krit.club logo

Coordinate Geometry - General Form: Ax + By + C = 0-advanced

Grade 9CBSE

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

🔑Concepts

•

The General Form Ax+By+C=0Ax + By + C = 0 represents a straight line on a Cartesian plane. The coefficients AA and BB determine the orientation (slope), while CC determines the position relative to the origin.

Graph of a linear equation in general form Ax + By + C = 0
•

The x-intercept is the point where the line crosses the x-axis (y=0y=0), calculated as x=−CAx = -\frac{C}{A}. The y-intercept is where it crosses the y-axis (x=0x=0), calculated as y=−CBy = -\frac{C}{B}.

Diagram showing x and y intercepts of a line
•

The slope mm of the line Ax+By+C=0Ax + By + C = 0 is given by m=−ABm = -\frac{A}{B}. If B=0B=0, the line is vertical (parallel to the y-axis). If A=0A=0, the line is horizontal (parallel to the x-axis).

Horizontal and vertical lines representing special cases of general form
•

The region formed by the line and the coordinate axes always creates a right-angled triangle. The area is 12×∣base∣×∣height∣\frac{1}{2} \times |\text{base}| \times |\text{height}|, where base and height are the absolute values of the intercepts.

📐Formulae

Ax+By+C=0Ax + By + C = 0

y=mx+cy = mx + c

Slope (m)=−AB\text{Slope } (m) = -\frac{A}{B}

y-intercept (c)=−CB\text{y-intercept } (c) = -\frac{C}{B}

x-intercept=−CA\text{x-intercept} = -\frac{C}{A}

Area of triangle formed by Ax+By+C=0 and the axes=12∣x-intercept×y-intercept∣=C22∣AB∣\text{Area of triangle formed by } Ax + By + C = 0 \text{ and the axes} = \frac{1}{2} |\text{x-intercept} \times \text{y-intercept}| = \frac{C^2}{2|AB|}

💡Examples

Problem 1:

Express the equation 3x−4y=123x - 4y = 12 in the general form Ax+By+C=0Ax + By + C = 0 and find its slope and y-intercept.

Solution:

Given equation: 3x−4y=123x - 4y = 12 To convert to general form, move 1212 to the left side: 3x−4y−12=03x - 4y - 12 = 0 Comparing with Ax+By+C=0Ax + By + C = 0, we get: A=3,B=−4,C=−12A = 3, B = -4, C = -12 Slope m=−AB=−3−4=34m = -\frac{A}{B} = -\frac{3}{-4} = \frac{3}{4} y-intercept c=−CB=−−12−4=−3c = -\frac{C}{B} = -\frac{-12}{-4} = -3

Explanation:

The general form requires all terms on one side. Slope and intercept are derived by rearranging the equation into y=mx+cy = mx + c or using the ratio of coefficients.

Problem 2:

Find the value of kk if the line 2x+ky−7=02x + ky - 7 = 0 passes through the point (1,−1)(1, -1).

Solution:

If the point (1,−1)(1, -1) lies on the line 2x+ky−7=02x + ky - 7 = 0, it must satisfy the equation. Substitute x=1x = 1 and y=−1y = -1 into the equation: 2(1)+k(−1)−7=02(1) + k(-1) - 7 = 0 2−k−7=02 - k - 7 = 0 −k−5=0-k - 5 = 0 −k=5-k = 5 k=−5k = -5

Explanation:

Any point on a line satisfies the algebraic equation of that line. Substitution allows us to solve for unknown parameters.

Problem 3:

Determine the area of the triangle formed by the line 2x+3y−6=02x + 3y - 6 = 0 and the coordinate axes.

Solution:

First, find the intercepts: For x-intercept, put y=0y = 0: 2x+3(0)−6=02x + 3(0) - 6 = 0 2x=6  ⟹  x=32x = 6 \implies x = 3 So, the base of the triangle is 33 units. For y-intercept, put x=0x = 0: 2(0)+3y−6=02(0) + 3y - 6 = 0 3y=6  ⟹  y=23y = 6 \implies y = 2 So, the height of the triangle is 22 units. Area of triangle = 12×base×height\frac{1}{2} \times \text{base} \times \text{height} Area=12×3×2=3 sq units\text{Area} = \frac{1}{2} \times 3 \times 2 = 3 \text{ sq units}

Explanation:

The line intersects the x-axis at (3,0)(3, 0) and the y-axis at (0,2)(0, 2), forming a right-angled triangle with the origin (0,0)(0, 0).

Problem 4:

Determine the value of mm if the line mx+2y−10=0mx + 2y - 10 = 0 has an x-intercept of 55.

Line passing through x-intercept (5,0)

Solution:

  1. At the x-intercept, the value of yy is 00.
  2. Substitute x=5x = 5 and y=0y = 0 into the equation mx+2y−10=0mx + 2y - 10 = 0.
  3. m(5)+2(0)−10=0m(5) + 2(0) - 10 = 0
  4. 5m−10=05m - 10 = 0
  5. 5m=10  ⟹  m=25m = 10 \implies m = 2.

Explanation:

Since the line passes through the point (5,0)(5, 0), we substitute these coordinates into the general equation to solve for the unknown coefficient mm.

Problem 5:

A line passes through the point (4,3)(4, 3) and is parallel to the x-axis. Express its equation in the form Ax+By+C=0Ax + By + C = 0.

Horizontal line y = 3 passing through (4,3)

Solution:

  1. A line parallel to the x-axis has a slope m=0m = 0.
  2. The equation of such a line is of the form y=ky = k.
  3. Since it passes through (4,3)(4, 3), the y-coordinate must be 33 everywhere on the line.
  4. So, y=3y = 3.
  5. Rearranging into Ax+By+C=0Ax + By + C = 0: 0x+1y−3=00x + 1y - 3 = 0.

Explanation:

Lines parallel to the x-axis have no xx term (coefficient A=0A = 0). The equation is simply yy equals the y-coordinate of the given point.