Find the ratio in which point Q will divide the join of a 3 2 and B 5 2 in two equal parts

Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

Let the vertices of quadrilateral are A(—4, -2), B(-3, -5), C(3, -2) and D(2, 3).

Area of ∆ABC

Here, we have
x1 = -4,    y1 = -2
x2 = -3,    y2 -5
x3 = 3,    y3 = -2
x3 = 3,    y3 = -2Now, area of ∆ABC

We know that, area of ∆ACD

Here, we have
x1 = -4,    y1 = -2
x2 = 3,    y2 = -2
and    x3 = 2,    y3 = 3Now, area of ∆ACD

Hence, the area of quadrilateralABCD = ar (∆ABC) + ar (∆ACD)= (10.5 + 17.5)

= 28 sq. units.

Suppose you have a line segment PQ ¯ on the coordinate plane, and you need to find the point on the segment 1 3 of the way from P to Q .

Let’s first take the easy case where P is at the origin and line segment is a horizontal one.

The length of the line is 6 units and the point on the segment 1 3 of the way from P to Q would be 2 units away from P , 4 units away from Q and would be at ( 2,0 ) .

Consider the case where the segment is not a horizontal or vertical line.

The components of the directed segment PQ ¯ are 〈 6,3 〉 and we need to find the point, say X on the segment 1 3 of the way from P to Q .

Then, the components of the segment PX ¯ are 〈 ( 1 3 )( 6 ),( 1 3 )( 3 ) 〉=〈 2,1 〉 .

Since the initial point of the segment is at origin, the coordinates of the point X are given by ( 0+2,0+1 )=( 2,1 ) .

Now let’s do a trickier problem, where neither P nor Q is at the origin.

Use the end points of the segment PQ ¯ to write the components of the directed segment.

〈 ( x 2 − x 1 ),( y 2 − y 1 ) 〉=〈 ( 7−1 ),( 2−6 ) 〉 =〈 6,−4 〉

Now in a similar way, the components of the segment PX ¯ where X is a point on the segment 1 3 of the way from P to Q are 〈 ( 1 3 )( 6 ),( 1 3 )( −4 ) 〉=〈 2,−1.25 〉 .

To find the coordinates of the point X add the components of the segment PX ¯ to the coordinates of the initial point P .

So, the coordinates of the point X are ( 1+2,6−1.25 )=( 3,4.75 ) .

Note that the resulting segments, PX ¯ and XQ ¯ , have lengths in a ratio of 1:2 .

In general: what if you need to find a point on a line segment that divides it into two segments with lengths in a ratio a:b ?

Consider the directed line segment XY ¯ with coordinates of the endpoints as X( x 1 , y 1 ) and Y( x 2 , y 2 ) .

Suppose the point Z divided the segment in the ratio a:b , then the point is a a+b of the way from X to Y .

So, generalizing the method we have, the components of the segment XZ ¯ are 〈 ( a a+b ( x 2 − x 1 ) ),( a a+b ( y 2 − y 1 ) ) 〉 .

Then, the X -coordinate of the point Z is

x 1 + a a+b ( x 2 − x 1 )= x 1 ( a+b )+a( x 2 − x 1 ) a+b = b x 1 +a x 2 a+b .

Similarly, the Y -coordinate is

y 1 + a a+b ( y 2 − y 1 )= y 1 ( a+b )+a( y 2 − y 1 ) a+b = b y 1 +a y 2 a+b .

Therefore, the coordinates of the point Z are ( b x 1 +a x 2 a+b , b y 1 +a y 2 a+b ) .

Example 1:

Find the coordinates of the point that divides the directed line segment MN ¯ with the coordinates of endpoints at M( −4,0 ) and M( 0,4 ) in the ratio 3:1 ?

Let L be the point that divides MN ¯ in the ratio 3:1 .

Here, ( x 1 , y 1 )=( −4,0 ),( x 2 , y 2 )=( 0,4 ) and a:b=3:1 .

Substitute in the formula. The coordinates of L are

( 1( −4 )+3( 0 ) 3+1 , 1( 0 )+3( 4 ) 3+1 ) .

Simplify.

( −4+0 4 , 0+12 4 )=( −1,3 )

Therefore, the point L( −1,3 ) divides MN ¯ in the ratio 3:1 .

Example 2:

What are the coordinates of the point that divides the directed line segment AB ¯ in the ratio 2:3 ?

Let C be the point that divides AB ¯ in the ratio 2:3 .

Here, ( x 1 , y 1 )=( −4,4 ),( x 2 , y 2 )=( 6,−5 ) and a:b=2:3 .

Substitute in the formula. The coordinates of C are