NCERT Class 11 Mathematics — Straight Lines, Exercise 9.1. All 11 questions solved.

Everything in this exercise comes from the slope:

$$m = \tan\theta = \frac{y_2 – y_1}{x_2 – x_1} \qquad (x_1 \ne x_2)$$

where $\theta$ is the angle the line makes with the positive $x$-axis, measured anticlockwise.

Three consequences do most of the work:

$$\text{parallel} \iff m_1 = m_2, \qquad \text{perpendicular} \iff m_1m_2 = -1$$

$$\tan\theta = \left|\frac{m_2 – m_1}{1 + m_1m_2}\right| \quad \text{for the angle between two lines}$$

The area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is

$$\frac12\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|$$

Key insight. Slope answers geometric questions without any distances. Question 6 asks for a right angle and question 8 for a parallelogram, and both forbid the tools you would reach for first — Pythagoras and the distance formula. The point is that $m_1m_2 = -1$ settles perpendicularity, and equal slopes on opposite sides settle a parallelogram, with no square roots anywhere.

Question 1

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

Solution. Plotting the four points in the given order traces a quadrilateral $\mathrm{ABCD}$ with

$$\mathrm A(-4,5), \quad \mathrm B(0,7), \quad \mathrm C(5,-5), \quad \mathrm D(-4,-2)$$

To find the area, split it along the diagonal $\mathrm{AC}$ into two triangles and add.

Triangle ABC with $(-4,5)$, $(0,7)$, $(5,-5)$:

$$\text{area} = \frac12\left|-4(7 – (-5)) + 0((-5) – 5) + 5(5 – 7)\right|$$ $$= \frac12\left|-48 + 0 – 10\right| = \frac{58}{2} = 29$$

Triangle ACD with $(-4,5)$, $(5,-5)$, $(-4,-2)$:

$$\text{area} = \frac12\left|-4((-5) – (-2)) + 5((-2) – 5) + (-4)(5 – (-5))\right|$$ $$= \frac12\left|12 – 35 – 40\right| = \frac{63}{2} = 31.5$$

Total:

$$29 + \frac{63}{2} = \frac{58 + 63}{2} = \frac{121}{2}$$

Question 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 the vertices of the triangle.

Solution. The base lies on the $y$-axis with its midpoint at the origin, so it runs from $(0, -a)$ to $(0, a)$ — a segment of length $2a$, as required.

The third vertex lies on the perpendicular bisector of the base, which is the $x$-axis, so it has the form $(h, 0)$. Its distance to $(0, a)$ must be $2a$:

$$\sqrt{h^2 + a^2} = 2a \quad\Longrightarrow\quad h^2 = 4a^2 – a^2 = 3a^2 \quad\Longrightarrow\quad h = \pm\sqrt3\,a$$

Both signs are valid — the triangle may point either way along the $x$-axis.

Question 3

Find the distance between $\mathrm P(x_1, y_1)$ and $\mathrm Q(x_2, y_2)$ when (i) $\mathrm{PQ}$ is parallel to the $y$-axis, (ii) $\mathrm{PQ}$ is parallel to the $x$-axis.

Solution.

(i) A segment parallel to the $y$-axis has both endpoints at the same abscissa, so $x_1 = x_2$ and the distance formula collapses:

$$\mathrm{PQ} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(y_2-y_1)^2} = \left|y_2 – y_1\right|$$

(ii) Similarly $y_1 = y_2$, and

$$\mathrm{PQ} = \left|x_2 – x_1\right|$$

The modulus is essential — a distance is never negative, whichever point is labelled first.

Question 4

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

Solution. A point on the $x$-axis has the form $(x, 0)$. Equate the two squared distances — squaring avoids the radicals entirely:

$$(x – 7)^2 + (0 – 6)^2 = (x – 3)^2 + (0 – 4)^2$$

$$x^2 – 14x + 49 + 36 = x^2 – 6x + 9 + 16$$

The $x^2$ terms cancel:

$$-14x + 85 = -6x + 25 \quad\Longrightarrow\quad 60 = 8x \quad\Longrightarrow\quad x = \frac{15}{2}$$

Question 5

Find the slope of a line which passes through the origin and the mid-point of the line segment joining the points $\mathrm P(0, -4)$ and $\mathrm B(8, 0)$.

Solution. The midpoint of $\mathrm{PB}$ is

$$\left(\frac{0 + 8}{2},\ \frac{-4 + 0}{2}\right) = (4, -2)$$

The slope of the line through the origin and $(4, -2)$:

$$m = \frac{-2 – 0}{4 – 0} = -\frac12$$

Question 6

Without using the Pythagoras theorem, show that the points $(4, 4)$, $(3, 5)$ and $(-1, -1)$ are the vertices of a right-angled triangle.

Solution. Label $\mathrm A(4,4)$, $\mathrm B(3,5)$, $\mathrm C(-1,-1)$ and compute the three slopes:

$$m_{\mathrm{AB}} = \frac{5-4}{3-4} = \frac{1}{-1} = -1$$

$$m_{\mathrm{BC}} = \frac{-1-5}{-1-3} = \frac{-6}{-4} = \frac32$$

$$m_{\mathrm{AC}} = \frac{-1-4}{-1-4} = \frac{-5}{-5} = 1$$

Now test the pairs. For $\mathrm{AB}$ and $\mathrm{AC}$:

$$m_{\mathrm{AB}} \times m_{\mathrm{AC}} = (-1)(1) = -1$$

So $\mathrm{AB} \perp \mathrm{AC}$, and the triangle has a right angle at $\mathrm A$. $\blacksquare$

(The other two products are $-\tfrac32$ and $\tfrac32$, neither equal to $-1$, so $\mathrm A$ is the only right angle.)

Question 7

Find the slope of the line which makes an angle of $30^\circ$ with the positive direction of the $y$-axis, measured anticlockwise.

Solution. Slope is $\tan$ of the angle with the positive $x$-axis, so the given angle must be converted first. The positive $y$-axis is at $90^\circ$; turning a further $30^\circ$ anticlockwise gives

$$\theta = 90^\circ + 30^\circ = 120^\circ$$

$$m = \tan 120^\circ = \tan(180^\circ – 60^\circ) = -\tan 60^\circ = -\sqrt3$$

The negative sign is the whole point of the question — a line inclined past the vertical slopes downwards.

Question 8

Without using the distance formula, show that the points $(-2,-1)$, $(4,0)$, $(3,3)$ and $(-3,2)$ are the vertices of a parallelogram.

Solution. Label them $\mathrm A(-2,-1)$, $\mathrm B(4,0)$, $\mathrm C(3,3)$, $\mathrm D(-3,2)$ and compute the slopes of the four sides:

$$m_{\mathrm{AB}} = \frac{0 – (-1)}{4 – (-2)} = \frac16, \qquad m_{\mathrm{CD}} = \frac{2 – 3}{-3 – 3} = \frac{-1}{-6} = \frac16$$

$$m_{\mathrm{BC}} = \frac{3 – 0}{3 – 4} = \frac{3}{-1} = -3, \qquad m_{\mathrm{DA}} = \frac{-1 – 2}{-2 – (-3)} = \frac{-3}{1} = -3$$

So $\mathrm{AB} \parallel \mathrm{CD}$ and $\mathrm{BC} \parallel \mathrm{DA}$: both pairs of opposite sides are parallel, which is the definition of a parallelogram. $\blacksquare$

Equal slopes give parallelism directly; no lengths are needed, which is exactly what the question demands.

Question 9

Find the angle between the $x$-axis and the line joining the points $(3, -1)$ and $(4, -2)$.

Solution. The slope of the line is

$$m = \frac{-2 – (-1)}{4 – 3} = \frac{-1}{1} = -1$$

So $\tan\theta = -1$. The angle of inclination is measured in $[0^\circ, 180^\circ)$, and in that range the solution of $\tan\theta = -1$ is

$$\theta = 180^\circ – 45^\circ = 135^\circ$$

Question 10

The slope of a line is double the slope of another line. If the tangent of the angle between them is $\dfrac13$, find the slopes of the lines.

Solution. Let the slopes be $m$ and $2m$. Using the angle formula:

$$\left|\frac{2m – m}{1 + m(2m)}\right| = \frac13 \quad\Longrightarrow\quad \left|\frac{m}{1 + 2m^2}\right| = \frac13$$

The modulus splits into two cases.

Case $\dfrac{m}{1+2m^2} = \dfrac13$:

$$3m = 1 + 2m^2 \quad\Longrightarrow\quad 2m^2 – 3m + 1 = 0 \quad\Longrightarrow\quad (2m-1)(m-1) = 0$$

so $m = \tfrac12$ or $m = 1$.

Case $\dfrac{m}{1+2m^2} = -\dfrac13$:

$$-3m = 1 + 2m^2 \quad\Longrightarrow\quad 2m^2 + 3m + 1 = 0 \quad\Longrightarrow\quad (2m+1)(m+1) = 0$$

so $m = -\tfrac12$ or $m = -1$.

Each value of $m$ gives a pair $(m, 2m)$:

Question 11

A line passes through $(x_1, y_1)$ and $(h, k)$. If the slope of the line is $m$, show that $k – y_1 = m(h – x_1)$.

Solution. By definition, the slope of the line through the two points is the change in ordinate over the change in abscissa:

$$m = \frac{k – y_1}{h – x_1}$$

Multiplying both sides by $(h – x_1)$, which is non-zero since the line is not vertical:

$$k – y_1 = m\,(h – x_1) \qquad \blacksquare$$

This is exactly the point-slope form of a line’s equation, and the next exercise uses it to write down equations directly.

Common mistakes

• Question 1, using the shoelace formula on badly ordered vertices. Splitting into triangles along a diagonal is safer: it fails visibly if the diagonal lies outside the shape, whereas a mis-ordered shoelace silently returns a wrong number.
• Question 2, placing the third vertex on the $y$-axis. The base already lies there; the apex must be on the perpendicular bisector, which is the $x$-axis.
• Question 3, dropping the modulus. $y_2 – y_1$ may be negative depending on how the points are labelled; a distance may not.
• Question 6, computing lengths anyway. The question forbids Pythagoras, so the intended route is the slope product $m_1m_2 = -1$. It is also faster.
• Question 7, answering $\tan 30^\circ$. The angle given is measured from the $y$-axis, so it must be converted to $120^\circ$ from the $x$-axis first. The answer is negative, which is the check that the conversion was done.
• Question 9, giving $-45^\circ$. The inclination of a line lies in $[0^\circ, 180^\circ)$, so the answer is $135^\circ$.
• Question 10, considering only one sign. The formula has a modulus, so both $+\tfrac13$ and $-\tfrac13$ must be solved, giving four pairs of slopes rather than two.
• Assuming a vertical line has slope $0$. A vertical line has undefined slope (the denominator $x_2 – x_1$ is zero); a horizontal line has slope $0$.

Practise next

• Exercise 9.2 — the various forms of a line’s equation, all built on the point-slope form derived in question 11.
• Exercise 9.3 — general form, distance from a point to a line, and distance between parallel lines.