Advertisements
Advertisements
Question
The midpoint of the line segment joining A (2a, 4) and B (-2, 3b) is C (1, 2a+1). Find the values of a and b.
Solution
The points are A (2a, 4) and B (-2, 3b).
Let C ( 1,2a + 1) be the mid-point of AB. Then:
` x = (x_1 +x_2) /2 , y = (y_1 +y_2)/2`
` ⇒ 1= (2a (-2))/2 , 2a +1 =(4+3b)/2`
`⇒ 2=2a-2, 4a +2 =4+3b`
⇒2a -2+2, 4a -3b = 4-2
`⇒ a = 4/2 , 4a -3b =2 `
⇒ a =2, 4a -3b =2
Putting the value of a in the equation 4a +3b =2 , we get:
4(2) - 3b =2
⇒ -3b = 2-8=-6
`⇒ b = 6/3 = 2`
Therefore, a =2 and b= 2
APPEARS IN
RELATED QUESTIONS
(Street Plan): A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction.
All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines.
There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find:
- how many cross - streets can be referred to as (4, 3).
- how many cross - streets can be referred to as (3, 4).
Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.
The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex.
Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle.
If the coordinates of the mid-points of the sides of a triangle be (3, -2), (-3, 1) and (4, -3), then find the coordinates of its vertices.
If (2, p) is the midpoint of the line segment joining the points A(6, -5) and B(-2,11) find the value of p.
Find the coordinates of the circumcentre of a triangle whose vertices are (–3, 1), (0, –2) and (1, 3).
The abscissa and ordinate of the origin are
Prove hat the points A (2, 3) B(−2,2) C(−1,−2), and D(3, −1) are the vertices of a square ABCD.
If (a,b) is the mid-point of the line segment joining the points A (10, - 6) , B (k,4) and a - 2b = 18 , find the value of k and the distance AB.
In \[∆\] ABC , the coordinates of vertex A are (0, - 1) and D (1,0) and E(0,10) respectively the mid-points of the sides AB and AC . If F is the mid-points of the side BC , find the area of \[∆\] DEF.
If the points A(1, –2), B(2, 3) C(a, 2) and D(– 4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.
If \[D\left( - \frac{1}{5}, \frac{5}{2} \right), E(7, 3) \text{ and } F\left( \frac{7}{2}, \frac{7}{2} \right)\] are the mid-points of sides of \[∆ ABC\] , find the area of \[∆ ABC\] .
The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ − b cos θ) is
If A (2, 2), B (−4, −4) and C (5, −8) are the vertices of a triangle, than the length of the median through vertex C is
f the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (−2, 5), then the coordinates of the other end of the diameter are
What is the nature of the line which includes the points (-5, 5), (6, 5), (-3, 5), (0, 5)?
Point (–3, 5) lies in the ______.
Assertion (A): The ratio in which the line segment joining (2, -3) and (5, 6) internally divided by x-axis is 1:2.
Reason (R): as formula for the internal division is `((mx_2 + nx_1)/(m + n) , (my_2 + ny_1)/(m + n))`
If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(–2, 3) and R(–3, –2), then the coordinates of its fourth vertex S are ______.