**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:**

The figure of the quadrilateral whose vertices are A(- 4, 5), B(0, 7), C(5, -5), and D(-4, -2) are shown in the below figure.

**Question 2.**

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

**Solution:**

Since base of an equilateral triangle lies along y-axis.

**Question 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:**

We are given that co-ordinates of P is (x_{1}, y_{1}) and Q is (x_{2}, y_{1}).

Distance between the points P(x_{1}, y_{1}) and Q(x_{2}, y_{1}) is

**Question 4.**

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

**Solution:**

Let the point be P(x, y). Since it lies on the x-axis ∴ y = 0 i.e., required point be (x, 0).

Since the required point is equidistant from points A(7, 6) and B(3, 4) ⇒ PA = PB

**Question 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:**

We are given that P(0, -4) and B(8, 0).

Let A be the midpoint of PB, then

**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:**

Let A(4, 4), B(3, 5) and C(-1, -1) be the vertices of ∆ABC.

Let m_{1} and m_{2} be the slopes of AB and AC respectively.

**Question 7.**

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

**Solution:**

The given line makes an angle of 90° + 30° = 120° with the positive direction of x-axis.

Hence, m = tan 120° = – .

**Question 8.**

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

**Solution:**

Let A(x, -1), B(2, 1) and C(4, 5) be the given collinear points. Then by collinearity of A, B, C, we have slope of AB = slope of BC

**Question 9.**

**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 A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2) be the vertices of the given quadrilateral ABCD. Then,

**Question 10.**

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

**Solution:**

We are given that the points are A(3, -1) and B(4, -2)

**Question 11.**

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

**Solution:**

Let m_{1} and m_{2} be the slopes of two lines.

**Question 12.**

**A line passes through (x _{1}, y_{2}) and (h, k). If slope of the line is m, show that k – y_{1} = m (h – x_{1}).**

**Solution:**

A line passes through (x_{1}, y_{1}) and (h, k). Also, the slope of the line is m.

**Question 13.**

**If three points (h, 0), (a, b) and (0, k) lie on a line, show that **

**Solution:**

Let A(h, 0), B(o, b) and C(0, k) be the given collinear points.

∴ Slope of AB = Slope of BC

**Question 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?**

**Solution:**

Slope of AB +

Let the population in year 2010 is y, and co-ordinate of C is (2010, y) then, slope of AB = slope of BC