Question

Please answer 12 to 16

Answer 1

Answer 12:

Refer to the attachment for the diagram.

We know that A median divides the side on which it is drawn into parts of equal length i.e into equal ratio.

Therefore, D is the mid-point of side BC.

Vertices:

$A(x_{1}, y_{1}) = (1, 1)$

$B(x_{2}, y_{2}) = (0, -4)$

$C(x_{3}, y_{3}) = (-5, 3)$

$D(x_{4}, y_{4}) = (\alpha , \beta)$

By Mid-point formula, we have:

$D(\alpha , \beta ) = \left( \frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2} \right)$

$\Rightarrow \qquad \left( \frac{0 + (-5)}{2} , \frac{(-4 + 3)}{2} \right) \\ \Rightarrow \qquad \left(-\frac{5}{2} , -\frac{1}{2} \right) \\ Hence, \bold{D\left( -\frac{5}{2} , -\frac{1}{2} \right)}$

We have median AD, by Distance Formula we get:

$</p><p>AD = \sqrt{{\left( \alpha - x_{1} \right)}^2 + {\left( \beta - y_{1} \right)}^2} \\ AD = \sqrt{{\left(-\frac{5}{2} - 1\right)}^2 + {\left(-\frac{1}{2} - 1\right)}^2} \\ AD = \sqrt{{\left( -\frac{7}{2} \right)}^2 + {\left( -\frac{3}{2} \right)}^2} \\ AD = \sqrt{\left( \frac{49}{4} \right) + \left( \frac{9}{4} \right)} \\ AD = \sqrt{\frac{58}{4}} \\ AD = \frac{\sqrt{58}}{2} units$

Answer 13:

Refer to the attachment for the diagram.

Since, the y-axis cuts the line segment, we get abscissa of that point 0. (1)

Let the ratio that the y-axis cuts the line segment be m : n .

Vertices:

$A(x_{1}, y_{1}) = (-3 , 4)$

$B(x_{2}, y_{2}) = (1, -2)$

Let the coordinates of the point where y-axis cuts the line segment be $(0, \alpha )$. [ x = 0 from eq(1) ]

By section formula , we get the formula of abscissa as $\frac{mx_{2} + nx_{1}}{m+n}$

But, from eq(1) , we have abscissa of the point 0 :

$\frac{mx_{2} + nx_{1}}{m+n} = 0 \\ mx_{2} + nx_{1} = 0 \\ m \cdot 1 + n \cdot -3 = 0 \\ m - 3n = 0 \\ m = 3n \qquad (2)$

Hence, the required ratio is 3 : 1

Answer 15:

We know that if the points are colinear then the area of the triangle formed by these points is 0 units.

Vertices:

• $A(x_{1}, y_{1}) = (3, -4)$
• $B(x_{2} , y_{2}) = (-1, -3)$
• $C(x_{3}, y_{3}) = (x, y)$

Area of triangle ABC = 0

$\mid \frac{1}{2}( x_{1}( y_{2} - y_{3}) \: + \: x_{2}( y_{3} - y_{1} ) \: + \: x_{3}( y_{1} - y_{2} ) ) \mid = 0 \\ \\ 3( - 3 - y) - 1(y - ( - 4)) + x( - 4 + 3) = 0 \\ \\ - 9 - 3y - y - 4 - 4x + 3x = 0 \\ \\ - x - 4y - 13 = 0 \\ x + 4y = 13 \: \: \: \: \: \: \: (1)$

Therefore, we get x + 4y = 13, putting this in the questioned equation.

$x + 4y - 13 = 0 \\ x + 4y = 13$

Hence, Proved.

Answer 16:

We know that if the points are colinear then the area of the triangle formed by these points is 0 units. (1)

a - b = 1 [ given ]

a = 1 + b (2)

Vertices:

• $A(x_{1}, y_{1}) = (-2, 1)$
• $B(x_{2} , y_{2}) = (a, b)$
• $C(x_{3}, y_{3}) = (4, 1)$

Area of triangle ABC = 0 [ from (1) ]

$</p><p>\mid \frac{1}{2} [ x_{1}(y_{2} - y{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2}) ] = 0 \mid \\ \\ -2(b - 1) + a(1 - (-1)) + 4(1 - b) = 0 \\ \\ -2b + 2 + 2a + 4 - 4b = 0 \\ 2a - 6b = -6 \\ a - 3b = -3 \\ 1 + b - 3b = -3 \qquad [ from (2) ] \\ -2b = -4 \\ b = \frac{4}{2} = 2</p><p>$

Therefore b = 2, putting this in eq(2) , we get:

=> a = 1 + b

=> a = 1 + 2

=> a = 3

Now,

a + b = 3 + 2

a + b = 5

Answer: 5