Chapter 9 — Straight Lines

Concept Used

In coordinate geometry, axes are reference lines:

• Points on the x-axis have zero y-coordinate
• Points on the y-axis have zero x-coordinate

Solution Roadmap

• Use coordinate definition of axes
• Translate into algebraic equation

Solution

Exam Significance

• Most fundamental concept in coordinate geometry (CBSE boards)
• Used implicitly in almost every graph-based question
• Essential for intercept form, distance, and slope problems
• Forms base for all higher-level coordinate geometry (JEE, NDA)

Concept Used

The equation of a line passing through a point \((x_1, y_1)\) with slope \(m\) is given by the point–slope form:

\[ y - y_1 = m(x - x_1) \]

Solution Roadmap

• Use point–slope formula
• Substitute given point and slope
• Simplify to standard form

Solution

Exam Significance

• Most standard line equation format in CBSE board exams
• Direct application of point–slope form (very common in JEE Main)
• Forms base for converting between different line forms
• Important for graph-based MCQs and analytical geometry problems

Concept Used

The equation of a line through \((x_1, y_1)\) with slope \(m\) is:

\[ y - y_1 = m(x - x_1) \]

When the line passes through the origin \((0,0)\), this reduces to the slope–intercept form:

\[ y = mx \]

Solution Roadmap

• Use point–slope form
• Substitute origin as point
• Simplify to standard form

Solution

Exam Significance

• Fundamental form of a straight line through origin (CBSE boards)
• Used extensively in JEE Main MCQs involving proportional relationships
• Important for understanding slope as rate of change
• Base concept for linear graphs and direct variation

Concept Used

The slope of a line inclined at an angle \(\theta\) with the positive x-axis is:

\[ m = \tan\theta \]

The equation of a line passing through a point is given by the point–slope form.

Solution Roadmap

• Convert angle into slope using \(m = \tan\theta\)
• Simplify \(\tan 75^\circ\) using identity
• Apply point–slope form
• Simplify to slope–intercept form

Solution

Exam Significance

• Very important: angle → slope conversion (CBSE + JEE Main)
• Standard use of trigonometric identities in coordinate geometry
• Common MCQ pattern involving special angles (30°, 45°, 60°, 75°)
• Strengthens connection between trigonometry and straight lines

Concept Used

A line intersecting the x-axis has y-coordinate zero at the point of intersection. The equation of a line with slope \(m\) passing through \((x_1, y_1)\) is:

\[ y - y_1 = m(x - x_1) \]

Solution Roadmap

• Identify x-intercept from given condition
• Use point–slope form with slope
• Simplify to standard form

Solution

Exam Significance

• Tests interpretation of geometric language (left/right of origin)
• Very common CBSE board pattern (intercept + slope)
• Frequent in JEE Main coordinate geometry MCQs
• Important for quick graph visualization and sign understanding

Concept Used

The slope of a line inclined at angle \(\theta\) with the positive x-axis is:

\[ m = \tan\theta \]

A line intersecting the y-axis at \((0,c)\) can be written using point–slope form.

Solution Roadmap

• Identify y-intercept point
• Convert angle to slope
• Apply point–slope form
• Simplify to standard form

Solution

Exam Significance

• Combines intercept + angle → slope conversion (CBSE core concept)
• Frequently asked in JEE Main as multi-concept MCQ
• Tests clarity in trigonometric slope evaluation
• Important for visualizing line direction and intercepts

Concept Used

The slope of a line passing through two points is:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

Once slope is known, the equation can be written using the point–slope form.

Solution Roadmap

• Find slope using two-point formula
• Apply point–slope form
• Simplify to standard form

Solution

Exam Significance

• Very common two-point form problem in CBSE boards
• Direct MCQ pattern in JEE Main
• Tests slope calculation accuracy (sign errors are common trap)
• Foundation for section formula and collinearity problems

Concept Used

A median of a triangle joins a vertex to the midpoint of the opposite side.

• Find midpoint using midpoint formula
• Join midpoint with given vertex
• Use slope + point–slope form to get equation

Solution Roadmap

• Find midpoint of \(PQ\)
• Find slope of line joining midpoint and \(R\)
• Use point–slope form
• Simplify equation

Solution

Exam Significance

• Classic CBSE problem combining midpoint + slope + line equation
• Frequently appears in JEE Main coordinate geometry section
• Tests multi-step reasoning (geometry + algebra)
• Important for centroid, medians, and triangle coordinate problems

Concept Used

Slopes of perpendicular lines satisfy:

\[ m_1 \cdot m_2 = -1 \]

Once slope is known, use point–slope form to find the equation.

Solution Roadmap

• Find slope of given line
• Use perpendicular condition to find required slope
• Apply point–slope form
• Simplify equation

Solution

Exam Significance

• Very important concept: perpendicular slopes (CBSE + JEE Main)
• Common multi-step MCQ (slope → perpendicular → equation)
• Tests sign handling and reciprocal understanding
• Foundation for normal form and shortest distance problems

Solution

Exam Significance

• High-level CBSE problem combining section formula + perpendicular slope
• Very common in JEE Main multi-step coordinate geometry questions
• Tests algebraic manipulation with parameters (n)
• Important for locus, ratio division, and advanced line problems

Sollution:

The line cuts equal intercepts on the coordinate axes and passes through the point \((2,3)\). Let each intercept be \(a\). Then, using the intercept form of the equation of a line, we have

\[ \begin{aligned} \frac{x}{a} + \frac{y}{a} = 1 \\ a = b \\ x + y = a \end{aligned} \]

Since both intercepts are equal, the equation simplifies to \(x + y = a\). This line must pass through the point \((2,3)\), so substituting these values gives

\[ \begin{aligned} x + y = a \\ 2 + 3 = a \\ a = 5 \end{aligned} \]

Substituting \(a = 5\) in the equation, we obtain

\[ \begin{aligned} \frac{x}{5} + \frac{y}{5} = 1 \\ x + y = 5 \end{aligned} \]

Hence, the equation of the required line is \(x + y = 5\)

Solution

The line cuts intercepts on the coordinate axes whose sum is 9. Let the x-intercept be \(a\), then the y-intercept will be \(9 - a\). Hence, the equation of the line in intercept form is

\[ \frac{x}{a} + \frac{y}{9 - a} = 1 \]

Since the line passes through the point \((2,2)\), substituting these values gives

\[ \begin{aligned} \frac{2}{a} + \frac{2}{9 - a} &= 1 \\ 2(9 - a) + 2a &= a(9 - a) \\ 18 - 2a + 2a &= 9a - a^2 \\ 18 &= 9a - a^2 \\ a^2 - 9a + 18 &= 0 \\ (a - 6)(a - 3) &= 0 \\ a &= 6 \quad \text{or} \quad a = 3 \end{aligned} \]

When \(a = 6\), the intercepts are \(6\) and \(3\), so the equation becomes

\[ \begin{aligned} \frac{x}{6} + \frac{y}{3} &= 1 \ x + 2y &= 6 \end{aligned} \]

When \(a = 3\), the intercepts are \(3\) and \(6\), so the equation becomes

\[ \begin{aligned} \frac{x}{3} + \frac{y}{6} &= 1 \ 2x + y &= 6 \end{aligned} \]

Hence, the required equations of the line are \(x + 2y = 6\) and \(2x + y = 6\)

Solution

The line makes an angle \(\frac{2\pi}{3}\) with the positive x-axis, so its slope is given by

\[ \begin{aligned} m &= \tan \frac{2\pi}{3} \\ &= \tan(180^\circ - 60^\circ) \\ &= -\tan 60^\circ \\ &= -\sqrt{3} \end{aligned} \]

Since the line passes through the point \((0,2)\), using point–slope form

\[ \begin{aligned} y - 2 &= -\sqrt{3}(x - 0) \\ y - 2 &= -\sqrt{3}x \\ y + \sqrt{3}x - 2 &= 0 \end{aligned} \]

For the required parallel line, the slope remains the same because parallel lines have equal slopes. It crosses the y-axis at a point 2 units below the origin, so the point is \((0,-2)\).

Using point–slope form again

\[ \begin{aligned} y + 2 &= -\sqrt{3}(x - 0) \\ y + 2 &= -\sqrt{3}x \\ y + \sqrt{3}x + 2 &= 0 \end{aligned} \]

Hence, the required equations are \(y + \sqrt{3}x - 2 = 0\) and \(y + \sqrt{3}x + 2 = 0\)

Solution

The perpendicular drawn from the origin meets the given line at the point \((-2,9)\). Hence, the line joining \((0,0)\) and \((-2,9)\) is perpendicular to the required line.

Slope of the line joining \((0,0)\) and \((-2,9)\) is

\[ \begin{aligned} m &= \frac{9 - 0}{-2 - 0} \\ &= -\frac{9}{2} \end{aligned} \]

Let the slope of the required line be \(M\). Since the two lines are perpendicular

\[ \begin{aligned} M \cdot \left(-\frac{9}{2}\right) &= -1 \\ M &= \frac{2}{9} \end{aligned} \]

The required line passes through the point \((-2,9)\), so using point–slope form

\[ \begin{aligned} y - 9 &= \frac{2}{9}(x + 2) \end{aligned} \]

Multiplying throughout by 9

\[ \begin{aligned} 9y - 81 &= 2x + 4 \\ 9y - 2x - 85 &= 0 \end{aligned} \]

Hence, the equation of the required line is \(9y - 2x - 85 = 0\)

Solution

Since the length \(L\) varies linearly with temperature \(C\), the relation between them can be written as

\[ \begin{aligned} L = mC + b \end{aligned} \]

The given data provides two points \((20, 124.942)\) and \((110, 125.134)\). The slope is

\[ \begin{aligned} m &= \frac{125.134 - 124.942}{110 - 20} \\ &= \frac{0.192}{90} \\ &= \frac{8}{375} \end{aligned} \]

Substituting the point \((20, 124.942)\) into the equation

\[ \begin{aligned} 124.942 &= \frac{8}{375} \cdot 20 + b \\ &= \frac{160}{375} + b \\ &= \frac{32}{75} + b \end{aligned} \]

Converting \(124.942 = \frac{62471}{500}\), we get

\[ \begin{aligned} b &= \frac{62471}{500} - \frac{32}{75} \\ &= \frac{187413 - 640}{1500} \\ &= \frac{186773}{1500} \end{aligned} \]

Hence, the required linear relation is

\[ \begin{aligned} L = \frac{8}{375}C + \frac{186773}{1500} \end{aligned} \]

Hence, \(L\) expressed in terms of \(C\) is \(L = \frac{8}{375}C + \frac{186773}{1500}\)

Solution

Let the weekly demand \(D\) (in litres) be a linear function of the selling price \(P\), so

\[ \begin{aligned} D = mP + b \end{aligned} \]

From the given data, the two points are \((14, 980)\) and \((16, 1220)\). The slope is

\[ \begin{aligned} m &= \frac{1220 - 980}{16 - 14} \\ &= \frac{240}{2} \\ &= 120 \end{aligned} \]

Substituting the value of \(m\) and the point \((14, 980)\)

\[ \begin{aligned} 980 &= 120 \cdot 14 + b \\ 980 &= 1680 + b \\ b &= -700 \end{aligned} \]

Thus, the demand function becomes

\[ \begin{aligned} D = 120P - 700 \end{aligned} \]

For \(P = 17\)

\[ \begin{aligned} D &= 120 \cdot 17 - 700 \\ &= 2040 - 700 \\ &= 1340 \end{aligned} \]

Hence, the owner can sell \(1340\) litres of milk per week at \(Rs\ 17\) per litre

Solution

Let the given line cut the x-axis at \(A(p,0)\) and the y-axis at \(B(0,q)\). Then its equation in intercept form is

\[ \begin{aligned} \frac{x}{p} + \frac{y}{q} = 1 \end{aligned} \]

Since \(P(a,b)\) is the midpoint of the segment joining \(A(p,0)\) and \(B(0,q)\), by midpoint formula

\[ \begin{aligned} a &= \frac{p}{2} \\ b &= \frac{q}{2} \end{aligned} \]

which gives

\[ \begin{aligned} p &= 2a \\ q &= 2b \end{aligned} \]

Substituting these values in the intercept form

\[ \begin{aligned} \frac{x}{2a} + \frac{y}{2b} &= 1 \end{aligned} \]

multiplying throughout by 2

\[ \begin{aligned} \frac{x}{a} + \frac{y}{b} &= 2 \end{aligned} \]

Hence, the equation of the line is \(\dfrac{x}{a} + \dfrac{y}{b} = 2\)

Solution

Let the required line intersect the x-axis at \(A(a,0)\) and the y-axis at \(B(0,b)\). Then its equation in intercept form is

\[ \begin{aligned} \frac{x}{a} + \frac{y}{b} = 1 \end{aligned} \]

The point \(R(h,k)\) divides the line segment joining \(A(a,0)\) and \(B(0,b)\) internally in the ratio \(1:2\). Using the section formula

\[ \begin{aligned} h &= \frac{1\cdot 0 + 2\cdot a}{1+2} = \frac{2a}{3} \\ k &= \frac{1\cdot b + 2\cdot 0}{1+2} = \frac{b}{3} \end{aligned} \]

which gives

\[ \begin{aligned} a &= \frac{3h}{2} \\ b &= 3k \end{aligned} \]

Substituting these values into the intercept form

\[ \begin{aligned} \frac{x}{\frac{3h}{2}} + \frac{y}{3k} &= 1 \\ \frac{2x}{3h} + \frac{y}{3k} &= 1 \end{aligned} \]

multiplying throughout by 3

\[ \begin{aligned} \frac{2x}{h} + \frac{y}{k} &= 3 \end{aligned} \]

Hence, the equation of the required line is \(\dfrac{2x}{h} + \dfrac{y}{k} = 3\)

Solution

Let the given points be \(A(3,0)\), \(B(-2,-2)\) and \(C(8,2)\). To prove that these points are collinear, we find the equation of the line passing through any two of them and check whether the third point satisfies it.

Slope of the line joining \(A\) and \(B\) is

\[ \begin{aligned} m &= \frac{-2 - 0}{-2 - 3} \\ &= \frac{-2}{-5} \\ &= \frac{2}{5} \end{aligned} \]

Equation of the line through \(A(3,0)\) with slope \(\frac{2}{5}\) is

\[ \begin{aligned} y - 0 &= \frac{2}{5}(x - 3) \\ 5y &= 2x - 6 \ 2x - 5y - 6 &= 0 \end{aligned} \]

Now substituting the coordinates of point \(C(8,2)\)

\[ \begin{aligned} 2(8) - 5(2) - 6 &= 16 - 10 - 6 \\ &= 0 \end{aligned} \]

Since point \(C\) satisfies the same equation, it lies on the line passing through \(A\) and \(B\)

Hence, the points \((3,0)\), \((-2,-2)\) and \((8,2)\) are collinear