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Coordinate Geometry - Straight Lines (Various Forms and General Equation)

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Inclination and Slope: The inclination of a line is the angle θ\theta (0θ<1800 \le \theta < 180^{\circ}) which the line makes with the positive direction of the x-axis, measured in the anticlockwise direction. Visually, a line leaning to the right has an acute angle and positive slope, while a line leaning to the left has an obtuse angle and negative slope. The slope mm is defined as m=tanθm = \tan \theta.

Slope-Intercept Form: A line with slope mm and y-intercept cc is represented by y=mx+cy = mx + c. Graphically, the y-intercept cc is the vertical distance from the origin where the line crosses the y-axis at point (0,c)(0, c). If the line passes through the origin, c=0c = 0 and the equation simplifies to y=mxy = mx.

Point-Slope and Two-Point Form: The Point-Slope form yy1=m(xx1)y - y_1 = m(x - x_1) represents a line passing through a fixed point (x1,y1)(x_1, y_1) with a specific tilt mm. When two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are given, the slope is first calculated as the 'rise over run' ratio, leading to the Two-Point form which defines the unique straight path connecting them.

Intercept Form: The equation xa+yb=1\frac{x}{a} + \frac{y}{b} = 1 represents a line that cuts the x-axis at (a,0)(a, 0) and the y-axis at (0,b)(0, b). Visually, aa and bb represent the signed lengths of the segments intercepted on the axes from the origin. This form is particularly useful for sketching lines quickly by marking two points on the axes.

Normal Form: A line can be defined by the length of the perpendicular (normal) drawn from the origin to the line, denoted by pp, and the angle α\alpha that this normal makes with the positive x-axis. The equation is xcosα+ysinα=px \cos \alpha + y \sin \alpha = p. Note that pp is always positive as it represents distance.

General Equation of a Line: Any first-degree equation in xx and yy of the form Ax+By+C=0Ax + By + C = 0 (where AA and BB are not both zero) represents a straight line. By rearranging this into slope-intercept form, we see the slope is AB-\frac{A}{B} and the y-intercept is CB-\frac{C}{B}.

Parallel and Perpendicular Lines: Two non-vertical lines are parallel if and only if their slopes are equal (m1=m2m_1 = m_2), meaning they have the same inclination and never meet. They are perpendicular if and only if the product of their slopes is 1-1 (m1m2=1m_1 \cdot m_2 = -1), which occurs when the lines intersect at a 9090^{\circ} angle.

📐Formulae

Slope: m=tanθm = \tan \theta

Slope from two points: m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}

Slope-Intercept Form: y=mx+cy = mx + c

Point-Slope Form: yy1=m(xx1)y - y_1 = m(x - x_1)

Two-Point Form: yy1=y2y1x2x1(xx1)y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)

InterceptForm:xa+yb=1Intercept Form: \frac{x}{a} + \frac{y}{b} = 1

Normal Form: xcosα+ysinα=px \cos \alpha + y \sin \alpha = p

General Equation: Ax+By+C=0Ax + By + C = 0

Distance of a point (x1,y1)(x_1, y_1) from a line: d=Ax1+By1+CA2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

💡Examples

Problem 1:

Find the equation of the line passing through the point (2,3)(2, 3) and perpendicular to the line 3x4y+5=03x - 4y + 5 = 0.

Solution:

Step 1: Find the slope of the given line. The equation is 3x4y+5=03x - 4y + 5 = 0. Rewriting in y=mx+cy = mx + c form: 4y=3x+5    y=34x+544y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}. Thus, m1=34m_1 = \frac{3}{4}.\Step 2: Find the slope (m2m_2) of the perpendicular line. Since m1m2=1m_1 \cdot m_2 = -1, we have 34m2=1    m2=43\frac{3}{4} \cdot m_2 = -1 \implies m_2 = -\frac{4}{3}.\Step 3: Use the point-slope form with point (2,3)(2, 3) and m=43m = -\frac{4}{3}.\y3=43(x2)y - 3 = -\frac{4}{3}(x - 2)\3(y3)=4(x2)3(y - 3) = -4(x - 2)\3y9=4x+8    4x+3y17=03y - 9 = -4x + 8 \implies 4x + 3y - 17 = 0.

Explanation:

To find the equation, we first determine the slope of the given line, then apply the perpendicularity condition (m1m2=1m_1 m_2 = -1) to find our required slope, and finally substitute the values into the point-slope formula.

Problem 2:

Reduce the equation 3x+y8=0\sqrt{3}x + y - 8 = 0 into normal form and find the values of pp and α\alpha.

Solution:

Step 1: Write the equation as 3x+y=8\sqrt{3}x + y = 8. Here, A=3A = \sqrt{3} and B=1B = 1.\Step 2: Calculate A2+B2=(3)2+12=3+1=2\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2.\Step 3: Divide the entire equation by 2.\32x+12y=82    32x+12y=4\frac{\sqrt{3}}{2}x + \frac{1}{2}y = \frac{8}{2} \implies \frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4.\Step 4: Compare with xcosα+ysinα=px \cos \alpha + y \sin \alpha = p.\We get cosα=32\cos \alpha = \frac{\sqrt{3}}{2}, sinα=12\sin \alpha = \frac{1}{2}, and p=4p = 4.\Since both sine and cosine are positive, α\alpha is in the first quadrant: α=30\alpha = 30^{\circ} or π6\frac{\pi}{6}.

Explanation:

Normal form reduction requires dividing the general equation by A2+B2\sqrt{A^2 + B^2} to normalize the coefficients into trigonometric values (sine and cosine), while pp represents the perpendicular distance from the origin.