# NCERT Solutions Class 11 Maths Chapter 09 Straight Lines

## NCERT Solutions Class 11 Maths Chapter 09 Straight Lines

The NCERT Solutions for Class 11 Maths Chapter 9, covering Straight Lines, are a helpful resource for students preparing for their Math board exams. The chapter covers important topics like slope, Horizontal and vertical Lines, Point-slope form, Two-point form, Slope-intercept form, Intercept – form, and Normal form. These solutions are designed to help students practice and improve their understanding using straightforward methods and formulas.

### NCERT Solutions for Class 11 Maths Chapter 09

**EXERCISE 9.1 PAGE NO: 158**

**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:**

Let ABCD be the given quadrilateral with vertices A (-4,5), B (0,7), C (5.-5) and D (-4,-2).

Now, let us plot the points on the Cartesian plane by joining the points AB, BC, CD, and AD, which give us the required quadrilateral.

To find the area, draw diagonal AC.

So, area (ABCD) = area (∆ABC) + area (∆ADC)

Then, area of triangle with vertices (x_{1},y_{1}) , (x_{2}, y_{2}) and (x_{3},y_{3}) is

Are of ∆ ABC = ½ [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= ½ [-4 (7 + 5) + 0 (-5 – 5) + 5 (5 – 7)] unit^{2}

= ½ [-4 (12) + 5 (-2)] unit^{2}

= ½ (58) unit^{2}

= 29 unit^{2}

Are of ∆ ACD = ½ [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= ½ [-4 (-5 + 2) + 5 (-2 – 5) + (-4) (5 – (-5))] unit^{2}

= ½ [-4 (-3) + 5 (-7) – 4 (10)] unit^{2}

= ½ (-63) unit^{2}

= -63/2 unit^{2}

Since area cannot be negative, area ∆ ACD = 63/2 unit^{2}

Area (ABCD) = 29 + 63/2

= 121/2 unit^{2}

**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:**

Let us consider ABC, the given equilateral triangle with side 2a.

Where, AB = BC = AC = 2a

In the above figure, assuming that the base BC lies on the x-axis such that the mid-point of BC is at the origin, i.e., BO = OC = a, where O is the origin.

The coordinates of point C are (0, a) and that of B are (0,-a).

The line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular.

So, vertex A lies on the y –axis.

By applying Pythagoras’ theorem,

(AC)^{2} = OA^{2} + OC^{2}

(2a)^{2}= a^{2} + OC^{2}

4a^{2} – a^{2} = OC^{2}

3a^{2 }= OC^{2}

OC =√3a

Co-ordinates of point C = **±** √3a, 0

∴ The vertices of the given equilateral triangle are (0, a), (0, -a), (√3a, 0)

Or (0, a), (0, -a) and (-√3a, 0)

**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.**

**Solution:**

Given:

Points P (x_{1}, y_{1}) and Q(x_{2}, y_{2})

**(i)** When PQ is parallel to the y-axis, then x_{1} = x_{2}

So, the distance between P and Q is given by

= |y_{2} – y_{1}|

**(ii)** When PQ is parallel to the x-axis, then y_{1} = y_{2}

So, the distance between P and Q is given by =

=

= |x_{2} – x_{1}|

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

**Solution:**

Let us consider (a, 0) to be the point on the x-axis that is equidistant from the point (7, 6) and (3, 4).

So,

Now, let us square on both sides; we get,

a^{2} – 14a + 85 = a^{2} – 6a + 25

-8a = -60

a = 60/8

= 15/2

∴ The required point is (15/2, 0)

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

**Solution:**

The co-ordinates of the mid-point of the line segment joining the points P (0, – 4) and B (8, 0) are (0+8)/2, (-4+0)/2 = (4, -2)

The slope ‘m’ of the line non-vertical line passing through the point (x_{1}, y_{1}) and

(x_{2}, y_{2}) is given by m = (y_{2} – y_{1})/(x_{2} – x_{1}) where, x ≠ x_{1}

The slope of the line passing through (0, 0) and (4, -2) is (-2-0)/(4-0) = -1/2

∴ The required slope is -1/2.

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

**Solution:**

The vertices of the given triangle are (4, 4), (3, 5) and (–1, –1).

The slope (m) of the line non-vertical line passing through the point (x_{1}, y_{1}) and

(x_{2}, y_{2}) is given by m = (y_{2} – y_{1})/(x_{2} – x_{1}) where, x ≠ x_{1}

So, the slope of the line AB (m_{1}) = (5-4)/(3-4) = 1/-1 = -1

The slope of the line BC (m_{2}) = (-1-5)/(-1-3) = -6/-4 = 3/2

The slope of the line CA (m_{3}) = (4+1)/(4+1) = 5/5 = 1

It is observed that m_{1}.m_{3} = -1.1 = -1

Hence, the lines AB and CA are perpendicular to each other.

∴ given triangle is right-angled at A (4, 4)

And the vertices of the right-angled ∆ are (4, 4), (3, 5) and (-1, -1)

**7. Find the slope of the line, which makes an angle of 30° with the positive direction of the y-axis measured anticlockwise.**

**Solution:**

We know that if a line makes an angle of 30° with the positive direction of the y-axis measured anti-clock-wise, then the angle made by the line with the positive direction of the x-axis measured anti-clock-wise is 90° + 30° = 120°

∴ The slope of the given line is tan 120° = tan (180° – 60°)

= – tan 60°

= –**√**3

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

**Solution:**

Let the given point be A (-2, -1) , B (4, 0) , C ( 3, 3) and D ( -3, 2)

So now, the slope of AB = (0+1)/(4+2) = 1/6

The slope of CD = (3-2)/(3+3) = 1/6

Hence, the Slope of AB = Slope of CD

∴ AB ∥ CD

Now,

The slope of BC = (3-0)/(3-4) = 3/-1 = -3

The slope of AD = (2+1)/(-3+2) = 3/-1 = -3

Hence, the Slope of BC = Slope of AD

∴ BC ∥ AD

Thus, the pair of opposite sides are quadrilateral are parallel, so we can say that ABCD is a parallelogram.

Hence, the given vertices, A (-2, -1), B (4, 0), C(3, 3) and D(-3, 2) are vertices of a parallelogram.

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 joining the points (3, -1) and (4, -2) is given by

m = (y_{2} – y_{1})/(x_{2} – x_{1}) where, x ≠ x_{1}

m = (-2 –(-1))/(4-3)

= (-2+1)/(4-3)

= -1/1

= -1

The angle of inclination of the line joining the points (3, -1) and (4, -2) is given by

tan θ = -1

θ = (90° + 45°) = 135°

∴ The angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

**10. The slope of a line is double the slope of another line. If the tangent of the angle between them is 1/3, find the slopes of the lines.**

**Solution:**

Let us consider ‘m_{1}’ and ‘m’ be the slope of the two given lines such that m_{1 }= 2m

We know that if θ is the angle between the lines l1 and l2 with slope m_{1} and m_{2}, then

1+2m^{2} = -3m

2m^{2} +1 +3m = 0

2m (m+1) + 1(m+1) = 0

(2m+1) (m+1)= 0

m = -1 or -1/2

If m = -1, then the slope of the lines are -1 and -2

If m = -1/2, then the slope of the lines are -1/2 and -1

Case 2:

2m^{2} – 3m + 1 = 0

2m^{2} – 2m – m + 1 = 0

2m (m – 1) – 1(m – 1) = 0

m = 1 or 1/2

If m = 1, then the slope of the lines are 1 and 2

If m = 1/2, then the slope of the lines are 1/2 and 1

∴ The slope of the lines are [-1 and -2] or [-1/2 and -1] or [1 and 2] or [1/2 and 1]

**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:**

Given: the slope of the line is ‘m’.

The slope of the line passing through (x_{1}, y_{1}) and (h, k) is (k – y_{1})/(h – x_{1})

So,

(k – y_{1})/(h – x_{1}) = m

(k – y_{1}) = m (h – x_{1})

Hence, proved.