Question

Can someone help me with number 7 and 8. Thank you

Answer 1

Answer:

Q7) ( 2,-2)

Q8) (- 0.5 , -2)

Step-by-step explanation:

As we have to find the mid-points of the line segments, the line segments will be divided in same ratio i.e. 1:1 (∵ A Mid-point divides a line segment into equal parts )

According to Section formula ,

$( x' , y') = (\frac{mx^2 + nx^1}{m+n} , \frac{my^2+ny^1}{m+n} )$

where (m,n) are the ratios in which a line segment is divided & ( x¹ , y¹ ) ( x², y²) are the co-ordinates of a line segment.

In Q7 ,

x¹ = 3 , y¹ = -9 , x² = 1 , y² = 5 & m = n = 1

So putting the section formula here gives :-

$( x' , y') = (\frac{1*1 + 1*3}{1+1} , \frac{1*5+1*(-9)}{1+1} ) = ( 2 , -2)$

In Q8 ,

x¹ = 5 , y¹ = -1 , x² = -6 , y² = -3

So putting the section formula here gives :-

$(x' , y') = (\frac{1*(-6) + 1*5}{1+1} , \frac{1*(-3) + 1*(-1)}{1+1}) = ( -0.5 , -2)$

Answer 2

Answer:

Let the midpoint be M(x,y).

7) x= (3+1)/2= 4/2= 2

y= (-9+5)/2= -4/2= -2

M(2,-2)

8) x= (5-6)/2= -1/2= -0.5

y= (-1-3)/2= -4/2= -2

M(-0.5,-2)