Question

Centonion Public School Society, which has so many schools in different cities of India. One of the branch of Centonion Public School is in Meerut. In that School hundreds of students are in a classroom. Out of them one of the girl is standing in the ground having coordinates (3, 4) facing towards west. She moves 5 units in straight line then take right and moves 4 units and stop. Now, she is at her coaching centre. The representation of the above situation on the coordinate axes is shown below: Based on the above information give the answer of the following questions.
1) What is the shortest distance between her school and coaching centre? a) √41 units b) 3 units c) 6 units d) 7 units 47. 4)Suppose point D(1, 4) divide the line segment AB in theratio k:1, then find the value of k a) 3 b) 3/2 c) 2 d) ½ 48.
3) If we draw perpendicular lines from points A and B to the x-axis, the region covered by these perpendicular lines is: a) 4 sq. units b) 5 sq. units c) 20 sq. units d) 10 sq. units 49. Find the area of triangle ABC a) 4 sq. units b) 5 sq. units c) 20 sq. units d) 10 sq. units
50. Find the image of the midpoint of AB with respect to x-axis a) ( 1 2 , −4) c) (− 1 2 , 4) b) ( 1 2 , 4) d) (− 1 2 , −4)​

Answer 1

Answer:

I don't know

Step-by-step explanation:

SEE ANSWER namaste

Answer 2

Tip:

• Section Formula:$x=\frac{mx_2+nx_1}{m+n} , where ~m~and~n ~are ~the~ ratios.\\y=\frac{my_2+ny_1}{m+n} ,where ~m~and~n ~are ~the~ ratios.$
• Area of triangle$=\frac{1}{2} \times base \times height$

Explanation:

• In the diagram $AB=5 cm$ , $BC=4 cm$ and$AC=x cm$.
• $D(1,4)$ divides the line segment $AB$ in the ratio $k:1$

Steps

Step 1 of 5:

1.) Shortest Distance $=AC =xcm$

    ∴ $AC = \sqrt{(AB)^{2} +(BC)^2}$

        $x = \sqrt{(5)^2+(4)^2}$

        $x= \sqrt{41}$$cm$    

So, the answer is a) $\sqrt{41}$ units

Step 2 of 5:

2.) Coordinates of $D$ $=(1,4)$

∴ By section Formula:

$1=\frac{3k-2}{k+1} \\4=\frac{4k+4}{k+1}$

∴ $k+1=3k-2\\2k=3\\k=\frac{3}{2}$

So, the answer is b) $\frac{3}{2}$

Step 3 of 5:

If  we draw the perpendicular lines from $A$ and $B$ to the x axis , the region covered by the perpendicular lines is the area of rectangle $ABCD$

$Area ~of~ rectangle = Length \times width$

Area of rectangle $= 5 \times 4 ~sq.~ units$

                             $=20~sq.~units$

So, the answer is c) $20 ~sq.~units$

Step 4 of 5:

Area of triangle $= \frac{1}{2} \times AB \times BC$

                          $=\frac{1}{2} \times 5 \times 4\\= 5\times 2\\=10 ~sq. units$

So, the answer is d) $10 ~sq.~units$

Step 5 of 5:

Midpoint of $AB= \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Midpoint of $AB =(1.2,4)$    

so, Image of the midpoint of $AB$ with respect to x-axis $= (-1.2,-4)$                    

So, the answer is d) $(-1.2,-4)$

Final answer:

1) - a) $\sqrt{41}$

2) - b) $\frac{3}{2}$

3) - c) $20 ~sq.~units$

4) - d) $10 ~sq.~units$

5) -  d) $(-1.2,-4)$