Question

Solve the following:
chapter: straight lines, class 11

Answer 1

Answer:

. 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, AD which gives us the required quadrilateral.

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 1

To find the area, draw diagonal AC

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

Then, area of triangle with vertices (x1,y1) , (x2, y2) and (x3,y3) is

Are of ∆ ABC = ½ [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]

= ½ [-4 (7 + 5) + 0 (-5 – 5) + 5 (5 – 7)] unit2

= ½ [-4 (12) + 5 (-2)] unit2

= ½ (58) unit2

= 29 unit2

Are of ∆ ACD = ½ [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]

= ½ [-4 (-5 + 2) + 5 (-2 – 5) + (-4) (5 – (-5))] unit2

= ½ [-4 (-3) + 5 (-7) – 4 (10)] unit2

= ½ (-63) unit2

= -63/2 unit2

Since area cannot be negative area ∆ ACD = 63/2 unit2

Area (ABCD) = 29 + 63/2

= 121/2 unit2

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.

Solution:

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 2

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

Where, AB = BC = AC = 2a

In the above figure, by 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 co-ordinates of point C are (0, a) and that of B are (0,-a)

Since 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 = OA2 + OC2

(2a)2= a2 + OC2

4a2 – a2 = OC2

3a2 = OC2

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 (x1, y1) and Q (x2, y2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Solution:

Given:

Points P (x1, y1) and Q(x2, y2)

(i) When PQ is parallel to y axis then x1 = x2

So, the distance between P and Q is given by

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 3

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 4

= |y2 – y1|

(ii) When PQ is parallel to the x-axis then y1 = y2

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

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 5

= NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 6

= |x2 – x1|

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

Solution:

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

So,

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 7

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

a2 – 14a + 85 = a2 – 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 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 (x1, y1) and

(x2, y2) is given by m = (y2 – y1)/(x2 – x1) where, x ≠ x1

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 the 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 (x1, y1) and

(x2, y2) is given by m = (y2 – y1)/(x2 – x1) where, x ≠ x1

So, the slope of the line AB (m1) = (5-4)/(3-4) = 1/-1 = -1

the slope of the line BC (m2) = (-1-5)/(-1-3) = -6/-4 = 3/2

the slope of the line CA (m3) = (4+1)/(4+1) = 5/5 = 1

It is observed that, m1.m3 = -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)

Answer 2

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