NCERT Solutions class-11 Maths Exercise 10.1

Exercise 10.1

1. Draw a quadrilateral in the Cartesian plane, whose vertices are and Also, find its area.

Ans. Area of quadrilateral ABCD = Area of + Area of trapezium ABED

=

=

= 26 + 34.8 = 60.8 sq. units

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2. The base of an equilateral triangle with side lies along the axis such that the mid-point of the base is at origin. Find the vertices of the triangle.

Ans. Given: Length of side of equilateral triangle = . The base of triangle lies along axis and the mid-point of base is at origin so that the coordinates of vertices are and

Let vertices of the third vertex be

The vertices of triangle are , and

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3. Find the distance between P and Q when (i) PQ is parallel to the axis (ii) PQ is parallel to the axis.

Ans. Given: P and Q are two points.

PQ =

(i) PQ is parallel to axis, then

PQ =

(ii) PQ is parallel to axis, then

PQ =

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

Ans. Let P be any point on the axis which is equidistant from Q (7, 6) and R (3, 4).

PQ =

=

=

And PR =

=

=

According to question, PQ = PR

=

Squaring both sides, =

Therefore, the coordinates of the point are

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

Ans. Here, mid-point of the line segment joining P and Q (8, 0) is

Since the line passes through points (0, 0) and .

Therefore, Slope of the line =

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

Ans. Let A (4, 4), B (3, 5) and C be three vertices of a

Slope of AB =

Slope of BC =

Slope of AC =

Now, Slope of AB x Slope of AC =

This shows that AB AC. Thus is right angled at point A.

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7. Find the slope of the line, which makes an angle of with the positive direction of axis measured anticlockwise.

Ans. The line makes an angle of with the positive direction of axis.

Now the line makes an angle of with the positive direction of axis.

Slope of the line =

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8. Find the value of for which the points and are collinear.

Ans. Let A B (2, 1) and C (4, 5) be three collinear points.

Slope of AB =

Slope of BC =

According to question, Slope of AB = Slope of BC

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9. Without using distance formula, show that the points and are the vertices of a parallelogram.

Ans. Let A B (4, 0), C (3, 3) and D be vertices of a quadrilateral ABCD.

Slope of AB = Slope of BC =

Slope of DC = Slope of AD =

Here Slope of AB = Slope of DC

AB DC

And Slope of BC = Slope of AD

BC AD

Therefore, ABCD is a parallelogram.

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10. Find the angle between the axis and the line joining the points and

Ans. Let A and B be two points. Let Q be the angle which AB makes with positive direction of axis.

Slope of AB =

Also Slope of AB =

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11. The slope of a line is double of the slope of the another line. If tangent of the angle between them is find the slopes of the lines.

Ans. Given: . Let the slopes of two lines be and

Taking

and

Taking

and

Therefore, the slopes of lines are and or 1 and

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12. A line passes through and If slope of the line is show that

Ans. Let A and B be two points. It is given that Slope of AB =

Slope of AB = = (given)

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13. Is three points and lies on a line, show that

Ans. Let A B and C be three points lie on the line.

Slope of AB = Slope of BC =

Slope of AB = Slope of BC (given)

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14. Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?

Ans. Given: The points on the line are A (1985, 92) and B (1995, 97).

Slope of AB =

Let the population in year 2010 be crores. Then C lies on the line.

Slope of BC =

Since points A, B and C lie on the line.

Slope of AB = Slope of BC

Therefore, population in 2010 will be 104.5 crores.