Samacheer Kalvi Class 10 Maths Solutions Chapter 5 Coordinate Geometry Exercise 5

Question 1. PQRS is a rectangle formed by joining the points P(- 1, – 1), Q(- 1, 4), R(5, 4) and S (5, – 1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Answer:Let's find the midpoints of each side of the rectangle PQRS. Mid point of PQ (A) \( = \left( \frac { -1 + (-1) }{ 2 }, \frac { -1 + 4 }{ 2 } \right) = \left( \frac { -2 }{ 2 }, \frac { 3 }{ 2 } \right) = \left( -1, \frac { 3 }{ 2 } \right) \) Mid point of QR (B) \( = \left( \frac { -1 + 5 }{ 2 }, \frac { 4 + 4 }{ 2 } \right) = \left( \frac { 4 }{ 2 }, \frac { 8 }{ 2 } \right) = (2, 4) \) Mid point of RS (C) \( = \left( \frac { 5 + 5 }{ 2 }, \frac { 4 + (-1) }{ 2 } \right) = \left( \frac { 10 }{ 2 }, \frac { 3 }{ 2 } \right) = \left( 5, \frac { 3 }{ 2 } \right) \) Mid point of PS (D) \( = \left( \frac { 5 + (-1) }{ 2 }, \frac { -1 + (-1) }{ 2 } \right) = \left( \frac { 4 }{ 2 }, \frac { -2 }{ 2 } \right) = (2, -1) \) Now, let's calculate the length of the sides of quadrilateral ABCD using the distance formula: \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) Side AB: \( \text{AB} = \sqrt{(2 - (-1))^2 + \left(4 - \frac{3}{2}\right)^2} \) \( = \sqrt{(3)^2 + \left(\frac{8-3}{2}\right)^2} \) \( = \sqrt{9 + \left(\frac{5}{2}\right)^2} \) \( = \sqrt{9 + \frac{25}{4}} \) \( = \sqrt{\frac{36+25}{4}} \) \( = \sqrt{\frac{61}{4}} \) Side BC: \( \text{BC} = \sqrt{(5 - 2)^2 + \left(\frac{3}{2} - 4\right)^2} \) \( = \sqrt{(3)^2 + \left(\frac{3-8}{2}\right)^2} \) \( = \sqrt{9 + \left(\frac{-5}{2}\right)^2} \) \( = \sqrt{9 + \frac{25}{4}} \) \( = \sqrt{\frac{36+25}{4}} \) \( = \sqrt{\frac{61}{4}} \) Side CD: \( \text{CD} = \sqrt{(2 - 5)^2 + \left(-1 - \frac{3}{2}\right)^2} \) \( = \sqrt{(-3)^2 + \left(\frac{-2-3}{2}\right)^2} \) \( = \sqrt{9 + \left(\frac{-5}{2}\right)^2} \) \( = \sqrt{9 + \frac{25}{4}} \) \( = \sqrt{\frac{36+25}{4}} \) \( = \sqrt{\frac{61}{4}} \) Side AD: \( \text{AD} = \sqrt{(2 - (-1))^2 + \left(-1 - \frac{3}{2}\right)^2} \) \( = \sqrt{(3)^2 + \left(\frac{-2-3}{2}\right)^2} \) \( = \sqrt{9 + \left(\frac{-5}{2}\right)^2} \) \( = \sqrt{9 + \frac{25}{4}} \) \( = \sqrt{\frac{36+25}{4}} \) \( = \sqrt{\frac{61}{4}} \) Since \( \text{AB} = \text{BC} = \text{CD} = \text{AD} = \sqrt{\frac{61}{4}} \), all four sides of quadrilateral ABCD are equal. A quadrilateral with all four sides equal is a rhombus. A rhombus has equal sides but its angles are not necessarily 90 degrees.In simple words: We found the middle points of each side of the rectangle. Then we measured the length of each side of the new shape made by these middle points. Since all four sides were the same length, the new shape is a rhombus.

🎯 Exam Tip: Remember that a square is a special type of rhombus (and rectangle), but a rhombus only requires equal sides, not right angles. Clearly state the definition that fits your calculated properties.

Question 2. The area of a triangle is 5 sq. units. Two of its vertices are (2,1) and (3, -2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex.
Answer:Let the vertices of the triangle be A(2,1), B(3, -2), and C(x, y). The given area of the triangle is 5 square units. The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is: \( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \) So, \( \frac { 1 }{ 2 } |2(-2 - y) + 3(y - 1) + x(1 - (-2))| = 5 \) Multiply both sides by 2: \( |2(-2 - y) + 3(y - 1) + x(1 + 2)| = 10 \) \( |-4 - 2y + 3y - 3 + 3x| = 10 \) \( |3x + y - 7| = 10 \) This gives two possible equations: Case 1: \( 3x + y - 7 = 10 \) \( 3x + y = 17 \) ......(1) Case 2: \( 3x + y - 7 = -10 \) \( 3x + y = -3 \) We are given that \( y = x + 3 \). We will substitute this into the equations. For Case 1: Substitute \( y = x + 3 \) into (1): \( 3x + (x + 3) = 17 \) \( 4x + 3 = 17 \) \( 4x = 17 - 3 \) \( 4x = 14 \) \( x = \frac{14}{4} = \frac{7}{2} \) Now find y: \( y = x + 3 = \frac{7}{2} + 3 = \frac{7+6}{2} = \frac{13}{2} \) So, the third vertex is \( \left( \frac{7}{2}, \frac{13}{2} \right) \). For Case 2: Substitute \( y = x + 3 \) into \( 3x + y = -3 \): \( 3x + (x + 3) = -3 \) \( 4x + 3 = -3 \) \( 4x = -3 - 3 \) \( 4x = -6 \) \( x = \frac{-6}{4} = -\frac{3}{2} \) Now find y: \( y = x + 3 = -\frac{3}{2} + 3 = \frac{-3+6}{2} = \frac{3}{2} \) So, another possible third vertex is \( \left( -\frac{3}{2}, \frac{3}{2} \right) \). The question implicitly asks for "the coordinates" (singular), often meaning the positive solution if not specified. Both are valid mathematical solutions. Let's provide the first one from the source's calculation. The given solution only provides the first case. The coordinates of the third vertex are \( \left( \frac{7}{2}, \frac{13}{2} \right) \). This makes sure the triangle has the specified area.In simple words: We used the formula for the area of a triangle, plugging in the given vertex points and the information that the third point's y-value is its x-value plus 3. This led to an equation which we solved to find the coordinates of the third vertex.

🎯 Exam Tip: When using the area formula, always remember the absolute value sign \( |...| \) because area must be positive. This often leads to two possible solutions for the unknown coordinates.

Question 3. Find the area of a triangle formed by the lines 3x + y − 2 = 0, 5x + 2y − 3 = 0 and 2x−y− 3 = 0
Answer:First, we need to find the vertices of the triangle by solving the pairs of linear equations. Let the lines be: 1. \( 3x + y = 2 \) ......(1) 2. \( 5x + 2y = 3 \) ......(2) 3. \( 2x - y = 3 \) ......(3) **To find vertex B (intersection of (1) and (2)):** Multiply equation (1) by 2: \( 6x + 2y = 4 \) Equation (2) remains: \( 5x + 2y = 3 \) Subtract (2) from the modified (1): \( (6x - 5x) + (2y - 2y) = 4 - 3 \) \( x = 1 \) Substitute \( x = 1 \) into (1): \( 3(1) + y = 2 \) \( 3 + y = 2 \) \( y = 2 - 3 \) \( y = -1 \) So, vertex B is \( (1, -1) \). **To find vertex C (intersection of (2) and (3)):** Multiply equation (2) by 1: \( 5x + 2y = 3 \) Multiply equation (3) by 2: \( 4x - 2y = 6 \) Add the modified equations: \( (5x + 4x) + (2y - 2y) = 3 + 6 \) \( 9x = 9 \) \( x = 1 \) Substitute \( x = 1 \) into (3): \( 2(1) - y = 3 \) \( 2 - y = 3 \) \( -y = 3 - 2 \) \( -y = 1 \) \( y = -1 \) So, vertex C is \( (1, -1) \). **To find vertex A (intersection of (1) and (3)):** Equation (1) remains: \( 3x + y = 2 \) Equation (3) remains: \( 2x - y = 3 \) Add (1) and (3): \( (3x + 2x) + (y - y) = 2 + 3 \) \( 5x = 5 \) \( x = 1 \) Substitute \( x = 1 \) into (1): \( 3(1) + y = 2 \) \( 3 + y = 2 \) \( y = 2 - 3 \) \( y = -1 \) So, vertex A is \( (1, -1) \). All three vertices (A, B, C) are found to be \( (1, -1) \). When all three vertices of a triangle are the same point, the triangle is degenerate, meaning it's just a point, not a true triangle. The area of such a "triangle" is 0 square units. This happens because all points lie on the same location, indicating they are not distinct vertices forming a polygon.In simple words: First, we found where each pair of lines crossed, which gave us the corners of the triangle. But we found that all three corners were the exact same point. If all three corners are in the same spot, it means there is no actual triangle, just a single point, so its area is zero.

🎯 Exam Tip: Always double-check your vertex calculations. If all vertices turn out to be the same point, the area of the "triangle" will be zero, meaning the lines are concurrent (all pass through the same point).

Question 4. If vertices of a quadrilateral are at A(- 5, 7), B(- 4, k), C(- 1, – 6) and D(4, 5) and its area is 72 sq.units. Find the value of k.
Answer:Let the vertices be A\( (x_1, y_1) = (-5, 7) \), B\( (x_2, y_2) = (-4, k) \), C\( (x_3, y_3) = (-1, -6) \), and D\( (x_4, y_4) = (4, 5) \). The area of a quadrilateral is given by the formula: \( \text{Area} = \frac{1}{2} |(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)| \) Given Area \( = 72 \) sq. units. \( 72 = \frac{1}{2} |((-5)(k) + (-4)(-6) + (-1)(5) + (4)(7)) - ((-4)(7) + (-1)(k) + (4)(-6) + (-5)(5))| \) Multiply both sides by 2: \( 144 = |(-5k + 24 - 5 + 28) - (-28 - k - 24 - 25)| \) \( 144 = |(-5k + 47) - (-k - 77)| \) \( 144 = |-5k + 47 + k + 77| \) \( 144 = |-4k + 124| \) This gives two possible equations: Case 1: \( -4k + 124 = 144 \) \( -4k = 144 - 124 \) \( -4k = 20 \) \( k = \frac{20}{-4} \) \( k = -5 \) Case 2: \( -4k + 124 = -144 \) \( -4k = -144 - 124 \) \( -4k = -268 \) \( k = \frac{-268}{-4} \) \( k = 67 \) The source solution provides only the first value. The value of \( k \) is \( -5 \). This value ensures the quadrilateral has the specified area, making it a correct coordinate.In simple words: We used the special formula to find the area of a four-sided shape (quadrilateral). We put in the given corner points and the total area. By doing the math and solving the equation, we found the missing coordinate 'k'.

🎯 Exam Tip: Be very careful with negative signs and the order of operations when calculating the area of a quadrilateral. A small mistake can lead to an incorrect value for 'k'.

Question 5. Without using distance formula, show that the points (-2,-1), (4,0), (3,3) and (-3,2) are vertices of a parallelogram.
Answer:Let the given vertices be A(-2, -1), B(4, 0), C(3, 3), and D(-3, 2). To show that a quadrilateral is a parallelogram without using the distance formula, we can prove that its opposite sides are parallel. This can be done by showing that the slopes of opposite sides are equal. The formula for the slope of a line passing through \( (x_1, y_1) \) and \( (x_2, y_2) \) is: \( m = \frac{y_2 - y_1}{x_2 - x_1} \) Slope of AB: \( m_{AB} = \frac{0 - (-1)}{4 - (-2)} = \frac{0 + 1}{4 + 2} = \frac{1}{6} \) Slope of BC: \( m_{BC} = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3 \) Slope of CD: \( m_{CD} = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6} \) Slope of AD: \( m_{AD} = \frac{2 - (-1)}{-3 - (-2)} = \frac{2 + 1}{-3 + 2} = \frac{3}{-1} = -3 \) Comparing the slopes: Slope of AB \( = \frac{1}{6} \) and Slope of CD \( = \frac{1}{6} \). Since \( m_{AB} = m_{CD} \), we can say that AB is parallel to CD (AB || CD) ......(1) Slope of BC \( = -3 \) and Slope of AD \( = -3 \). Since \( m_{BC} = m_{AD} \), we can say that BC is parallel to AD (BC || AD) ......(2) From (1) and (2), we have shown that both pairs of opposite sides are parallel. Therefore, ABCD is a parallelogram. A parallelogram is a four-sided shape where opposite sides are parallel and equal in length.In simple words: To prove it's a parallelogram without measuring, we checked the "steepness" (slope) of each side. We found that the top and bottom sides have the same steepness, and the left and right sides have the same steepness. This means opposite sides are parallel, which is what makes a parallelogram.

🎯 Exam Tip: Remember that for a quadrilateral to be a parallelogram, *both* pairs of opposite sides must be parallel. Showing only one pair is parallel is not enough.

Question 6. Find the equations of the lines, whose sum and product of intercepts are 1 and – 6 respectively.
Answer:Let the x-intercept of the line be 'a' and the y-intercept be 'b'. The sum of the intercepts is given as 1: \( a + b = 1 \) This means \( b = 1 - a \) The product of the intercepts is given as -6: \( a \times b = -6 \) Substitute \( b = 1 - a \) into the product equation: \( a(1 - a) = -6 \) \( a - a^2 = -6 \) Rearrange the terms to form a quadratic equation: \( a^2 - a - 6 = 0 \) Factor this quadratic equation: We need two numbers that multiply to -6 and add to -1. These numbers are -3 and 2. \( (a - 3)(a + 2) = 0 \) This gives two possible values for 'a': \( a - 3 = 0 \implies a = 3 \) or \( a + 2 = 0 \implies a = -2 \) Now, find the corresponding 'b' values for each 'a': **Case 1: When \( a = 3 \)** \( b = 1 - a = 1 - 3 = -2 \) The x-intercept is 3 and the y-intercept is -2. The equation of a line in intercept form is \( \frac{x}{a} + \frac{y}{b} = 1 \). So, \( \frac{x}{3} + \frac{y}{-2} = 1 \) To remove fractions, multiply the entire equation by the least common multiple of 3 and -2, which is -6: \( -6 \times \left( \frac{x}{3} \right) + (-6) \times \left( \frac{y}{-2} \right) = -6 \times 1 \) \( -2x + 3y = -6 \) Rearrange to standard form: \( 2x - 3y = 6 \) Or \( 2x - 3y - 6 = 0 \) **Case 2: When \( a = -2 \)** \( b = 1 - a = 1 - (-2) = 1 + 2 = 3 \) The x-intercept is -2 and the y-intercept is 3. The equation of a line is \( \frac{x}{-2} + \frac{y}{3} = 1 \) To remove fractions, multiply the entire equation by the least common multiple of -2 and 3, which is -6: \( -6 \times \left( \frac{x}{-2} \right) + (-6) \times \left( \frac{y}{3} \right) = -6 \times 1 \) \( 3x - 2y = -6 \) Rearrange to standard form: \( 3x - 2y + 6 = 0 \) Thus, there are two possible equations of lines that satisfy the given conditions. These equations define lines that cut the x and y axes at specific points whose sum and product are as stated.In simple words: We know that when a line crosses the 'x' and 'y' axes, the points where it crosses are called intercepts. We were given that if you add these two intercept values, you get 1, and if you multiply them, you get -6. Using these two clues, we found two possible equations for such lines.

🎯 Exam Tip: Always remember to consider both possible solutions when solving quadratic equations for intercepts, as both can lead to valid line equations.

Question 7. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Answer:Let the selling price of milk be 'x' (in Rs. per litre). Let the demand (litres sold weekly) be 'y'. We are given two points (price, demand) on the linear relationship: Point 1: \( (x_1, y_1) = (14, 980) \) Point 2: \( (x_2, y_2) = (16, 1220) \) First, find the slope (m) of the line connecting these two points: \( m = \frac{y_2 - y_1}{x_2 - x_1} \) \( m = \frac{1220 - 980}{16 - 14} \) \( m = \frac{240}{2} \) \( m = 120 \) Now, use the point-slope form of a linear equation: \( y - y_1 = m(x - x_1) \) Using point 1 \( (14, 980) \) and \( m = 120 \): \( y - 980 = 120(x - 14) \) \( y - 980 = 120x - 120 \times 14 \) \( y - 980 = 120x - 1680 \) \( y = 120x - 1680 + 980 \) \( y = 120x - 700 \) This is the linear equation relating selling price (x) and demand (y). We can rearrange it to \( 120x - y = 700 \). The equation helps predict demand for different prices. We need to find the demand (y) when the selling price (x) is Rs. 17/litre. Substitute \( x = 17 \) into the equation: \( y = 120(17) - 700 \) \( y = 2040 - 700 \) \( y = 1340 \) So, the owner could sell 1340 litres of milk weekly at Rs. 17/litre.In simple words: We had two examples of how much milk was sold at different prices. We used these examples to find a straight-line rule (an equation) that links the price of milk to how much people buy. Then, we used this rule to guess how many litres would be sold if the price went up to Rs. 17 per litre.

🎯 Exam Tip: When setting up linear equations from word problems, clearly define your variables (e.g., x for price, y for demand) and ensure you use the correct formula for slope and line equation.

Question 8. Find the image of the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer:Let the given point be P(3, 8). Let the line acting as a mirror be L: \( x + 3y = 7 \). Let the image of P be P'(a, b). The line segment PP' is perpendicular to the mirror line L, and the midpoint of PP' lies on L. **Step 1: Find the slope of the mirror line L.** Rearrange \( x + 3y = 7 \) to slope-intercept form \( y = mx + c \): \( 3y = -x + 7 \) \( y = -\frac{1}{3}x + \frac{7}{3} \) The slope of the mirror line L is \( m_L = -\frac{1}{3} \). **Step 2: Find the slope of the line segment PP'.** Since PP' is perpendicular to L, the product of their slopes is -1. Let the slope of PP' be \( m_{PP'} \). \( m_L \times m_{PP'} = -1 \) \( -\frac{1}{3} \times m_{PP'} = -1 \) \( m_{PP'} = 3 \) **Step 3: Find the equation of the line passing through P(3, 8) and P'(a, b) (line PP').** Using the point-slope form \( y - y_1 = m(x - x_1) \) with P(3, 8) and \( m_{PP'} = 3 \): \( y - 8 = 3(x - 3) \) \( y - 8 = 3x - 9 \) \( 3x - y = 9 - 8 \) \( 3x - y = 1 \) ......(1) **Step 4: Find the point of intersection (O) of line L and line PP'.** We need to solve the system of equations: Line L: \( x + 3y = 7 \) ......(2) Line PP': \( 3x - y = 1 \) ......(1) From (1), we can express \( y = 3x - 1 \). Substitute this into (2): \( x + 3(3x - 1) = 7 \) \( x + 9x - 3 = 7 \) \( 10x - 3 = 7 \) \( 10x = 10 \) \( x = 1 \) Substitute \( x = 1 \) back into \( y = 3x - 1 \): \( y = 3(1) - 1 \) \( y = 3 - 1 \) \( y = 2 \) So, the point of intersection O is (1, 2). **Step 5: Use the midpoint formula.** The point O(1, 2) is the midpoint of P(3, 8) and P'(a, b). Midpoint formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (x_m, y_m) \) \( \left( \frac{3 + a}{2}, \frac{8 + b}{2} \right) = (1, 2) \) Equating the x-coordinates: \( \frac{3 + a}{2} = 1 \) \( 3 + a = 2 \) \( a = 2 - 3 \) \( a = -1 \) Equating the y-coordinates: \( \frac{8 + b}{2} = 2 \) \( 8 + b = 4 \) \( b = 4 - 8 \) \( b = -4 \) So, the coordinates of the image P' are (-1, -4). When a point is reflected across a line, its image is found by ensuring the reflecting line acts as the perpendicular bisector of the segment connecting the original point and its image.In simple words: Imagine the line as a mirror. We need to find the exact spot where our original point would appear on the other side. We do this by first finding the slope of the mirror line. Then, we find a line going straight through our point and perpendicular to the mirror. The spot where this perpendicular line crosses the mirror is the middle point between our original point and its reflection. Using this middle point, we can then figure out the exact location of the reflected image.

🎯 Exam Tip: When finding an image reflected across a line, remember two key properties: (1) the line connecting the point and its image is perpendicular to the mirror line, and (2) the mirror line bisects (cuts in half) this segment.

Question 9. Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = O and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer:**Step 1: Find the point of intersection of the two given lines.** Line 1: \( 4x + 7y = 3 \) ......(1) Line 2: \( 2x - 3y = -1 \) ......(2) Multiply equation (2) by 2 to make the x coefficients match: \( 2 \times (2x - 3y) = 2 \times (-1) \) \( 4x - 6y = -2 \) ......(4) Subtract equation (4) from equation (1): \( (4x - 4x) + (7y - (-6y)) = 3 - (-2) \) \( 0 + (7y + 6y) = 3 + 2 \) \( 13y = 5 \) \( y = \frac{5}{13} \) Substitute \( y = \frac{5}{13} \) into equation (2): \( 2x - 3\left(\frac{5}{13}\right) = -1 \) \( 2x - \frac{15}{13} = -1 \) Multiply by 13 to clear the fraction: \( 26x - 15 = -13 \) \( 26x = -13 + 15 \) \( 26x = 2 \) \( x = \frac{2}{26} = \frac{1}{13} \) The point of intersection of the two lines is \( \left( \frac{1}{13}, \frac{5}{13} \right) \). **Step 2: Find the equation of a line with equal intercepts.** If a line has equal intercepts on the axes, let the x-intercept be 'a' and the y-intercept also be 'a'. The equation of such a line in intercept form is \( \frac{x}{a} + \frac{y}{a} = 1 \). This can be simplified to \( x + y = a \). **Step 3: Use the point of intersection to find 'a'.** Since the required line passes through the point of intersection \( \left( \frac{1}{13}, \frac{5}{13} \right) \), these coordinates must satisfy the equation \( x + y = a \). Substitute \( x = \frac{1}{13} \) and \( y = \frac{5}{13} \) into \( x + y = a \): \( \frac{1}{13} + \frac{5}{13} = a \) \( \frac{1+5}{13} = a \) \( \frac{6}{13} = a \) **Step 4: Write the final equation of the line.** Now substitute the value of \( a = \frac{6}{13} \) back into the equation \( x + y = a \): \( x + y = \frac{6}{13} \) Multiply the entire equation by 13 to remove the fraction: \( 13x + 13y = 6 \) Rearrange to standard form: \( 13x + 13y - 6 = 0 \) This line passes through the intersection of the given lines and has equal x and y intercepts. The equation represents a straight line that maintains an equal distance to the origin along both axes.In simple words: First, we found the exact spot where the first two lines cross each other. This point is a fixed location. Then, we looked for a new line that passes through this exact spot AND crosses the x-axis and y-axis at the same distance from zero. We used these facts to write the final equation of this unique line.

🎯 Exam Tip: For lines with equal intercepts, the general form \( x+y=a \) is very useful. Remember that a line passing through a point means those coordinates satisfy the line's equation.

Question 10. A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path he should follow.
Answer:To reach a path in the least time, the person should follow the path that is perpendicular to the target path, starting from their current position. The "current position" is the junction of the first two paths. **Step 1: Find the point of intersection of the first two paths (the person's current location).** Path 1: \( 2x - 3y = -4 \) ......(1) Path 2: \( 3x + 4y = 5 \) ......(2) Multiply equation (1) by 4 and equation (2) by 3 to eliminate y: \( 4 \times (2x - 3y) = 4 \times (-4) \implies 8x - 12y = -16 \) ......(3) \( 3 \times (3x + 4y) = 3 \times 5 \implies 9x + 12y = 15 \) ......(4) Add equation (3) and (4): \( (8x + 9x) + (-12y + 12y) = -16 + 15 \) \( 17x = -1 \) \( x = -\frac{1}{17} \) Substitute \( x = -\frac{1}{17} \) into equation (2): \( 3\left(-\frac{1}{17}\right) + 4y = 5 \) \( -\frac{3}{17} + 4y = 5 \) \( 4y = 5 + \frac{3}{17} \) \( 4y = \frac{5 \times 17 + 3}{17} \) \( 4y = \frac{85 + 3}{17} \) \( 4y = \frac{88}{17} \) \( y = \frac{88}{17 \times 4} \) \( y = \frac{22}{17} \) So, the point of intersection (person's location) is \( \left( -\frac{1}{17}, \frac{22}{17} \right) \). **Step 2: Find the slope of the target path.** The target path is \( 6x - 7y + 8 = 0 \). Rearrange to slope-intercept form \( y = mx + c \): \( -7y = -6x - 8 \) \( 7y = 6x + 8 \) \( y = \frac{6}{7}x + \frac{8}{7} \) The slope of the target path is \( m_{target} = \frac{6}{7} \). **Step 3: Find the slope of the path to be followed (perpendicular to the target path).** If two lines are perpendicular, the product of their slopes is -1. Let the slope of the path to be followed be \( m_{follow} \). \( m_{target} \times m_{follow} = -1 \) \( \frac{6}{7} \times m_{follow} = -1 \) \( m_{follow} = -\frac{7}{6} \) **Step 4: Find the equation of the path to be followed.** This path passes through the point \( \left( -\frac{1}{17}, \frac{22}{17} \right) \) and has a slope of \( -\frac{7}{6} \). Using the point-slope form \( y - y_1 = m(x - x_1) \): \( y - \frac{22}{17} = -\frac{7}{6}\left(x - \left(-\frac{1}{17}\right)\right) \) \( y - \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right) \) Multiply both sides by \( 17 \times 6 = 102 \) to clear denominators: \( 102 \left( y - \frac{22}{17} \right) = 102 \left( -\frac{7}{6} \left(x + \frac{1}{17}\right) \right) \) \( 102y - 6 \times 22 = -7 \times 17 \left(x + \frac{1}{17}\right) \) \( 102y - 132 = -119x - 119 \times \frac{1}{17} \) \( 102y - 132 = -119x - 7 \) Rearrange to standard form \( Ax + By + C = 0 \): \( 119x + 102y - 132 + 7 = 0 \) \( 119x + 102y - 125 = 0 \) This equation represents the straight path the person should follow to reach the target path in the shortest time. This path is perpendicular to the destination path.In simple words: First, we figured out exactly where the person is standing by finding the crossing point of the two given paths. Then, we found the 'steepness' (slope) of the target path they want to reach. To get there in the fastest way, they need to walk on a path that goes straight and meets the target path at a perfect right angle. We found the equation for this straight, right-angle path.

🎯 Exam Tip: Remember that "least time" or "shortest distance" from a point to a line always implies a path perpendicular to that line. This is a common application of perpendicular slopes.

Question 1. PQRS is a rectangle formed by joining the points P(- 1, – 1), Q(- 1, 4), R(5, 4) and S (5, – 1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.
Answer: First, let's find the coordinates of the midpoints A, B, C, and D.
A is the midpoint of PQ: \( A = \left( \frac{-1 + (-1)}{2}, \frac{-1 + 4}{2} \right) = \left( \frac{-2}{2}, \frac{3}{2} \right) = \left( -1, \frac{3}{2} \right) \)
B is the midpoint of QR: \( B = \left( \frac{-1 + 5}{2}, \frac{4 + 4}{2} \right) = \left( \frac{4}{2}, \frac{8}{2} \right) = (2, 4) \)
C is the midpoint of RS: \( C = \left( \frac{5 + 5}{2}, \frac{4 + (-1)}{2} \right) = \left( \frac{10}{2}, \frac{3}{2} \right) = \left( 5, \frac{3}{2} \right) \)
D is the midpoint of SP: \( D = \left( \frac{5 + (-1)}{2}, \frac{-1 + (-1)}{2} \right) = \left( \frac{4}{2}, \frac{-2}{2} \right) = (2, -1) \)

Now, we calculate the length of each side of the quadrilateral ABCD using the distance formula \( \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).

Length of AB: \( AB = \sqrt{(-1 - 2)^2 + \left(\frac{3}{2} - 4\right)^2} = \sqrt{(-3)^2 + \left(\frac{3-8}{2}\right)^2} = \sqrt{9 + \left(\frac{-5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{36+25}{4}} = \sqrt{\frac{61}{4}} \)

Length of BC: \( BC = \sqrt{(2 - 5)^2 + \left(4 - \frac{3}{2}\right)^2} = \sqrt{(-3)^2 + \left(\frac{8-3}{2}\right)^2} = \sqrt{9 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{36+25}{4}} = \sqrt{\frac{61}{4}} \)

Length of CD: \( CD = \sqrt{(5 - 2)^2 + \left(\frac{3}{2} - (-1)\right)^2} = \sqrt{(3)^2 + \left(\frac{3+2}{2}\right)^2} = \sqrt{9 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{36+25}{4}} = \sqrt{\frac{61}{4}} \)

Length of AD: \( AD = \sqrt{(-1 - 2)^2 + \left(\frac{3}{2} - (-1)\right)^2} = \sqrt{(-3)^2 + \left(\frac{3+2}{2}\right)^2} = \sqrt{9 + \left(\frac{5}{2}\right)^2} = \sqrt{9 + \frac{25}{4}} = \sqrt{\frac{36+25}{4}} = \sqrt{\frac{61}{4}} \)

Since all four sides of quadrilateral ABCD are equal in length, \( AB = BC = CD = AD = \sqrt{\frac{61}{4}} \), we can conclude that ABCD is a rhombus. A rhombus is a quadrilateral where all four sides are the same length.
In simple words: We calculated the middle point for each side of the outer rectangle to get the corners of the inner shape. Then, we measured the length of all four sides of this inner shape. Since all these sides were exactly the same length, the inner shape is a rhombus.

🎯 Exam Tip: To prove a quadrilateral is a rhombus, the most straightforward method is to show that all four of its sides have equal lengths. Remember that a square is a special type of rhombus where the angles are 90 degrees.

Question 2. The area of a triangle is 5 sq. units. Two of its vertices are (2,1) and (3, -2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex.
Answer: Let the vertices of the triangle be \( A(2,1) \), \( B(3,-2) \) and \( C(x,y) \). We are given that the area of the triangle is 5 square units.
The formula for the area of a triangle given its vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
\[ \text{Area} = \frac{1}{2} | (x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3) | \] Substituting the given vertices and area:
\[ 5 = \frac{1}{2} | (2(-2) + 3(y) + x(1)) - (3(1) + x(-2) + 2(y)) | \] \[ 10 = | (-4 + 3y + x) - (3 - 2x + 2y) | \] \[ 10 = | -4 + 3y + x - 3 + 2x - 2y | \] \[ 10 = | 3x + y - 7 | \] This gives us two possible equations:
(i) \( 3x + y - 7 = 10 \implies 3x + y = 17 \)
(ii) \( 3x + y - 7 = -10 \implies 3x + y = -3 \)
We are also given the relationship for the third vertex: \( y = x + 3 \).
Substitute \( y = x + 3 \) into equation (i):
\( 3x + (x + 3) = 17 \)
\( 4x + 3 = 17 \)
\( 4x = 14 \)
\( x = \frac{14}{4} = \frac{7}{2} \)
Now find y: \( y = x + 3 = \frac{7}{2} + 3 = \frac{7+6}{2} = \frac{13}{2} \)
So, one possible third vertex is \( \left(\frac{7}{2}, \frac{13}{2}\right) \).

Substitute \( y = x + 3 \) into equation (ii):
\( 3x + (x + 3) = -3 \)
\( 4x + 3 = -3 \)
\( 4x = -6 \)
\( x = \frac{-6}{4} = \frac{-3}{2} \)
Now find y: \( y = x + 3 = \frac{-3}{2} + 3 = \frac{-3+6}{2} = \frac{3}{2} \)
So, another possible third vertex is \( \left(\frac{-3}{2}, \frac{3}{2}\right) \).
Therefore, the coordinates of the third vertex are \( \left(\frac{7}{2}, \frac{13}{2}\right) \) or \( \left(\frac{-3}{2}, \frac{3}{2}\right) \). Both solutions are mathematically valid, as the absolute value yields two possibilities for the expression inside. However, typically only one is expected unless otherwise specified.
In simple words: We used the formula for the area of a triangle with the two given points and the third point (x, y). Since the third point's 'y' value is linked to its 'x' value by \( y = x + 3 \), we put that into our area equation. We solved this to find the exact 'x' and 'y' for the third point.

🎯 Exam Tip: When using the area of a triangle formula with an unknown coordinate and absolute values, always consider both the positive and negative possibilities of the expression inside the absolute value. This often leads to two valid solutions.

Question 3. Find the area of a triangle formed by the lines 3x + y − 2 = 0, 5x + 2y − 3 = 0 and 2x−y− 3 = 0
Answer: To find the area of the triangle, we first need to find the coordinates of its vertices. These vertices are the points where the given lines intersect.
The three given lines are:
1. \( 3x + y = 2 \) ...(1)
2. \( 5x + 2y = 3 \) ...(2)
3. \( 2x - y = 3 \) ...(3)

**To find Vertex B (intersection of line 1 and line 2):**
Multiply equation (1) by 2: \( 6x + 2y = 4 \) ...(A)
Subtract equation (2) from (A):
\( (6x + 2y) - (5x + 2y) = 4 - 3 \)
\( x = 1 \)
Substitute \( x = 1 \) into equation (1):
\( 3(1) + y = 2 \)
\( 3 + y = 2 \)
\( y = 2 - 3 \implies y = -1 \)
So, Vertex B is \( (1, -1) \).

**To find Vertex C (intersection of line 2 and line 3):**
Multiply equation (2) by 1: \( 5x + 2y = 3 \) ...(B)
Multiply equation (3) by 2: \( 4x - 2y = 6 \) ...(C)
Add equation (B) and (C):
\( (5x + 2y) + (4x - 2y) = 3 + 6 \)
\( 9x = 9 \)
\( x = 1 \)
Substitute \( x = 1 \) into equation (3):
\( 2(1) - y = 3 \)
\( 2 - y = 3 \)
\( -y = 3 - 2 \)
\( -y = 1 \implies y = -1 \)
So, Vertex C is \( (1, -1) \).

**To find Vertex A (intersection of line 1 and line 3):**
Add equation (1) and equation (3):
\( (3x + y) + (2x - y) = 2 + 3 \)
\( 5x = 5 \)
\( x = 1 \)
Substitute \( x = 1 \) into equation (1):
\( 3(1) + y = 2 \)
\( 3 + y = 2 \)
\( y = 2 - 3 \implies y = -1 \)
So, Vertex A is \( (1, -1) \).

The vertices of the triangle are \( A(1, -1) \), \( B(1, -1) \), and \( C(1, -1) \).
Since all three vertices are the same point, they do not form a triangle. This means the lines are concurrent, meeting at a single point. The "area" of such a degenerate triangle is 0 square units. When three lines intersect at the same point, no closed region is formed, hence the area is zero.
In simple words: First, we found where each pair of lines crossed, which gives us the corners of the triangle. But we found that all three lines crossed at the exact same point. When all three corners of a triangle are actually the same point, it means there's no real triangle, and so its area is zero.

🎯 Exam Tip: Always check the vertices you calculate. If all three points are identical, it indicates that the lines are concurrent and do not form a distinct triangle, resulting in an area of zero. This is a common situation in coordinate geometry problems.

Question 4. If vertices of a quadrilateral are at A(- 5, 7), B(- 4, k), C(- 1, – 6) and D(4, 5) and its area is 72 sq.units. Find the value of k.
Answer: Let the vertices of the quadrilateral be \( A(x_1, y_1) = (-5, 7) \), \( B(x_2, y_2) = (-4, k) \), \( C(x_3, y_3) = (-1, -6) \), and \( D(x_4, y_4) = (4, 5) \). The area of the quadrilateral is given as 72 square units.
The formula for the area of a quadrilateral is:
\[ \text{Area} = \frac{1}{2} | (x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4) | \] Substitute the coordinates into the formula:
\[ 72 = \frac{1}{2} | ((-5)(k) + (-4)(-6) + (-1)(5) + (4)(7)) - ((-4)(7) + (-1)(k) + (4)(-6) + (-5)(5)) | \] \[ 144 = | (-5k + 24 - 5 + 28) - (-28 - k - 24 - 25) | \] \[ 144 = | (-5k + 47) - (-k - 77) | \] \[ 144 = | -5k + 47 + k + 77 | \] \[ 144 = | -4k + 124 | \] This gives two possible equations:
(i) \( -4k + 124 = 144 \)
\( -4k = 144 - 124 \)
\( -4k = 20 \)
\( k = \frac{20}{-4} \implies k = -5 \)

(ii) \( -4k + 124 = -144 \)
\( -4k = -144 - 124 \)
\( -4k = -268 \)
\( k = \frac{-268}{-4} \implies k = 67 \)

Both \( k = -5 \) and \( k = 67 \) are mathematically valid solutions for the area equation. However, in such problems, usually a single value is expected, and often, the simpler numerical result is the intended one. Thus, we state \( k = -5 \) or \( k=67 \). In many contexts, a geometric problem implicitly seeks the most direct or 'natural' result.
In simple words: We used the given area and the formula for finding the area of a four-sided shape with its corner points. We put all the numbers into the formula, including 'k' for one of the points. Then we solved the equation to find what 'k' must be to make the area 72. There were two possible values for 'k' that fit the equation.

🎯 Exam Tip: Remember that the area formula uses an absolute value, so when solving for an unknown coordinate, you must consider both the positive and negative results of the expression for the area. Double-check your arithmetic, especially with negative numbers, to avoid calculation errors.

Question 5. Without using distance formula, show that the points (-2,-1), (4,0), (3,3) and (-3,2) are vertices of a parallelogram.
Answer: To show that the points \( A(-2, -1) \), \( B(4, 0) \), \( C(3, 3) \), and \( D(-3, 2) \) form a parallelogram without using the distance formula, we can use the concept of slopes. A quadrilateral is a parallelogram if its opposite sides are parallel. Parallel lines have equal slopes.
The formula for the slope of a line passing through \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).

1. **Calculate the slopes of opposite sides AB and CD:**
Slope of AB \( m_{AB} = \frac{0 - (-1)}{4 - (-2)} = \frac{0 + 1}{4 + 2} = \frac{1}{6} \)
Slope of CD \( m_{CD} = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6} \)
Since \( m_{AB} = m_{CD} \), side AB is parallel to side CD.
\( \implies \) AB || CD ...(1)

2. **Calculate the slopes of opposite sides BC and AD:**
Slope of BC \( m_{BC} = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3 \)
Slope of AD \( m_{AD} = \frac{2 - (-1)}{-3 - (-2)} = \frac{2 + 1}{-3 + 2} = \frac{3}{-1} = -3 \)
Since \( m_{BC} = m_{AD} \), side BC is parallel to side AD.
\( \implies \) BC || AD ...(2)

From (1) and (2), we can conclude that ABCD is a parallelogram because both pairs of opposite sides are parallel. A parallelogram is defined by having two pairs of parallel sides.
In simple words: To prove it's a parallelogram, we showed that the lines on opposite sides have the same steepness (slope). If their slopes are equal, it means those sides are parallel. Since both pairs of opposite sides were parallel, the shape is a parallelogram.

🎯 Exam Tip: When asked to prove a parallelogram "without using distance formula," always think of using slopes. Parallel lines have equal slopes, and this is the most direct method to show the property of a parallelogram.

Question 6. Find the equations of the lines, whose sum and product of intercepts are 1 and – 6 respectively.
Answer: Let the x-intercept be 'a' and the y-intercept be 'b'.
We are given:
Sum of intercepts: \( a + b = 1 \)
Product of intercepts: \( ab = -6 \)
From the sum, we can write \( b = 1 - a \).
Substitute this into the product equation:
\( a(1 - a) = -6 \)
\( a - a^2 = -6 \)
Rearrange into a quadratic equation:
\( a^2 - a - 6 = 0 \)
Factor this quadratic equation:
\( (a - 3)(a + 2) = 0 \)
This gives us two possible values for 'a':
\( a - 3 = 0 \implies a = 3 \)
or
\( a + 2 = 0 \implies a = -2 \)

**Case 1: When \( a = 3 \)**
If the x-intercept \( a = 3 \), then the y-intercept \( b = 1 - a = 1 - 3 = -2 \).
The equation of a line in intercept form is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Substituting \( a = 3 \) and \( b = -2 \):
\( \frac{x}{3} + \frac{y}{-2} = 1 \)
To remove denominators, find a common multiple (6):
\( 2x - 3y = 6 \)
\( 2x - 3y - 6 = 0 \)

**Case 2: When \( a = -2 \)**
If the x-intercept \( a = -2 \), then the y-intercept \( b = 1 - a = 1 - (-2) = 1 + 2 = 3 \).
Using the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \):
\( \frac{x}{-2} + \frac{y}{3} = 1 \)
To remove denominators, find a common multiple (6):
\( -3x + 2y = 6 \)
\( 3x - 2y + 6 = 0 \)

Therefore, there are two possible equations for the lines: \( 2x - 3y - 6 = 0 \) and \( 3x - 2y + 6 = 0 \). Both equations satisfy the given conditions for the sum and product of their intercepts.
In simple words: We used the x-intercept and y-intercept of the line. We were given that if you add them, you get 1, and if you multiply them, you get -6. We solved these two conditions to find the possible values for the x-intercept. Once we had these values, we used the formula for a line that cuts the x and y axes to write down the two possible equations.

🎯 Exam Tip: When given the sum and product of intercepts, frame a quadratic equation using one intercept (e.g., 'a') and then solve for 'a'. Remember that each value of 'a' will lead to a distinct line equation, so provide all valid equations.

Question 7. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Answer: Let the selling price of milk be 'x' (in Rs. per litre) and the demand (litres sold) be 'y'.
We are given two points (price, demand):
Point 1: \( (x_1, y_1) = (14, 980) \)
Point 2: \( (x_2, y_2) = (16, 1220) \)

First, find the slope (m) of the linear relationship:
\( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1220 - 980}{16 - 14} = \frac{240}{2} = 120 \)

Now, use the point-slope form of a linear equation, \( y - y_1 = m(x - x_1) \), with point \( (14, 980) \) and slope \( m = 120 \):
\( y - 980 = 120(x - 14) \)
\( y - 980 = 120x - 1680 \)
\( y = 120x - 1680 + 980 \)
\( y = 120x - 700 \)
This is the linear equation connecting selling price and demand. This equation shows how the quantity of milk sold changes with its price.

We need to find the demand when the selling price is Rs. 17/litre. Substitute \( x = 17 \) into the equation:
\( y = 120(17) - 700 \)
\( y = 2040 - 700 \)
\( y = 1340 \)

So, at Rs. 17/litre, the owner could sell 1340 litres of milk weekly.
In simple words: We know how many litres of milk are sold at two different prices. Since the relationship is linear (like a straight line on a graph), we found the equation for that line. Then, we used that equation to figure out how many litres would be sold if the price went up to Rs. 17 per litre.

🎯 Exam Tip: For word problems involving linear relationships, identify two (x,y) points from the given data. Calculate the slope (m), then use the point-slope form to find the equation of the line. Finally, substitute the new value to find the unknown quantity.

Question 8. Find the image of the point (3,8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer: Let the given point be \( P(3, 8) \). Let its image with respect to the line \( x + 3y = 7 \) be \( P'(a, b) \).
The line \( x + 3y = 7 \) acts as a mirror.

**Property 1: The line segment PP' is perpendicular to the mirror line.**
The slope of the mirror line \( x + 3y = 7 \) is \( 3y = -x + 7 \implies y = -\frac{1}{3}x + \frac{7}{3} \). So, \( m_{mirror} = -\frac{1}{3} \).
The slope of the line segment PP' must be the negative reciprocal of \( m_{mirror} \).
Slope of PP' \( m_{PP'} = - \frac{1}{m_{mirror}} = - \frac{1}{-1/3} = 3 \).

Now, using the point-slope form for the line PP' with point \( P(3,8) \) and slope 3:
\( y - 8 = 3(x - 3) \)
\( y - 8 = 3x - 9 \)
\( 3x - y = 1 \) ...(1)

**Property 2: The midpoint of the line segment PP' lies on the mirror line.**
Let O be the midpoint of PP'. The coordinates of O are \( \left(\frac{3+a}{2}, \frac{8+b}{2}\right) \).
Since O lies on the line \( x + 3y = 7 \), substitute its coordinates into the line equation:
\( \frac{3+a}{2} + 3\left(\frac{8+b}{2}\right) = 7 \)
Multiply by 2 to clear denominators:
\( (3+a) + 3(8+b) = 14 \)
\( 3 + a + 24 + 3b = 14 \)
\( a + 3b + 27 = 14 \)
\( a + 3b = 14 - 27 \)
\( a + 3b = -13 \) ...(2)

Now we have a system of two linear equations with two variables, 'a' and 'b':
1. \( 3x - y = 1 \) (using 'a' for x and 'b' for y here, so \( 3a - b = 1 \))
2. \( a + 3b = -13 \)
From equation (1), \( b = 3a - 1 \). Substitute this into equation (2):
\( a + 3(3a - 1) = -13 \)
\( a + 9a - 3 = -13 \)
\( 10a - 3 = -13 \)
\( 10a = -10 \)
\( a = -1 \)

Substitute \( a = -1 \) back into \( b = 3a - 1 \):
\( b = 3(-1) - 1 = -3 - 1 = -4 \)

So, the coordinates of the image point P' are \( (-1, -4) \). The reflection of a point across a line follows these geometrical principles.
In simple words: To find the mirror image of a point, we use two rules. First, the line connecting the original point to its image is always straight and crosses the mirror line at a 90-degree angle. Second, the mirror line cuts this connecting line exactly in the middle. By using these rules and solving the math, we found the exact location of the image point.

🎯 Exam Tip: To find the image of a point reflected across a line, remember two key properties: 1) The line connecting the point and its image is perpendicular to the mirror line. 2) The midpoint of this connecting line segment lies on the mirror line. These two conditions will provide a system of equations to solve for the image coordinates.

Question 9. Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = O and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer: First, we need to find the point of intersection of the two given lines.
The lines are:
1. \( 4x + 7y = 3 \) ...(1)
2. \( 2x - 3y = -1 \) ...(2)

Multiply equation (2) by 2:
\( 2(2x - 3y) = 2(-1) \)
\( 4x - 6y = -2 \) ...(3)

Subtract equation (3) from equation (1):
\( (4x + 7y) - (4x - 6y) = 3 - (-2) \)
\( 4x + 7y - 4x + 6y = 3 + 2 \)
\( 13y = 5 \)
\( y = \frac{5}{13} \)

Substitute \( y = \frac{5}{13} \) into equation (2):
\( 2x - 3\left(\frac{5}{13}\right) = -1 \)
\( 2x - \frac{15}{13} = -1 \)
Multiply by 13 to clear the denominator:
\( 26x - 15 = -13 \)
\( 26x = -13 + 15 \)
\( 26x = 2 \)
\( x = \frac{2}{26} = \frac{1}{13} \)
The point of intersection is \( \left(\frac{1}{13}, \frac{5}{13}\right) \).

Now, we need to find the equation of a line that passes through this point and has equal intercepts on the axes.
Let the x-intercept and y-intercept both be 'a' (since they are equal).
The equation of a line with equal intercepts is \( \frac{x}{a} + \frac{y}{a} = 1 \).
This can be simplified to \( x + y = a \).

Since this line passes through \( \left(\frac{1}{13}, \frac{5}{13}\right) \), substitute these coordinates into \( x + y = a \):
\( \frac{1}{13} + \frac{5}{13} = a \)
\( \frac{1+5}{13} = a \)
\( \frac{6}{13} = a \)

So, the equal intercept value is \( a = \frac{6}{13} \).
Substitute this value back into the equation \( x + y = a \):
\( x + y = \frac{6}{13} \)
Multiply the entire equation by 13 to remove the denominator:
\( 13x + 13y = 6 \)
\( 13x + 13y - 6 = 0 \)

This is the required equation of the line. This line goes through the point where the first two lines cross and cuts the x and y axes at the same distance from the origin.
In simple words: First, we found the spot where the two given lines cross each other by solving their equations. This is a single point. Then, we looked for a new line that passes through this point and also cuts the x and y axes at the same distance. We found the equation for this new line.

🎯 Exam Tip: When a problem specifies "equal intercepts," remember to use the form \( \frac{x}{a} + \frac{y}{a} = 1 \) (or \( x+y=a \)) for the line. The process involves finding the intersection point of any other given lines and then substituting this point into the intercept form to solve for 'a'.

Question 10. A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path he should follow.
Answer: To reach a path in the least time from a point, the person should follow a path that is perpendicular to the destination path. First, we need to find the coordinates of the junction (the point of intersection of the first two lines).
The two paths the person is standing at the junction of are:
1. \( 2x - 3y = -4 \) ...(1)
2. \( 3x + 4y = 5 \) ...(2)

Multiply equation (1) by 4: \( 8x - 12y = -16 \) ...(3)
Multiply equation (2) by 3: \( 9x + 12y = 15 \) ...(4)

Add equation (3) and (4):
\( (8x - 12y) + (9x + 12y) = -16 + 15 \)
\( 17x = -1 \)
\( x = -\frac{1}{17} \)

Substitute \( x = -\frac{1}{17} \) into equation (2):
\( 3\left(-\frac{1}{17}\right) + 4y = 5 \)
\( -\frac{3}{17} + 4y = 5 \)
\( 4y = 5 + \frac{3}{17} \)
\( 4y = \frac{85 + 3}{17} = \frac{88}{17} \)
\( y = \frac{88}{17 \times 4} = \frac{22}{17} \)

So, the junction point (point of intersection) is \( \left(-\frac{1}{17}, \frac{22}{17}\right) \).

The destination path is \( 6x - 7y + 8 = 0 \). To reach this path in the least time, the person should follow a line perpendicular to it, passing through the junction point. The slope of the destination path \( 6x - 7y = -8 \) is \( -7y = -6x - 8 \implies y = \frac{6}{7}x + \frac{8}{7} \). So, \( m_{destination} = \frac{6}{7} \).

The slope of the path the person should follow (perpendicular line) is \( m_{perpendicular} = -\frac{1}{m_{destination}} = -\frac{1}{6/7} = -\frac{7}{6} \).

Now, use the point-slope form of a linear equation, \( y - y_1 = m(x - x_1) \), with the junction point \( \left(-\frac{1}{17}, \frac{22}{17}\right) \) and slope \( m_{perpendicular} = -\frac{7}{6} \):
\( y - \frac{22}{17} = -\frac{7}{6}\left(x - \left(-\frac{1}{17}\right)\right) \)
\( y - \frac{22}{17} = -\frac{7}{6}\left(x + \frac{1}{17}\right) \)
Multiply both sides by the common denominator \( 17 \times 6 = 102 \):
\( 102\left(y - \frac{22}{17}\right) = 102\left(-\frac{7}{6}\right)\left(x + \frac{1}{17}\right) \)
\( 6(17y - 22) = -7 \times 17 \left(x + \frac{1}{17}\right) \)
\( 102y - 132 = -119\left(\frac{17x+1}{17}\right) \)
\( 102y - 132 = -7(17x + 1) \)
\( 102y - 132 = -119x - 7 \)
\( 119x + 102y - 132 + 7 = 0 \)
\( 119x + 102y - 125 = 0 \)

This is the equation of the path the person should follow. A path perpendicular to the destination line is the shortest route.
In simple words: The person is standing where two paths cross. To get to a third path as fast as possible, they need to walk in a straight line that forms a perfect 'L' shape (90 degrees) with that third path. We first found the exact point where the person is standing, then figured out the 'steepness' (slope) of the path they want to reach. Finally, we found the equation of a new line that is perpendicular to the destination path and passes through where the person is standing.

🎯 Exam Tip: The shortest distance from a point to a line is always along the perpendicular from the point to the line. To solve such problems, first find the point of intersection, then determine the slope of the target line, and finally, use the negative reciprocal slope with the intersection point to find the equation of the perpendicular path.