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Q : 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?

Given point A(1985,92) and B(1995,97) Now, we know that  Therefore, the slope of line AB is   Now, the equation of the line passing through the point (1985,92) and with slope =    is given by  Now, in the year 2010 the population is Therefore, the population in the year 2010 is 104.5 crore

Q: 13         If three points  $(h,0),(a,b)$  and  $(0,k)$  lie on a line, show that   $\frac{a}{h}+\frac{b}{k}=1.$

Points    and    lie on a line so by this we can say that their slopes are also equal We know that Slope of AB =  Slope of AC =  Now, Slope of AB = slope of AC  Now divide both the sides by hk Hence proved

Q: 12         A line passes through  $(x_1,y_1)$  and  $(h,k)$ .  If slope of the line is $m$, show that

$k-y_1=m(h-x_1)$.

Given that  A line passes through    and    and slope of the line is m Now, Hence proved

Q : 11        The slope of a line is double of the slope of another line. If tangent of the angle between them is    $\frac{1}{3},$  find the slopes of the lines

Let   are the slopes of lines and  is the angle between them Then, we know that It is given that        and      Now, Now,  Now, According to which value of  Therefore,

Q : 10        Find the angle between the x-axis and the line joining the points $(3,-1)$ and  $(4,-2)$

We know that So, we need to find the slope of line joining points (3,-1) and (4,-2) Now, Therefore, angle made by line with positive x-axis when measure in anti-clockwise direction is

Q: 9         Without using the distance formula, show that points  $(-2,-1),(4,0),(3,3)$  and $(-3,2)$  are  the vertices of a parallelogram.

Given points are    and We know the pair of the opposite side are parallel to each other in a parallelogram Which means their slopes are also equal The slope of AB =   The slope of BC =   The slope of CD =   The slope of AD =  We can clearly see that The slope of AB = Slope of CD               (which means they are parallel) and The slope of BC = Slope of AD               (which...

Q : 8         Find the value of $x$ for which the points  $(x,-1),(2,1)$ and  $(4,5)$  are collinear.

Point is collinear which means they lie on the same line by this we can say that their slopes are equal Given points are A(x,-1) , B(2,1) and C(4,5) Now, The slope of AB = Slope of BC Therefore, the value of x is 1

Q: 7         Find the slope of the line, which makes an angle of  $30^{\circ}$ with the positive direction of  $y$-axis measured anticlockwise.

It is given that the line makes an angle of   with the positive direction of  -axis measured anticlockwise Now, we know that  line makes an angle of   with the positive direction of  -axis  Therefore, the angle made by line with the positive x-axis is =  Now, Therefore, the slope of the line is

Q ; 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.

It is given that  point A(4,4) , B(3,5) and C(-1,-1) are the vertices of a right-angled triangle Now, We know that the distance between two points is given by Length of AB   Length of BC  Length of AC  Now, we know that  Pythagoras theorem is Is clear that Hence proved

Q: 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  $B(8,0)$.

Mid-point of the line joining  the points and  . is It is given that line also passes through origin which means passes through the point (0, 0) Now, we have two points on the line so we can now find the slope of a line by using formula  Therefore, the slope of the line is

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

Point is on the x-axis, therefore, y coordinate is 0 Let's assume the point is (x, 0) Now, it is given that the given point (x, 0) is equidistance from point  (7, 6) and (3, 4) We know that Distance between two points is given by Now, and Now, according to the given condition Squaring both the sides Therefore, the point is

Q : 3         Find the distance between  $P(x_1,y_1)$  and  $Q(x_2,y_2)$  when :

(ii) PQ is parallel to the $x$-axis.

When PQ is parallel to the x-axis  then, x coordinates are equal i.e.  Now, we know that the distance between two points is given by Now, in this case  Therefore, Therefore, the distance between and    when  PQ is parallel to the x-axis  is

Q : 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.

When PQ is parallel to the y-axis  then, x coordinates are equal i.e.  Now, we know that the distance between two points is given by Now, in this case  Therefore, Therefore, the distance between and    when  PQ is parallel to y-axis  is

Q : 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.

it is given that it is an equilateral triangle and length of all sides is 2a The base of the triangle lies on y-axis such origin is the midpoint Therefore, Coordinates of point  A and B are   respectively Now, Apply Pythagoras theorem in triangle AOC Therefore, coordinates of vertices of the triangle are

Q : 1     Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4,5),(0,7),(5,-5)$  and  $(-4,-2)$. Also, find its area.

Area of ABCD = Area of ABC + Area of ACD Now, we know that the area of a triangle with vertices   is given by Therefore, Area of triangle ABC  Similarly, Area of triangle ACD  Now, Area of ABCD = Area of ABC + Area of ACD
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