Find the angle between the lines x - 2y + 2 = 0 an – Class 11 Mathematics

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Q. Find the angle between the lines x - 2y + 2 = 0 and x + 3y + 4 = 0.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2019Subjective· 2mImportance★★★★★

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Concept understanding — Coordinate Geometry

Coordinate Geometry: Where Algebra Meets Geometry

Imagine you're telling a friend where you left your book in a library. You don't say "near the window" — you say "third shelf, second row, fourth book from the left." You're using numbers to pin down an exact location.

Coordinate geometry does the same thing, but for points on a flat surface. It gives every point a precise address — a pair of numbers — so we can describe shapes, distances, and positions using algebra.

The Big Idea

Before coordinate geometry, geometry was about drawing shapes and proving things with logic alone. Algebra was about numbers and equations. These two worlds seemed separate.

Then René Descartes (a French mathematician) had a simple but revolutionary idea: draw two perpendicular number lines that cross at zero. Now every point on the plane has a unique pair of numbers — its coordinates.

That's it. That's the entire foundation.

The Coordinate System

Take a horizontal line — call it the x-axis. Take a vertical line — call it the y-axis. They cross at a point called the origin, labelled $O$ .

Any point $P$ is located by two numbers:

Its x-coordinate: how far right (positive) or left (negative) from the origin
Its y-coordinate: how far up (positive) or down (negative) from the origin

We write this as an ordered pair: $(x, y)$ .

Note

The order matters. $(3, 5)$ is not the same point as $(5, 3)$ . The first number is always the horizontal position; the second is always the vertical.

A Concrete Example

Plot the point $A(2, 3)$ :

Start at the origin $(0, 0)$ .
Move 2 units to the right along the x-axis.
From there, move 3 units up (parallel to the y-axis).
Mark the point.

Now plot $B(-1, 4)$ :

Start at the origin.
Move 1 unit left (negative x-direction).
Move 4 units up.
Mark the point.

Every point on the plane has exactly one such address. And every pair of numbers corresponds to exactly one point. This one-to-one matching is what makes coordinate geometry powerful.

The Four Quadrants

The axes divide the plane into four regions, called quadrants:

Quadrant	x-sign	y-sign	Example
I	$+$	$+$	$(2, 3)$
II	$-$	$+$	$(-1, 4)$
III	$-$	$-$	$(-3, -2)$
IV	$+$	$-$	$(5, -1)$

Points on the axes themselves (where either coordinate is zero) don't belong to any quadrant.

Why This Matters

Once every point has a number address, we can:

Calculate distances between points using the Pythagorean theorem
Find midpoints by averaging coordinates
Describe lines with equations like $y = mx + c$
Solve geometric problems using algebra instead of drawing

Important

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

This is just the Pythagorean theorem in disguise.

The Precise Statement

Coordinate geometry (also called analytic geometry) is the study of geometry using a coordinate system. It establishes a correspondence between:

Points on a plane and ordered pairs of real numbers
Geometric figures (lines, circles, curves) and algebraic equations …

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Convert both lines to slope form, then apply the angle-between-two-lines formula.

Line 1: $x-2y+2=0 \Rightarrow y=\dfrac{x}{2}+1$ , so $m_1=\dfrac12$ .

Line 2: $x+3y+4=0 \Rightarrow y=-\dfrac{x}{3}-\dfrac43$ , so $m_2=-\dfrac13$ .

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 23 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$(-4, 3)$
(a) $x - 2y + 10 = 0$
(b) $x - 2y - 10 = 0$
(c) $2x - y + 10 = 0$
(d) $x + 2y + 10 = 0$
›Reveal solutionSolution

Apply the point-slope equation $y-y_1 = m(x-x_1)$ and simplify to general form.

Point-slope form: $y - y_1 = m(x - x_1)$ with $(x_1,y_1)=(-4,3)$ and $m=\dfrac12$ :

$y - 3 = \dfrac12(x-(-4)) = \dfrac12(x+4)$

Multiply both sides by 2: …

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CBSE 2026Set ANNUAL1 mark
$a$
›Reveal solutionSolution

A line cutting intercepts $a$ and $b$ on the x- and y-axis respectively has equation $\dfrac{x}{a}+\dfrac{y}{b}=1$ .

The line passes through $(a, 0)$ on the x-axis and $(0, b)$ on the y-axis.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The co-ordinates of points in the xy-plane are in the form of ______.$
›Reveal solutionSolution

Every point in the Cartesian xy-plane is represented as an ordered pair $(x, y)$ .

In the Cartesian coordinate system, two perpendicular number lines (the x-axis and y-axis) intersect at the origin.

…

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CBSE 2025Set ANNUAL1 markMCQ
$2x - 3y + 6 = 0$
(a) 2
(b) 8
(c) $\dfrac{2}{3}$
(d) $\dfrac{2\tan x}{1 - \tan^2 x}$
(e) $\sin 2x$
(f) 10
(g) 20
(h) $\dfrac{2\tan x}{1 + \tan^2 x}$
(i) 4
›Reveal solutionSolution

Substituting $x = 0$ into the line equation $2x - 3y + 6 = 0$ gives $y = 2$ , the $y$ -intercept.

For the line $2x - 3y + 6 = 0$ , the $y$ -intercept is found by setting $x = 0$ :

$2(0) - 3y + 6 = 0 \Rightarrow -3y = -6 \Rightarrow y = 2$ . …

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CBSE 2024Set ANNUAL1 markMCQ
$(2,1)$
(a) $x+y-3=0$
(b) $x+y+3=0$
(c) $5x+3y+2=0$
(d) None of these
›Reveal solutionSolution

Find the slope from the two given points, then use the point-slope form of a line.

Slope through $(2,1)$ and $(5,-2)$ :

$m = \frac{-2-1}{5-2} = \frac{-3}{3} = -1$

Using point-slope form $y - y_1 = m(x-x_1)$ with point $(2,1)$ :

$y - 1 = -1(x - 2)$ …

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CBSE 2024Set ANNUAL1 markMCQ
$(x-y+2)+k(2x+3y+5)=0$
(a) $k=1$
(b) $k=\dfrac{4}{7}$
(c) $k=\dfrac{7}{4}$
(d) None of these
›Reveal solutionSolution

Expand the family of lines to standard form $Ax+By+C=0$ , write its slope $-A/B$ , and set it equal to the slope of the given parallel line.

Expand $(x-y+2)+k(2x+3y+5)=0$ :

$x - y + 2 + 2kx + 3ky + 5k = 0$

$(1+2k)x + (3k-1)y + (2+5k) = 0$

The slope of a line $Ax+By+C=0$ is $-A/B$ , so this line's slope is:

$m_1 = -\frac{1+2k}{3k-1}$

The line $3x+y=0$ has slope $m_2 = -3$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$(4, 3)$
(a) $3x+4y=10$
(b) $3x+4y=24$
(c) $3x+4y-20=0$
(d) $3x-4y+24=0$
›Reveal solutionSolution

Parallel lines share the same $x,y$ coefficients; substituting the given point fixes the constant as 24.

A line parallel to $3x+4y=12$ has the same left-hand-side coefficients, differing only in the constant:

$3x+4y=c$

…

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CBSE 2023Set ANNUAL1 mark
$(-2, 3)$
›Reveal solutionSolution

The equation of the line is $4x+y+5=0$ .

Using point-slope form:

$y-3 = -4(x-(-2)) = -4(x+2)$ …

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CBSE 2023Set ANNUAL1 mark
$(3, -2)$
›Reveal solutionSolution

The equation of the line through $(3,-2)$ and $(-1,4)$ is $3x+2y-5=0$ .

Slope: $m=\dfrac{4-(-2)}{-1-3}=\dfrac6{-4}=-\dfrac32$ .

Using point-slope form through $(3,-2)$ : …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Case study: Due to Covid-19 pandemic situation, people are maintaining social distances. Three friends Along, Mharemo and Senti are sitting on the vertices of a triangle whose coordinates are A(3, 1), M(2, -3) and S(-3, 3). The equation of the line AM is$
(a) $4x+y-11=0$
(b) $4x-y-11=0$
(c) $4x+y+11=0$
(d) $4x-y+11=0$
›Reveal solutionSolution

Find the slope of AM, then use point-slope form through A.

A $(3,1)$ , M $(2,-3)$ .

Slope of AM $=\dfrac{-3-1}{2-3}=\dfrac{-4}{-1}=4$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Case study (continued — triangle with vertices A(3,1), M(2,-3), S(-3,3)): The equation of a line passing through A and parallel to SM is$
(a) $6x+5y-23=0$
(b) $6x-5y-23=0$
(c) $6x-5y+23=0$
(d) $6x+5y+23=0$
›Reveal solutionSolution

Find slope of SM, then write the parallel line through A with that same slope.

S $(-3,3)$ , M $(2,-3)$ .

Slope of SM $=\dfrac{-3-3}{2-(-3)}=\dfrac{-6}{5}$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Case study (continued — triangle with vertices A(3,1), M(2,-3), S(-3,3)): Equation of median through A is$
(a) $2x+7y+1=0$
(b) $2x-7y+1=0$
(c) $2x+7y-1=0$
(d) $2x-7y-1=0$
›Reveal solutionSolution

The median from A passes through the midpoint of the opposite side SM.

Midpoint of S $(-3,3)$ and M $(2,-3)$ : $\left(\dfrac{-3+2}{2},\dfrac{3-3}{2}\right)=\left(-\dfrac12,\,0\right)$ .

Slope of the median (through A $(3,1)$ and this midpoint): $\dfrac{0-1}{-1/2-3}=\dfrac{-1}{-7/2}=\dfrac{2}{7}$ .

…

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Q.Find the angle between the lines x - 2y + 2 = 0 and x + 3y + 4 = 0.

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