GSEB Class 11 Maths Solutions Chapter 10 Straight Lines Exercise 10.1

Question 1. Draw a quadrilateral in the cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, – 5) and (- 4, - 2). Also, find its area?
Answer: The given points \( (-4, 5) \), \( (0, 7) \), \( (5, -5) \), and \( (-4, -2) \) are shown on the graph. We call these points A, B, C, and D, in that specific order.We now split the quadrilateral into two triangles: \( \triangle ABD \) and \( \triangle BCD \). The area of any triangle can be calculated using the formula \( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \). The corners of \( \triangle ABD \) are \( (-4, 5) \), \( (0, 7) \), and \( (-4, -2) \). So, the area of \( \triangle ABD \) is calculated as: \( \frac{1}{2} |-4(7 + 2) + 0(-2 - 5) + (-4)(5 - 7)| \) This simplifies to \( \frac{1}{2} |-36 + 8| = \frac{28}{2} = 14 \). The corners of \( \triangle BCD \) are \( (0, 7) \), \( (5, -5) \), and \( (-4, -2) \). Next, for the area of \( \triangle BCD \): We use the formula again to get \( \frac{1}{2} |0(-5 + 2) + 5(-2 - 7) - 4(7 + 5)| \) This further simplifies to \( \frac{1}{2} |-45 - 48| = \frac{93}{2} \). Thus, the area of quadrilateral ABCD is the sum of the areas of \( \triangle ABD \) and \( \triangle BCD \). This gives us \( 14 + \frac{93}{2} \). Adding these values, we get \( \frac{28+93}{2} = \frac{121}{2} = 60.5 \) square units.
In simple words: We draw the quadrilateral by plotting its four points. To find its area, we split it into two triangles. We calculate the area of each triangle using a specific formula and then add those areas together to get the total area of the quadrilateral.

Exam Tip: Remember to clearly label your points and show the division into triangles. Double-check your calculations, especially with negative numbers, to avoid common errors.

Question 2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.
Answer: Let BC be the base of \( \triangle ABC \). Since the midpoint is at the origin and the side length is \( 2a \), we have \( BO = OC = a \). Also, because it is an equilateral triangle, \( AB = AC = 2a \). Now, using Pythagoras theorem in \( \triangle AOB \) (where O is the origin), \( AO^2 = AB^2 - BO^2 \). Substituting the values, \( AO^2 = (2a)^2 - a^2 = 4a^2 - a^2 = 3a^2 \). So, \( AO = \sqrt{3a^2} = \sqrt{3}a \). Therefore, point A, which lies on the x-axis, will have coordinates \( (\sqrt{3}a, 0) \). Point B is at \( (0, a) \) and point C is at \( (0, -a) \). If point A is on the left side of the y-axis, then the triangle's vertices will be \( (-\sqrt{3}a, 0) \), \( (0, a) \), and \( (0, -a) \).
In simple words: An equilateral triangle means all sides are equal. The base is on the y-axis, centered at the origin. We use this information and the Pythagorean theorem to find the x-coordinate of the third point, which helps us determine all three corner points of the triangle.

Exam Tip: For problems involving geometric figures on coordinate axes, drawing a simple sketch can help visualize the problem and correctly apply formulas like the distance formula or Pythagoras theorem.

Question 3. Find the distance between \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), when
(i) PQ is parallel to y-axis
(ii) PQ is parallel to x-axis.
Answer:
(i) If line segment PQ is parallel to the y-axis, all points along this line will share the same x-coordinate. So, we can say \( x_2 = x_1 \). Therefore, points P and Q are \( P(x_1, y_1) \) and \( Q(x_1, y_2) \). The distance PQ is then given by \( |y_2 - y_1| \).
(ii) If PQ is parallel to the x-axis, every point on this line will have the same y-coordinate. Thus, \( y_2 = y_1 \). Hence, the coordinates of P and Q are \( P(x_1, y_1) \) and \( Q(x_2, y_1) \). The distance PQ is then \( |x_2 - x_1| \).
In simple words: When a line is vertical (parallel to the y-axis), only the y-coordinates change, so the distance is the difference in y-values. When a line is horizontal (parallel to the x-axis), only the x-coordinates change, and the distance is the difference in x-values. We use absolute values to ensure the distance is always positive.

Exam Tip: Remember that distance is always a non-negative value, so use the absolute value when calculating distances between points, especially when coordinates can be negative.

Question 4. Find a point on the x-axis which is equidistant from the points (7, 6) and (3, 4).
Answer: Assume the point on the x-axis is \( (x_1, 0) \). The other two points are A\( (7, 6) \) and B\( (3, 4) \). We know that PA = PB, which means \( PA^2 = PB^2 \). So, \( (x_1 - 7)^2 + 6^2 = (x_1 - 3)^2 + 4^2 \). Expanding this gives \( x_1^2 - 14x_1 + 49 + 36 = x_1^2 - 6x_1 + 9 + 16 \). Simplifying, we get \( -14x_1 + 85 = -6x_1 + 25 \). Rearranging terms results in \( -14x_1 + 6x_1 = 25 - 85 \). Which is \( -8x_1 = -60 \). Thus, \( x_1 = \frac{-60}{-8} = \frac{60}{8} = \frac{15}{2} \). Therefore, the point P on the x-axis, which is equally distant from A and B, is \( (\frac{15}{2}, 0) \).
In simple words: We are looking for a point on the x-axis that is the same distance from two other points. We use the distance formula, setting the squared distances equal to each other. Solving the resulting equation helps us find the x-coordinate of our desired point.

Exam Tip: When a point lies on the x-axis, its y-coordinate is 0. Similarly, for a point on the y-axis, its x-coordinate is 0. This simplification helps reduce variables in distance calculations.

Question 5. Find the slope of a line which passes through the origin and mid-point of the line segment joining the points P(0, – 4) and B(8, 0).
Answer: The coordinates of the mid-point of two points \( (x_1, y_1) \) and \( (x_2, y_2) \) are given by the formula \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \). The mid-point of the line segment joining A\( (0, -4) \) and B\( (8, 0) \) is \( (\frac{8+0}{2}, \frac{0-4}{2}) \), which calculates to \( (4, -2) \). The slope of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is found using the formula \( \frac{y_2-y_1}{x_2-x_1} \). Here, the two points are O\( (0, 0) \) (the origin) and M\( (4, -2) \) (the midpoint we just found). So, the slope is \( \frac{-2-0}{4-0} = \frac{-2}{4} = - \frac{1}{2} \).
In simple words: First, we find the middle point of the line segment by averaging its x-coordinates and y-coordinates. Then, we calculate the slope of the line that connects this middle point to the origin (0,0) using the slope formula.

Exam Tip: Remember the midpoint formula \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \) and the slope formula \( m = \frac{y_2-y_1}{x_2-x_1} \). Make sure to simplify fractions correctly for the final slope value.

Question 6. Without using the pythagoras theorem, show that the points (4, 4), (3, 5) and (- 1, – 1) are the vertices of a right angled traingle.
Answer: We have three points A\( (4, 4) \), B\( (3, 5) \) and C\( (-1, -1) \). The slope of a line joining two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( \frac{y_2-y_1}{x_2-x_1} \). Slope of AB \( = \frac{5-4}{3-4} = \frac{1}{-1} = -1 \). Slope of BC \( = \frac{-1-5}{-1-3} = \frac{-6}{-4} = \frac{3}{2} \). Slope of AC \( = \frac{-1-4}{-1-4} = \frac{-5}{-5} = 1 \). The product of slopes of AB and AC is \( (-1) \times (1) = -1 \).
\( \implies \) Since the product of the slopes of AB and AC is -1, this means that line AB is perpendicular to line AC.
\( \implies \) Therefore, \( \triangle ABC \) is a right-angled triangle with the right angle at A.
In simple words: To show a triangle is right-angled without using Pythagoras, we check the slopes of its sides. If two sides are perpendicular, their slopes will multiply to give -1. We calculated the slopes for AB, BC, and AC. Since the slopes of AB and AC multiply to -1, these two lines are perpendicular, making the triangle a right-angled one.

Exam Tip: Two lines are perpendicular if and only if the product of their slopes is -1. This is a key condition to prove right angles in coordinate geometry without using distance-based theorems.

Question 7. Find the slope of the line which makes an angle of 30° with the positive direction of y-axis (measured anticlockwise).
Answer: The line OP makes an angle of 30° with the y-axis, measured in an anti-clockwise direction.
\( \implies \) Therefore, OP makes an angle of \( 90^\circ + 30^\circ \) with the positive direction of the x-axis. This equals \( 120^\circ \) with the positive direction of the x-axis. So, the slope of OP \( = \tan 120^\circ \). We know that \( \tan 120^\circ = \tan (180^\circ - 60^\circ) = -\tan 60^\circ \). Thus, the slope of OP \( = -\sqrt{3} \).
In simple words: The slope of a line is found using the tangent of the angle it makes with the positive x-axis. If the line forms a 30° angle with the y-axis, then with the x-axis, it forms \( 90^\circ + 30^\circ = 120^\circ \). The tangent of \( 120^\circ \) is \( -\sqrt{3} \), which is our slope.

Exam Tip: Remember that the standard angle for slope is measured counter-clockwise from the positive x-axis. If given an angle with the y-axis, convert it to an angle with the x-axis by adding or subtracting \( 90^\circ \) as appropriate.

Question 8. Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.
Answer: We have the points A\( (x, -1) \), B\( (2, 1) \) and C\( (4, 5) \). For three points to be collinear, the slope of the line segment AB must be equal to the slope of the line segment BC. The slope of AB \( = \frac{1-(-1)}{2-x} = \frac{1+1}{2-x} = \frac{2}{2-x} \). The slope of BC \( = \frac{5-1}{4-2} = \frac{4}{2} = 2 \). Setting the slopes equal: \( \frac{2}{2-x} = 2 \).
\( \implies \) Cross-multiplying, we get \( 2 = 2(2-x) \).
\( \implies 2 = 4 - 2x \).
\( \implies 2x = 4 - 2 \).
\( \implies 2x = 2 \).
\( \implies x = 1 \). So, the value of x for which the points are collinear is 1.
In simple words: For three points to lie on the same straight line, the slope calculated between the first two points must be the same as the slope calculated between the next two points. We set these two slope expressions equal to each other and solve for the unknown 'x'.

Exam Tip: The concept of collinearity is fundamental. Always use the slope formula to check if three points lie on the same straight line, as it directly tests if the 'steepness' between consecutive points is consistent.

Question 9. Without using distance formula, show that the points (- 2, 1), (4, 0), (3, 3) and (- 3, 2) are the vertices of a parallelogram.
Answer: The given points are A\( (-2, -1) \), B\( (4, 0) \), C\( (3, 3) \) and D\( (-3, 2) \). A quadrilateral is a parallelogram if its opposite sides are parallel. We can show sides are parallel by checking if their slopes are equal. Slope of AB \( = \frac{0-(-1)}{4-(-2)} = \frac{1}{4+2} = \frac{1}{6} \). Slope of DC \( = \frac{2-3}{-3-3} = \frac{-1}{-6} = \frac{1}{6} \). Since the slope of AB is equal to the slope of DC, we have AB parallel to DC. Slope of AD \( = \frac{2-(-1)}{-3-(-2)} = \frac{3}{-3+2} = \frac{3}{-1} = -3 \). Slope of BC \( = \frac{3-0}{3-4} = \frac{3}{-1} = -3 \). Since the slope of AD is equal to the slope of BC, we have AD parallel to BC.
\( \implies \) As both pairs of opposite sides are parallel, ABCD is a parallelogram.
In simple words: To prove a shape is a parallelogram without using distances, we need to show that its opposite sides are parallel. We do this by calculating the slopes of all four sides. If the slopes of opposite sides are equal, then those sides are parallel, confirming it's a parallelogram.

Exam Tip: For proving geometric properties using coordinates, relying on slopes for parallelism/perpendicularity and midpoints for bisection are often more efficient than the distance formula for all segments.

Question 10. Find the angle between x-axis and the line joining the points (3, – 1) and (4, – 2).
Answer: Let the two points be P\( (3, -1) \) and Q\( (4, -2) \). The slope of the line joining these points is given by \( m = \frac{y_2-y_1}{x_2-x_1} \). Slope \( = \frac{-2-(-1)}{4-3} = \frac{-2+1}{1} = \frac{-1}{1} = -1 \). We know that the slope \( m \) is also equal to \( \tan \theta \), where \( \theta \) is the angle the line makes with the positive x-axis. So, \( \tan \theta = -1 \). The angle \( \theta \) for which \( \tan \theta = -1 \) in the range \( [0^\circ, 180^\circ) \) is \( 135^\circ \). Therefore, the required angle is \( 135^\circ \).
In simple words: To find the angle a line makes with the x-axis, we first calculate the slope of the line using the given points. The slope is equal to the tangent of the angle. Once we have the tangent value, we find the angle using inverse tangent, making sure to pick the correct angle within the standard range.

Exam Tip: Remember that the angle \( \theta \) is typically measured from the positive x-axis in a counter-clockwise direction. If \( \tan \theta \) is negative, \( \theta \) will be in the second quadrant (between \( 90^\circ \) and \( 180^\circ \)).

Question 11. The slope of a line is double of the slope of another line. If tangent of the angle between them is \( \frac{1}{3} \), find the slopes of the lines?
Answer: Let \( m_1 \) and \( m_2 \) be the slopes of the two lines. Given that \( m_1 = 2m_2 \). Let \( \theta \) be the angle between the lines. The formula for the tangent of the angle between two lines is \( \tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| \). We are given \( \tan \theta = \frac{1}{3} \). So, \( \frac{1}{3} = |\frac{m_1 - m_2}{1 + m_1 m_2}| \). Putting \( m_1 = 2m_2 \) into the equation, we get: \( \frac{1}{3} = |\frac{2m_2 - m_2}{1 + (2m_2)m_2}| \)
\( \implies \frac{1}{3} = |\frac{m_2}{1 + 2m_2^2}| \). This implies \( \frac{m_2}{1 + 2m_2^2} = \frac{1}{3} \) or \( \frac{m_2}{1 + 2m_2^2} = -\frac{1}{3} \). Case 1: \( \frac{m_2}{1 + 2m_2^2} = \frac{1}{3} \)
\( \implies 3m_2 = 1 + 2m_2^2 \)
\( \implies 2m_2^2 - 3m_2 + 1 = 0 \). Factoring the quadratic equation:
\( \implies (2m_2 - 1)(m_2 - 1) = 0 \). This gives us \( m_2 = 1 \) or \( m_2 = \frac{1}{2} \). If \( m_2 = 1 \), then \( m_1 = 2m_2 = 2(1) = 2 \). The slopes are 2 and 1. If \( m_2 = \frac{1}{2} \), then \( m_1 = 2m_2 = 2(\frac{1}{2}) = 1 \). The slopes are 1 and \( \frac{1}{2} \). Case 2: \( \frac{m_2}{1 + 2m_2^2} = -\frac{1}{3} \)
\( \implies -3m_2 = 1 + 2m_2^2 \)
\( \implies 2m_2^2 + 3m_2 + 1 = 0 \). Factoring the quadratic equation:
\( \implies (2m_2 + 1)(m_2 + 1) = 0 \). This gives us \( m_2 = -1 \) or \( m_2 = -\frac{1}{2} \). If \( m_2 = -1 \), then \( m_1 = 2m_2 = 2(-1) = -2 \). The slopes are -2 and -1. If \( m_2 = -\frac{1}{2} \), then \( m_1 = 2m_2 = 2(-\frac{1}{2}) = -1 \). The slopes are -1 and \( -\frac{1}{2} \). Thus, the possible pairs of slopes for these lines are \( (2, 1) \), \( (1, \frac{1}{2}) \), \( (-2, -1) \), or \( (-1, -\frac{1}{2}) \).
In simple words: We are given that one line's slope is twice the other's, and the tangent of the angle between them is \( \frac{1}{3} \). We use the formula for the tangent of the angle between two lines, substitute the given relationship between the slopes, and then solve the resulting quadratic equation to find the possible values for the slopes. Since the formula involves an absolute value, we consider both positive and negative cases.

Exam Tip: When using the formula \( \tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}| \), remember the absolute value. This means you need to solve for both \( \frac{m_1 - m_2}{1 + m_1 m_2} = \tan \theta \) and \( \frac{m_1 - m_2}{1 + m_1 m_2} = -\tan \theta \) to find all possible solutions.

Question 12. A line passes through the points \( (x_1, y_1) \) and \( (h, k) \). If the slope of the line is m, show that \( k - y_1 = m(h – x_1) \).
Answer: The slope of the line joining the points A\( (x_1, y_1) \) and B\( (h, k) \) is given by the formula \( m = \frac{k-y_1}{h-x_1} \). We are given that the slope is \( m \). So, we have \( m = \frac{k-y_1}{h-x_1} \). By cross multiplying \( (h-x_1) \) with \( m \), we get: \( m(h-x_1) = k-y_1 \).
\( \implies k - y_1 = m(h – x_1) \). This matches the equation we were asked to show.
In simple words: The slope of a line is defined as the change in y-coordinates divided by the change in x-coordinates between any two points on the line. By rearranging this definition, we can easily show the given equation, which connects the change in y, change in x, and the slope.

Exam Tip: This question demonstrates the point-slope form of a linear equation. Understanding how to derive this from the basic slope formula is essential for working with lines in coordinate geometry.

Question 13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that \( \frac{a}{h} + \frac{b}{k} = 1 \).
Answer: The given points are A\( (h, 0) \), B\( (a, b) \) and C\( (0, k) \). Since they lie on the same line, the slope of AB must be equal to the slope of BC. Slope of AB \( = \frac{b-0}{a-h} = \frac{b}{a-h} \). Slope of BC \( = \frac{k-b}{0-a} = \frac{k-b}{-a} \). Setting the slopes equal: \( \frac{b}{a-h} = \frac{k-b}{-a} \). By cross-multiplication, we get: \( -ab = (a-h)(k-b) \).
\( \implies -ab = ak - ab - hk + hb \).
\( \implies 0 = ak - hk + hb \). Rearranging the terms, we get: \( ak + hb = hk \). Now, dividing the entire equation by \( hk \) (assuming \( h \ne 0 \) and \( k \ne 0 \)): \( \frac{ak}{hk} + \frac{hb}{hk} = \frac{hk}{hk} \).
\( \implies \frac{a}{h} + \frac{b}{k} = 1 \). This proves the required relationship.
In simple words: For three points to be on the same line, the slope between the first two points must equal the slope between the second and third points. We set these slope formulas equal, simplify the equation, and then divide by \( hk \) to arrive at the desired result \( \frac{a}{h} + \frac{b}{k} = 1 \).

Exam Tip: This problem is a classic demonstration of proving geometric properties using collinearity and slope. Remember that dividing by \( hk \) is only valid if \( h \) and \( k \) are non-zero, which is implicit for distinct intercept points.

Question 14. Consider the following population and year graph. Find the slope of the line AB and using it, find what will be the population in the year 2010?
Answer:The slope of the line joining the points A\( (1985, 92) \) and B\( (1995, 97) \) is calculated as: \( \frac{97-92}{1995-1985} = \frac{5}{10} = \frac{1}{2} \). Let \( p \) be the population in the year 2010. The point P\( (2010, p) \) lies on the same line. This implies that the slope of BP must be equal to the slope of AB. Slope of BP \( = \frac{p-97}{2010-1995} = \frac{p-97}{15} \). Setting the slopes equal: \( \frac{p-97}{15} = \frac{1}{2} \).
\( \implies 2(p-97) = 15 \).
\( \implies 2p - 194 = 15 \).
\( \implies 2p = 15 + 194 \).
\( \implies 2p = 209 \).
\( \implies p = \frac{209}{2} = 104.5 \). Therefore, the population in the year 2010 will be 104.5 crores.
In simple words: We calculate the 'steepness' (slope) of the line by using the population change between 1985 and 1995. Since the population is expected to follow the same line, we use this slope to predict the population in 2010 by setting the slope from 1995 to 2010 equal to the initial slope and solving for the unknown population.

Exam Tip: When dealing with real-world data represented on a graph, the slope often represents a rate of change (e.g., population growth rate). Using this constant rate of change for prediction is a common application of linear equations.