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MathematicsClass 11Straight Lines

Straight Lines | Class 11 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 4 min read

Straight Lines | Class 11 Mathematics Notes

Straight Lines – this guide gives you a concise, exam-ready overview of Straight Lines from Class 11 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

9.3 Various Forms of the Equation of a Line

A line in the coordinate plane can be represented algebraically by an equation involving x and y. To determine if a point lies on a line, its coordinates must satisfy the line's equation. Different forms of line equations are used depending on the given information. Horizontal lines at distance a from x-axis have equations y = a or y = -a. Vertical lines at distance b from y-axis have equations x = b or x = -b. The point-slope form is y - y0 = m(x - x0), where m is the slope and (x0, y0) a known point on the line. The two-point form uses two points (x1, y1) and (x2, y2) on the line: (y - y1) = ((y2 - y1)/(x2 - x1))(x - x1). The slope-intercept form y = mx + c uses slope m and y-intercept c. If the line cuts x-axis at d, equation is y = m(x - d). The intercept form uses intercepts a and b on x- and y-axes: (x/a) + (y/b) = 1. The general form is Ax + By + C = 0, where A and B are not both zero. These forms facilitate solving geometric problems and graphing lines.

📊 Diagram: Fig 9.8 (a) and (b) showing horizontal and vertical lines; Fig 9.9 showing lines parallel to axes through (-2, 3); Fig 9.10 point-slope form illustration; Fig 9.11 two-point form illustration; Fig 9.12 slope-intercept form illustration; Fig 9.13 intercept form illustration.

🧪 Activity: Derive equations of lines under various given conditions; plot lines using different forms.

🔗 Connection: Prepares for understanding distance of a point from a line and related geometric properties.

Frequently asked questions

1. Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4, 5)$ , $(0, 7)$ , $(5, -5)$ and $(-4, -2)$ . Also, find its area.

To draw the quadrilateral, plot the points (-4,5), (0,7), (5,-5), and (-4,-2) on the Cartesian plane and join them in order. To find the area, use the shoelace formula:

Vertices in order: A(-4,5), B(0,7), C(5,-5), D(-4,-2)

Area = (1/2) |x1y2 + x2y3 + x3y4 + x4y1 - (y1x2 + y2x3 + y3x4 + y4x1)| = (1/2) |(-4)(7) + 0(-5) + 5(-2) + (-4)(5) - [5(0) + 7(5) + (-5)(-4) + (-2)(-4)]| = (1/2) |(-28) + 0 + (-10) + (-20) - [0 + 35 + 20 + 8]| = (1/2) |(-58) - 63| = (1/2) |-121| = 60.5 units²

So, the area of

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 vertices of the triangle.

Given an equilateral triangle with side length 2a, base lies along y-axis with midpoint at origin.

Base endpoints: Since base length = 2a along y-axis, and midpoint at origin, the base vertices are at (0, a) and (0, -a).

To find the third vertex, let it be at (x, 0) on x-axis (since base is vertical, the apex lies on x-axis).

Distance from apex to each base vertex = side length = 2a.

Using distance formula: Distance between (x,0) and (0,a) = 2a \Rightarrow \sqrt{(x-0)^2 + (0 - a)^2} = 2a \Ri

3. Find the distance between P(x_1, y_1) and Q(x_2, y_2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

(i) If PQ is parallel to y-axis, then x-coordinates of P and Q are the same, i.e., x_1 = x_2. Distance = |y_2 - y_1|

(ii) If PQ is parallel to x-axis, then y-coordinates of P and Q are the same, i.e., y_1 = y_2. Distance = |x_2 - x_1|

4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Let the point on x-axis be (x, 0). Distance from (7,6) to (x,0): \sqrt{(x - 7)^2 + (0 - 6)^2} = \sqrt{(x - 7)^2 + 36} Distance from (3,4) to (x,0): \sqrt{(x - 3)^2 + (0 - 4)^2} = \sqrt{(x - 3)^2 + 16}

Equate distances: \sqrt{(x - 7)^2 + 36} = \sqrt{(x - 3)^2 + 16} Square both sides: (x - 7)^2 + 36 = (x - 3)^2 + 16 Expand: (x^2 - 14x + 49) + 36 = (x^2 - 6x + 9) + 16 Simplify: x^2 - 14x + 85 = x^2 - 6x + 25 Cancel x^2: -14x + 85 = -6x + 25 Bring variables to one side: -14x + 6x = 25 - 85 -8x = -6

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