* **Q-1. Construct a quadrilateral in the Cartesian plane with vertices (-2, 5), (0, 6), (4, -4) and (-3, -1). Also, find the area of the quadrilateral.*

*Solution. *

Let, MNOP be the given quadrilateral having vertices M (-2, 5), N (0, 6), O (4, -4) and P (-3, -1).

Now, plot M, N, O and P on the Cartesian plane and join MN, NO, OP and PM.

The quadrilateral, thus, formed will be

As we need to find the area of the quadrilateral MNOP, so draw a diagonal, say, MO.

Now,

Area of (MNOP) = area of (

Area of a triangle with vertices (a_{1}, b_{1}), (a_{2}, b_{2}), (a_{3}, b_{3}) and (a_{4}, b_{4}) is given by:

_{1 }(b_{2} – b_{3 }) + a_{2 }(b_{3} – b_{1 }) + a_{3 }(b_{1} – b_{2 }) |

Now,

Vertices of

Area of

=

=

= 12 unit^{2}

Vertices of

Area of

=

=

= 22.5 unit^{2 }

*Area of MNOP = Area of ΔMNO + Area of ΔMOP*

*= 12 + 22.5 = 34.5 22.5 unit ^{2 }*

*Q-2. Consider an equilateral triangle, each of whose sides are 2b which lies on the y- axis in such a manner that, the mid- point of each of its base is at the origin. Obtain all the vertices of the equilateral triangle. *

*Solution.*

Let, XYZ be the given equilateral triangle each of whose side is 2b.

Since, XYZ is an equilateral triangle,

So, XY = YZ = ZX = 2b

Now, consider that the base YZ lies along y-axis in such a way that it’s mid- point is at the origin.

So, YA = AZ = b, where A is the origin.

Thus, by our observation, the coordinates of the point Y are (0, b), while the coordinates of the point Z is (0, -a).

We know that, the line joining the vertex of an equilateral triangle having the mid- point of its opposite sides is perpendicular.

Thus, the vertex X lies on the y- axis.

By applying the Pythagoras theorem to

( XZ )^{2} = ( AX )^{2} + ( AZ )^{2}

The coordinate of the point X = (

*Hence, the vertices of the equilateral triangle XYZ are X( 3–√b , 0 ), Y( 0, b ) and Z( 0, -b ) or X( −3–√b , 0 ), Y( 0, b ) and Z( 0, -b )*

*Q-3. What is the distance between X (a _{1}, b_{1}) and Y (a_{2}, b_{2}) if:*

*(a) XY is parallel to x- axis*

*(b) XY is parallel to y- axis*

*Solution.*

Given points are X (a_{1}, b_{1}) and Y (a_{2}, b_{2})

**(a)** The given condition is:

XY is parallel to the x- axis, i.e., b_{1 }= b_{2}

Therefore, the distance hence measured between X and Y in this case =

=

=

= | a_{2 – }a_{1 }|

**(b)** The given condition is:

XY is parallel to the y- axis, i.e., a_{1 }= a_{2}

Therefore, the distance hence measured between X and Y in this case =

=

=

= | b_{2 – }b_{1 }|

*Q-4. Consider two points (6, 5) and (4, 2). Get a point on the y- axis which is equivalent from the given two points.*

*Solution. *

Let us assume that, (0, y) be the given point on y- axis which is equivalent to the two given points (6, 5) and (4, 2).

Now,

Squaring both side, we will get

6a = 41

a =

**a = 6.8 unit**

*Therefore, (0, 416 ) is the point on Y – axis which is equivalent to the given points (6, 5) and (4, 2).*

*Q-5. What is the slope of a line passing through the origin and, the mid- point of the line- segment joining the two points O (0, -5) and A (9, 0)?*

*Solution.*

The two points of the line- segment is O (0, -5) and A (9, 0).

Then, the co-ordinates of the mid-point of line-segment OA are-

We know that the slope (say, m) of the non- vertical line which passes through the points (a_{1}, b_{1}) and (a_{2}, b_{2}) is:

m = _{2 }≠ a_{1}

Hence, the slope of the line- segment passing through (0, 0) and (4.5, -2.5) is given by:

*Therefore, the required slope of the line-segment = −59*

*Q-6. Prove that the points (5, 5), (4, 6) and (-2, -2) are the vertices of the right- angled triangle, without using Pythagoras theorem.*

*Solution. *

There are three points given in the question. Let, those points be the vertices of the triangle XYZ, namely, X (5, 5), Y (4, 6) and Z (-2, -2).

We know that, the slope (say, m) of the non- vertical line which passes through the points (a_{1}, b_{1}) and (a_{2}, b_{2}) is:

m = _{2 }≠ a_{1}

The slope of XY( m_{1} ) =

The slope of YZ( m_{2} ) =

The slope of ZX( m_{3} ) =

Here, we can see that, m_{1}. m_{3} = -1

This proves that, the line- segments XY and ZX is perpendicular to each other, i.e., the triangle XYZ is right- angled at X (5, 5).

*Hence, proved that the given points, i.e., (5, 5), (4, 6) and (-2, -2) are the vertices of the right- angled triangle.*

* *

*Q-7. What is the slope of the line which makes an angle 60∘ along the positive direction of the Y- axis which is measured in anticlockwise sequence.*

*Solution.*

By mathematical calculation rule,

If a line makes an angle of

Then, if the angle measured by the line in the x- axis measured in positive direction of the x- axis, measured in anticlockwise direction is-

*Therefore, the slope of the given line is tan 150∘ = tan( 180∘–30∘) = – tan 30∘ = 13√.*

*Q-8. What will be the value of a so that the points (a, -2), (3, 2) and (5, 6) get collinear to each other?*

*Solution. *

Let us assume that, the points X (a, -2), Y (3, 2) and Z (5, 6) are collinear.

Now,

Slope of XY = Slope of YZ

4 = 6 – 2a

2a = 2

a = 1

*Therefore, the value of a = 1.*

*Q-9. Prove that the points (-3, -2), (5, 0), (4, 4) and (-4, 2) are the vertices of a parallelogram without using the distance formula.*

*Solution.*

Let us denote the given points as P (-3, -2), Q (5, 0), R (4, 4) and S (-4, 2).

Now,

Slope of the side PQ =

Slope of the side RS =

Thus,

Slope of the side PQ = Slope of the side RS

Hence,

PQ and RS are parallel to each other.

Now,

Slope of the side QR =

Slope of the side PS =

Thus,

Slope of the side QR = Slope of the side PS

Hence,

QR and PS are parallel to each other.

Hence, both of the opposite side‘s pairs of the quadrilateral PQRS are parallel. Thus, PQRS is a parallelogram.

*Therefore, the given points P (-3, -2), Q (5, 0), R (4, 4) and S (-4, 2) are the vertices of the parallelogram.*

*Q-10. Consider two points (4, -2) and (5, -3). What is the angle between the x- axis and the line joining the given two points?*

*Solution.*

The slope of the line joining the given two points, namely, X (4, -2) and Y (5, -3) is given by:

M =

Thus,

The angle of inclination, say

tan

Since,

*Therefore, between the x- axis and the line joining the given two points, the angle of inclination is 135∘.*

*Q-11. Assume that the slope of a line AB is twice the slope of the other line PQ. Get all the slopes of the lines when the tangent of the angle between them is 13.*

*Solution.*

Let us assume that, the slope of the given two lines be m_{1} and m_{2}, so that

m_{2 }= 2 m_{1}

Now,

It is known that if _{1} and m_{2}, then

tan

According to the question, the tangent of the angle between the two given lines is

*CASE- I:*

Let, m_{1} = -1, then the slopes of the lines are -1 and -2.

Let,

*CASE- II:*

Let, m_{1} = 1, then the slopes of the lines are 1 and 2.

Let,

Therefore, the slopes of the lines AB and PQ are -1 and -2 or,

*Q-12. Consider a line passing through two points (a _{1}, b_{1}) and (j, k). Assume that the slope of the line passing through these points is m. Prove that:*

*k – b _{1} = m (j – a_{1}).*

*Solution. *

As we know that:

The slope of the line, say AB, passing through the two given points (a_{1}, b_{1}) and (j, k) is given by:

Therefore,

Hence, proved.

*Q-13. Assume that the three points (a, 0), (p, r) and (0, h) lies on the same line. Prove that:*

*Solution.*

As per the given condition in the question,

Points X( a, 0 ), Y( p, r ) and Z( 0, h ) lies on the same line, then

The slope of XY = The slope of YZ

-pr = (h – r) (p – a)

-pr = hp – ha – rp + ra

hp + ra = ha

Dividing both side by ha, we will get

Hence, proved.

*Q-14. Take the records of population and year graph given below. What will be the slope of the line XY? By using this, find the population in the year 2005.*

*Solution.*

Here we can observe that the line XY passes through points X (1985, 90) and Y (1995, 95).

Then, the slope of the line XY is given by

Let, a be the population of the year 2005.

Then, as per the data in the graph, the line XY will exactly pass through the point Z( 2005, y ).

Slope of XY = Slope of YZ

So, y – 95 = 5

So, y = 100

Hence, the slope of the line XY is