) and \((x_2,y_2)\) are the coordinates of the two points.

Step 3: Put coordinates of point \(B(8,-9)\) into the line equation:

\(L(B) = 2(8) + 3(-9) - 5 = 16 - 27 - 5 = -16\).

Absolute value → \(|-16| = 16\).

Step 4: Put coordinates of point \(C(2,1)\):

\(L(C) = 2(2) + 3(1) - 5 = 4 + 3 - 5 = 2\).

Absolute value → \(|2| = 2\).

Step 5: Therefore, the ratio is

\(16 : 2 = 8 : 1\).

Step 6: To find the coordinates of the point dividing \(BC\) in the ratio \(8:1\), use the section formula:

If point divides \((x_1,y_1)\) and \((x_2,y_2)\) in ratio \(m:n\), then

\[ P = \left( \dfrac{mx_2 + nx_1}{m+n}, \dfrac{my_2 + ny_1}{m+n} \right) \]

Step 7: Here, \(m=8, n=1, x_1=8, y_1=-9, x_2=2, y_2=1\).

So,

\[ x = \dfrac{8(2) + 1(8)}{8+1} = \dfrac{16+8}{9} = \dfrac{24}{9} = \dfrac{8}{3} \]

\[ y = \dfrac{8(1) + 1(-9)}{9} = \dfrac{8-9}{9} = \dfrac{-1}{9} \]

Step 8: Thus, the point of division is

\(\big(\tfrac{8}{3}, -\tfrac{1}{9}\big)\).

Final Answer: Ratio = \(8:1\), Point = \(\left( \tfrac{8}{3}, -\tfrac{1}{9} \right)\).