# NCERT Solutions for Class 11th Mathematics

## Chapter 10. Straight Lines

### Exercise 10.1

**Exercise 10.1**

**Q1. 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, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as joining Point A to C forms a diagonal AC.

Accordingly, area (ABCD) = area(∆ABC) + area (∆ACD)

Using area of triangle formula.

Therefore, Area of ∆ACD where A (-4, 5), C (5, -5), D (-4, -2)

**Q2. The base of an equilateral triangle with side 2 a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.**

**Solution: **

**Q3. Find the dis tance 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 are : 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}

Using Distance formula for distance between P and Q

**Q4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).**

**Solution: **

Let the point on x-axis be P (x, 0) which is equidistance from points A (7, 6) and B (3, 4).

Accordingly,

AP = BP

Or AP^{2} = BP^{2} [Squaring both sides]

⇒ (x – 7)^{2} + (0 – 6)^{2} = (x – 3)^{2} + (0 – 4)^{2}

⇒ x^{2} – 14x + 49 + 36 = x^{2} – 6x + 9 + 16

⇒ x^{2} – 14x + 85 = x^{2} – 6x + 25

⇒ 85 – 25 = 14x – 6x

**Q5. 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 coordinates of the mid-point of the line segment joining the points P (0, *– *4) and B (8, 0).

**Q6. 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

A (4, 4), B (3, 5), and C (–1, –1).

If given vertices are of a right angle triangle.

m_{1}m_{3} = -1

Slope of AB (m_{1}) × Slope of AC (m_{3}) = -1

It means side AB and AC are perpendicular to each other.

Here, given triangle is right-angled at point A (4, 4).

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

**Q7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.**

**Q8. Find the value of x for which the points (x, – 1), (2,1) and (4, 5) are collinear.**

**Solution: **

Let point be A (*x*, –1), B (2, 1), and C (4, 5).

If points A (*x*, –1), B (2, 1), and C (4, 5) are collinear, then

Slope of AB = Slope of BC

**Q9. Without using distance formula, show that points ( – 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.**

**Solution: **

To be the points A(*– *2, *– *1), B(4, 0), C(3, 3) and D(*–*3, 2) of the vertices of a parallelogram.

There must be AB || CD or BC || AD

Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

**Q10. Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).**

**Solution: **

Let be points A(3,*–*1) and B(4,*–*2) are given for a line.

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

**Solution:**

**Q12. A line passes through ( x_{1}, y_{1}) and (h, k). If slope of the line is m, show that k – y_{1} = m (h – x_{1}).**

**Solution: **

Line passes through points (*x*_{1}*, y*_{1}) and (*h, k*).

Q14. 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?

**Solution: **

Line AB passes through points A(1985, 92) and B(1995, 97).

Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y).