NCERT Solutions for Class 11 Maths: Straight Lines Ex 10.1

Revisiting Coordinate Geometry Basics

Before we dive into the new concepts of Chapter 10, let's have a quick but thorough review of the tools you already have from Class 10. These formulas are the building blocks for everything in this exercise.

1. The Distance Formula: To find the distance 'd' between two points P(x₁, y₁) and Q(x₂, y₂), we use the Pythagorean theorem:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

1. The Section Formula (Internal Division): This helps find the coordinates of a point R(x, y) that divides the line segment joining P(x₁, y₁) and Q(x₂, y₂) in the ratio m:n.

R(x, y) = ( (mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n) )
A special case is the Mid-point Formula, where the ratio is 1:1 (m=n=1):
Mid-point = ( (x₁ + x₂)/2, (y₁ + y₂)/2 )

1. Area of a Triangle: For a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃), the area is given by:

Area = ½ |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
A crucial application of this is to check if three points are collinear (lie on the same straight line). If the area of the triangle formed by them is zero, the points are collinear.

Understanding the Slope of a Line

Angle of Inclination (θ)

The angle 'θ' that a line makes with the positive direction of the x-axis, measured in the anti-clockwise direction. The value of θ is always between 0° and 180° (i.e., 0° ≤ θ < 180°).

Slope or Gradient (m)

The slope 'm' of a non-vertical line is the tangent of its angle of inclination. It measures the steepness and direction of the line. Formula: m = tan(θ). A horizontal line has a slope of 0 (since θ=0°), and a vertical line has an undefined slope (since θ=90°).

Slope of a Line Joining Two Points

If a line passes through two points P(x₁, y₁) and Q(x₂, y₂), its slope can be calculated directly without finding the angle. The formula is m = (y₂ - y₁) / (x₂ - x₁). This is often remembered as 'rise over run' – the change in the y-coordinate divided by the change in the x-coordinate.

Worked Examples for Ex 10.1

• Example 1: Finding the area of a quadrilateral. Find the area of the quadrilateral whose vertices, taken in order, are A(-4, -2), B(-3, -5), C(3, -2), and D(2, 3). Step 1: Divide the quadrilateral into two triangles. We can find the area of the quadrilateral ABCD by adding the areas of two triangles, ΔABC and ΔACD. Step 2: Calculate the area of ΔABC. Using the area formula with vertices A(-4, -2), B(-3, -5), and C(3, -2): Area(ΔABC) = ½ |(-4)(-5 - (-2)) + (-3)(-2 - (-2)) + 3(-2 - (-5))| = ½ |(-4)(-3) + (-3)(0) + 3(3)| = ½ |12 + 0 + 9| = ½ |21| = 10.5 sq. units. Step 3: Calculate the area of ΔACD. Using vertices A(-4, -2), C(3, -2), and D(2, 3): Area(ΔACD) = ½ |(-4)(-2 - 3) + 3(3 - (-2)) + 2(-2 - (-2))| = ½ |(-4)(-5) + 3(5) + 2(0)| = ½ |20 + 15 + 0| = ½ |35| = 17.5 sq. units. Step 4: Add the areas. Area(ABCD) = Area(ΔABC) + Area(ΔACD) = 10.5 + 17.5 = 28 sq. units. Final Answer: The area of the quadrilateral is 28 square units.
• Example 2: Using the concept of slope for collinearity. Show that the points P(1, -1), Q(2, 1), and R(4, 5) are collinear. Step 1: Understand the condition for collinearity using slopes. If three points lie on the same line, the slope of the line segment joining the first two points must be equal to the slope of the line segment joining the second and third points. That is, Slope(PQ) must be equal to Slope(QR). Step 2: Calculate the slope of PQ. Using the formula m = (y₂ - y₁) / (x₂ - x₁) for P(1, -1) and Q(2, 1): Slope(PQ) = (1 - (-1)) / (2 - 1) = 2 / 1 = 2. Step 3: Calculate the slope of QR. Using the formula for Q(2, 1) and R(4, 5): Slope(QR) = (5 - 1) / (4 - 2) = 4 / 2 = 2. Step 4: Compare the slopes. Since Slope(PQ) = Slope(QR) = 2, and they share a common point Q, the points P, Q, and R must lie on the same straight line. Final Answer: The points are collinear as the slopes of the segments PQ and QR are equal.

Exam Tips and Common Traps

When solving problems from Ex 10.1, keep these points in mind to avoid common errors:

• Collinearity Check: You have three methods to prove collinearity for points A, B, and C:

1. Distance Formula: Show that AB + BC = AC (or another combination). This can be calculation-intensive.
2. Area of Triangle: Show that the area of ΔABC is 0. This is reliable.
3. Slope Method: Show that slope(AB) = slope(BC). This is often the fastest and most elegant method.

• Angle of Inclination: Always remember that the angle of inclination θ is measured from the positive x-axis in an anti-clockwise direction. For example, if a line has a slope m = -1, then tan(θ) = -1. The correct inclination is θ = 135°, not -45° or 315°, because 0° ≤ θ < 180°.

• Parallel and Perpendicular Lines: Two non-vertical lines are parallel if and only if their slopes are equal (m₁ = m₂). Two non-vertical lines are perpendicular if and only if the product of their slopes is -1 (m₁ * m₂ = -1).

Practice Questions with Solutions

• Q: The vertices of a triangle are P(2, 3), Q(4, -1), and R(1, 2). Find the length of the median from the vertex Q. A: Step 1: Find the midpoint of the side PR. The median from Q will connect Q to this midpoint. Let's call the midpoint S. Using the midpoint formula for P(2, 3) and R(1, 2): S = ((2+1)/2, (3+2)/2) = (3/2, 5/2). Step 2: Find the length of the median QS using the distance formula. The coordinates are Q(4, -1) and S(3/2, 5/2). Length(QS) = √[(3/2 - 4)² + (5/2 - (-1))²] = √[(-5/2)² + (7/2)²] = √[25/4 + 49/4] = √[74/4] = (√74)/2. Final answer: The length of the median from vertex Q is (√74)/2 units.
• Q: Find the slope of the line passing through the origin and the midpoint of the line segment joining the points A(0, -4) and B(8, 0). A: Step 1: Find the midpoint of the line segment AB. Let the midpoint be M. Using the midpoint formula for A(0, -4) and B(8, 0): M = ((0+8)/2, (-4+0)/2) = (4, -2). Step 2: The line passes through the origin O(0, 0) and the midpoint M(4, -2). Find the slope of OM. Using the slope formula m = (y₂ - y₁) / (x₂ - x₁): Slope(OM) = (-2 - 0) / (4 - 0) = -2 / 4 = -1/2. Final answer: The slope of the line is -1/2.
• Q: A line passes through (x₁, 7) and (2, 3) and its slope is 2. Find the value of x₁. A: Step 1: Set up the slope formula with the given points and slope. The points are P₁(x₁, 7) and P₂(2, 3). The slope m = 2. m = (y₂ - y₁) / (x₂ - x₁) 2 = (3 - 7) / (2 - x₁) Step 2: Solve the equation for x₁. 2 = -4 / (2 - x₁) 2 * (2 - x₁) = -4 4 - 2x₁ = -4 -2x₁ = -8 x₁ = 4. Final answer: The value of x₁ is 4.
• Q: Without using the distance formula, show that points A(-2, -1), B(4, 0), C(3, 3), and D(-3, 2) are the vertices of a parallelogram. A: Step 1: Recall the property of a parallelogram related to slopes. A quadrilateral is a parallelogram if its opposite sides are parallel. This means the slopes of opposite sides must be equal. We need to show: Slope(AB) = Slope(CD) and Slope(BC) = Slope(DA). Step 2: Calculate the slopes of all four sides. Slope(AB) = (0 - (-1)) / (4 - (-2)) = 1 / 6. Slope(BC) = (3 - 0) / (3 - 4) = 3 / -1 = -3. Slope(CD) = (2 - 3) / (-3 - 3) = -1 / -6 = 1 / 6. Slope(DA) = (-1 - 2) / (-2 - (-3)) = -3 / 1 = -3. Step 3: Compare the slopes of opposite sides. We see that Slope(AB) = Slope(CD) = 1/6. We also see that Slope(BC) = Slope(DA) = -3. Since both pairs of opposite sides have equal slopes, they are parallel. Final answer: The points A, B, C, and D are the vertices of a parallelogram because the slopes of opposite sides are equal.

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